2007PascalSolution滑铁卢竞赛题答案

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Galois滑铁卢数学竞赛(Grade 10)-数学Mathematics-2007-试题 exam

2007Galois Contest (Grade 10)Wednesday,April 18,20071.Jim shops at a strange fruit store.Instead of putting prices on each item,the mathematical store owner will answer questions about combinations of items.(a)In Aisle 1,Jim receives the following answers to his questions:Jim’s Question AnswerWhat is the sum of the prices of an Apple and a Cherry?62centsWhat is the sum of the prices of a Banana and a Cherry?66centsWhat is diﬀerence between the prices of an Apple and a Banana?Which has a higher price?Explain how you obtained your answer.(b)In Aisle 2,Jim receives the following answers to his questions:Jim’s Question AnswerWhat is the sum of the prices of a Mango and a Nectarine?60centsWhat is the sum of the prices of a Pear and a Nectarine?60centsWhat is the sum of the prices of a Mango and a Pear?68centsWhat is the price of a Pear?Explain how you obtained your answer.(c)In Aisle 3,Jim receives the following answers to his questions:Jim’s Question AnswerWhat is the sum of the prices of a Tangerine and a Lemon?60centsHowmuch more does a Tangerine cost than a Grapefruit?6centsWhat is the sum of the prices of Grapefruit,a Tangerine and a Lemon?94centsWhat is the price of a Lemon?Explain how you obtained youranswer.2.(a)In the diagram,what is the perimeter of the sector of the circle with radius 12?Explain how you obtained your answer.(b)Two sectors of a circle of radius 12are placed side by side,as shown.Determine the area of ﬁgure ABCD .Explain how you obtainedyour answer.A (c)In the diagram,AOB is a sector of a circle with ∠AOB =60◦.OY is drawn perpendicular to AB and intersects AB at X .What is the length of XY ?Explain how you obtained your answer.A O BX Y1212(d)See over...2007Galois Contest Page2(d)Two sectors of a circle of radius12overlap as shown.Determine the area of the shaded region.Explain how youobtained your answer.R3.(a)Each face of a5by5by5wooden cube is divided into1by1squares.Each square is painted black or white,asshown.Next,the cube is cut into1by1by1cubes.Howmany of these cubes have at least two painted faces?Explain how you obtained youranswer.(b)A(2k+1)by(2k+1)by(2k+1)cube,where k is a in thesame manner as the5by5by5cube with white squares in the corners.Again,the cube is cut into1by1by1cubes.i.In terms of k,how many of these cubes have exactly two white faces?Explain howyou obtained your answer.ii.Prove that there is no value of k for which the number of cubes having at least two white faces is2006.4.Jill has a container of small cylindrical rods in six diﬀerent colours.Each colour of rod has adiﬀerent length as summarized in the chart.Colour LengthGreen3cmPink4cmYellow5cmBlack7cmViolet8cmRed9cmThese rods can be attached together to form a pole.There are2ways to choose a set of yellow and green rods that will form a pole29cm in length: 8green rods and1yellow rod OR3green rods and4yellow rods.(a)How many diﬀerent sets of yellow and green rods can be chosen that will form a pole62cm long?Explain how you obtained your answer.(b)Among the green,yellow,black and red rods,ﬁnd,with justiﬁcation,two colours for whichit is impossible to make a pole62cm in length using only rods of those two colours.(c)If at least81rodsof each of the colours green,pink,violet,and red must be used,howmany diﬀerent sets of rods of these four colours can be chosen that will form a pole2007cm in length?Explain how you got your answer.。

2011EuclidSolution

1.(a)Since (x +1)+(x +2)+(x +3)=8+9+10,then 3x +6=27or 3x =21and so x =7.(b)Since 25+√x =6,then squaring both sides gives 25+√x =36or √x =11.Since √x =11,then squaring both sides again,we obtain x =112=121.Checking, 25+√121=√25+11=√36=6,as required.(c)Since (a,2)is the point of intersection of the lines with equations y =2x −4and y =x +k ,then the coordinates of this point must satisfy both equations.Using the ﬁrst equation,2=2a −4or 2a =6or a =3.Since the coordinates of the point (3,2)satisfy the equation y =x +k ,then 2=3+k or k =−1.2.(a)Since the side length of the original square is 3and an equilateral triangle of side length 1is removed from the middle of each side,then each of the two remaining pieces of each side of the square has length 1.Also,each of the two sides of each of the equilateral triangles that are shown has length 1.1111Therefore,each of the 16line segments in the ﬁgure has length 1,and so the perimeter of the ﬁgure is 16.(b)Since DC =DB ,then CDB is isosceles and ∠DBC =∠DCB =15◦.Thus,∠CDB =180◦−∠DBC −∠DCB =150◦.Since the angles around a point add to 360◦,then∠ADC =360◦−∠ADB −∠CDB =360◦−130◦−150◦=80◦.(c)By the Pythagorean Theorem in EAD ,we have EA 2+AD 2=ED 2or 122+AD 2=132,and so AD =√169−144=5,since AD >0.By the Pythagorean Theorem in ACD ,we have AC 2+CD 2=AD 2or AC 2+42=52,and so AC =√25−16=3,since AC >0.(We could also have determined the lengths of AD and AC by recognizing 3-4-5and 5-12-13right-angled triangles.)By the Pythagorean Theorem in ABC ,we have AB 2+BC 2=AC 2or AB 2+22=32,and so AB =√9−4=√5,since AB >0.3.(a)Solution 1Since we want to make 15−y x as large as possible,then we want to subtract as little as possible from 15.In other words,we want to make y x as small as possible.To make a fraction with positive numerator and denominator as small as possible,wemake the numerator as small as possible and the denominator as large as possible.Since 2≤x ≤5and 10≤y ≤20,then we make x =5and y =10.Therefore,the maximum value of 15−y x is 15−105=13.Solution2Since y is positive and2≤x≤5,then15−yx≤15−y5for any x with2≤x≤5andpositive y.Since10≤y≤20,then15−y5≤15−105for any y with10≤y≤20.Therefore,for any x and y in these ranges,15−yx≤15−105=13,and so the maximumpossible value is13(which occurs when x=5and y=10).(b)Solution1First,we add the two given equations to obtain(f(x)+g(x))+(f(x)−g(x))=(3x+5)+(5x+7)or2f(x)=8x+12which gives f(x)=4x+6.Since f(x)+g(x)=3x+5,then g(x)=3x+5−f(x)=3x+5−(4x+6)=−x−1.(We could alsoﬁnd g(x)by subtracting the two given equations or by using the second of the given equations.)Since f(x)=4x+6,then f(2)=14.Since g(x)=−x−1,then g(2)=−3.Therefore,2f(2)g(2)=2×14×(−3)=−84.Solution2Since the two given equations are true for all values of x,then we can substitute x=2to obtainf(2)+g(2)=11f(2)−g(2)=17Next,we add these two equations to obtain2f(2)=28or f(2)=14.Since f(2)+g(2)=11,then g(2)=11−f(2)=11−14=−3.(We could alsoﬁnd g(2)by subtracting the two equations above or by using the second of these equations.)Therefore,2f(2)g(2)=2×14×(−3)=−84.4.(a)We consider choosing the three numbers all at once.We list the possible sets of three numbers that can be chosen:{1,2,3}{1,2,4}{1,2,5}{1,3,4}{1,3,5}{1,4,5}{2,3,4}{2,3,5}{2,4,5}{3,4,5} We have listed each in increasing order because once the numbers are chosen,we arrange them in increasing order.There are10sets of three numbers that can be chosen.Of these10,the4sequences1,2,3and1,3,5and2,3,4and3,4,5are arithmetic sequences.Therefore,the probability that the resulting sequence is an arithmetic sequence is410or25.(b)Solution 1Join B to D .AConsider CBD .Since CB =CD ,then ∠CBD =∠CDB =12(180◦−∠BCD )=12(180◦−60◦)=60◦.Therefore, BCD is equilateral,and so BD =BC =CD =6.Consider DBA .Note that ∠DBA =90◦−∠CBD =90◦−60◦=30◦.Since BD =BA =6,then ∠BDA =∠BAD =12(180◦−∠DBA )=12(180◦−30◦)=75◦.We calculate the length of AD .Method 1By the Sine Law in DBA ,we have AD sin(∠DBA )=BA sin(∠BDA ).Therefore,AD =6sin(30◦)sin(75◦)=6×12sin(75◦)=3sin(75◦).Method 2If we drop a perpendicular from B to P on AD ,then P is the midpoint of AD since BDA is isosceles.Thus,AD =2AP .Also,BP bisects ∠DBA ,so ∠ABP =15◦.Now,AP =BA sin(∠ABP )=6sin(15◦).Therefore,AD =2AP =12sin(15◦).Method 3By the Cosine Law in DBA ,AD 2=AB 2+BD 2−2(AB )(BD )cos(∠ABD )=62+62−2(6)(6)cos(30◦)=72−72(√32)=72−36√3Therefore,AD = 36(2−√3)=6 2−√3since AD >0.Solution 2Drop perpendiculars from D to Q on BC and from D to R on BA .AThen CQ =CD cos(∠DCQ )=6cos(60◦)=6×12=3.Also,DQ =CD sin(∠DCQ )=6sin(60◦)=6×√32=3√3.Since BC =6,then BQ =BC −CQ =6−3=3.Now quadrilateral BQDR has three right angles,so it must have a fourth right angle and so must be a rectangle.Thus,RD =BQ =3and RB =DQ =3√3.Since AB =6,then AR =AB −RB =6−3√3.Since ARD is right-angled at R ,then using the Pythagorean Theorem and the fact that AD >0,we obtain AD =√RD 2+AR 2= 32+(6−3√3)2= 9+36−36√3+27= 72−36√3which we can rewrite as AD = 36(2−√3)=6 2−√3.5.(a)Let n be the original number and N be the number when the digits are reversed.Sincewe are looking for the largest value of n ,we assume that n >0.Since we want N to be 75%larger than n ,then N should be 175%of n ,or N =74n .Suppose that the tens digit of n is a and the units digit of n is b .Then n =10a +b .Also,the tens digit of N is b and the units digit of N is a ,so N =10b +a .We want 10b +a =74(10a +b )or 4(10b +a )=7(10a +b )or 40b +4a =70a +7b or 33b =66a ,and so b =2a .This tells us that that any two-digit number n =10a +b with b =2a has the required property.Since both a and b are digits then b <10and so a <5,which means that the possible values of n are 12,24,36,and 48.The largest of these numbers is 48.(b)We “complete the rectangle”by drawing a horizontal line through C which meets they -axis at P and the vertical line through B at Q .x A (0,Since C has y -coordinate 5,then P has y -coordinate 5;thus the coordinates of P are (0,5).Since B has x -coordinate 4,then Q has x -coordinate 4.Since C has y -coordinate 5,then Q has y -coordinate 5.Therefore,the coordinates of Q are (4,5),and so rectangle OP QB is 4by 5and so has area 4×5=20.Now rectangle OP QB is made up of four smaller triangles,and so the sum of the areas of these triangles must be 20.Let us examine each of these triangles:• ABC has area 8(given information)• AOB is right-angled at O ,has height AO =3and base OB =4,and so has area 12×4×3=6.• AP C is right-angled at P ,has height AP =5−3=2and base P C =k −0=k ,and so has area 1×k ×2=k .• CQB is right-angled at Q ,has height QB =5−0=5and base CQ =4−k ,andso has area 12×(4−k )×5=10−52k .Since the sum of the areas of these triangles is 20,then 8+6+k +10−52k =20or 4=32k and so k =83.6.(a)Solution 1Suppose that the distance from point A to point B is d km.Suppose also that r c is the speed at which Serge travels while not paddling (i.e.being carried by just the current),that r p is the speed at which Serge travels with no current (i.e.just from his paddling),and r p +c his speed when being moved by both his paddling and the current.It takes Serge 18minutes to travel from A to B while paddling with the current.Thus,r p +c =d 18km/min.It takes Serge 30minutes to travel from A to B with just the current.Thus,r c =d 30km/min.But r p =r p +c −r c =d 18−d 30=5d 90−3d 90=2d 90=d 45km/min.Since Serge can paddle the d km from A to B at a speed of d 45km/min,then it takes him 45minutes to paddle from A to B with no current.Solution 2Suppose that the distance from point A to point B is d km,the speed of the current of the river is r km/h,and the speed that Serge can paddle is s km/h.Since the current can carry Serge from A to B in 30minutes (or 12h),then d r =12.When Serge paddles with the current,his speed equals his paddling speed plus the speed of the current,or (s +r )km/h.Since Serge can paddle with the current from A to B in 18minutes (or 310h),then d r +s =310.The time to paddle from A to B with no current would be d sh.Since d r =12,then r d =2.Since d r +s =310,then r +s d =103.Therefore,s d =r +s d −r d =103−2=43.Thus,d s =34,and so it would take Serge 34of an hour,or 45minutes,to paddle from A to B with no current.Solution 3Suppose that the distance from point A to point B is d km,the speed of the current of the river is r km/h,and the speed that Serge can paddle is s km/h.Since the current can carry Serge from A to B in 30minutes (or 12h),then d r =12or d =1r .When Serge paddles with the current,his speed equals his paddling speed plus the speed of the current,or (s +r )km/h.Since Serge can paddle with the current from A to B in 18minutes (or 310h),then d r +s =310or d =310(r +s ).Since d =12r and d =310(r +s ),then 12r =310(r +s )or 5r =3r +3s and so s =23r .To travel from A to B with no current,the time in hours that it takes is d s =12r 2r =34,or 45minutes.(b)First,we note that a =0.(If a =0,then the “parabola”y =a (x −2)(x −6)is actuallythe horizontal line y =0which intersects the square all along OR .)Second,we note that,regardless of the value of a =0,the parabola has x -intercepts 2and 6,and so intersects the x -axis at (2,0)and (6,0),which we call K (2,0)and L (6,0).This gives KL =4.Third,we note that since the x -intercepts of the parabola are 2and 6,then the axis ofsymmetry of the parabola has equation x =12(2+6)=4.Since the axis of symmetry of the parabola is a vertical line of symmetry,then if theparabola intersects the two vertical sides of the square,it will intersect these at the same height,and if the parabola intersects the top side of the square,it will intersect it at two points that are symmetrical about the vertical line x =4.Fourth,we recall that a trapezoid with parallel sides of lengths a and b and height h hasarea 12h (a +b ).We now examine three cases.Case1:a<0Here,the parabola opens downwards.Since the parabola intersects the square at four points,it must intersect P Q at points M and N.(The parabola cannot intersect the vertical sides of the square since it gets “narrower”towards the vertex.)xx =4Since the parabola opens downwards,then MN<KL=4.Since the height of the trapezoid equals the height of the square(or8),then the area of the trapezoid is1h(KL+MN)which is less than1(8)(4+4)=32.But the area of the trapezoid must be36,so this case is not possible.Case2:a>0;M and N on P QWe have the following conﬁguration:xx =4Here,the height of the trapezoid is8,KL=4,and M and N are symmetric about x=4.Since the area of the trapezoid is36,then12h(KL+MN)=36or12(8)(4+MN)=36or4+MN=9or MN=5.Thus,M and N are each52units from x=4,and so N has coordinates(32,8).Since this point lies on the parabola with equation y=a(x−2)(x−6),then8=a(32−2)(32−6)or8=a(−12)(−92)or8=94a or a=329.Case3:a>0;M and N on QR and P Oxx =4Here,KL=4,MN=8,and M and N have the same y-coordinate.Since the area of the trapezoid is36,then12h(KL+MN)=36or12h(4+8)=36or6h=36or h=6.Thus,N has coordinates(0,6).Since this point lies on the parabola with equation y=a(x−2)(x−6),then 6=a(0−2)(0−6)or6=12a or a=12.Therefore,the possible values of a are329and12.7.(a)Solution1Consider a population of100people,each of whom is75years old and who behave ac-cording to the probabilities given in the question.Each of the original100people has a50%chance of living at least another10years,so there will be50%×100=50of these people alive at age85.Each of the original100people has a20%chance of living at least another15years,so there will be20%×100=20of these people alive at age90.Since there is a25%(or14)chance that an80year old person will live at least another10years(that is,to age90),then there should be4times as many of these people alive at age80than at age90.Since there are20people alive at age90,then there are4×20=80of the original100 people alive at age80.In summary,of the initial100people of age75,there are80alive at age80,50alive at age85,and20people alive at age90.Because50of the80people alive at age80are still alive at age85,then the probability that an80year old person will live at least5more years(that is,to age85)is50=5,or 62.5%.Solution2Suppose that the probability that a75year old person lives to80is p,the probability that an80year old person lives to85is q,and the probability that an85year old person lives to90is r.We want to the determine the value of q.For a75year old person to live at least another10years,they must live another5years (to age80)and then another5years(to age85).The probability of this is equal to pq. We are told in the question that this is equal to50%or0.5.Therefore,pq=0.5.For a75year old person to live at least another15years,they must live another5years (to age80),then another5years(to age85),and then another5years(to age90).The probability of this is equal to pqr.We are told in the question that this is equal to20% or0.2.Therefore,pqr=0.2Similarly,since the probability that an80year old person will live another10years is25%,then qr=0.25.Since pqr=0.2and pq=0.5,then r=pqrpq=0.20.5=0.4.Since qr=0.25and r=0.4,then q=qrr=0.250.4=0.625.Therefore,the probability that an80year old man will live at least another5years is0.625,or62.5%.(b)Using logarithm rules,the given equation is equivalent to22log10x=3(2·2log10x)+16or(2log10x)2=6·2log10x+16.Set u=2log10x.Then the equation becomes u2=6u+16or u2−6u−16=0.Factoring,we obtain(u−8)(u+2)=0and so u=8or u=−2.Since2a>0for any real number a,then u>0and so we can reject the possibility that u=−2.Thus,u=2log10x=8which means that log10x=3.Therefore,x=1000.8.(a)First,we determine theﬁrst entry in the50th row.Since theﬁrst column is an arithmetic sequence with common diﬀerence3,then the50th entry in theﬁrst column(theﬁrst entry in the50th row)is4+49(3)=4+147=151.Second,we determine the common diﬀerence in the50th row by determining the second entry in the50th row.Since the second column is an arithmetic sequence with common diﬀerence5,then the 50th entry in the second column(that is,the second entry in the50th row)is7+49(5) or7+245=252.Therefore,the common diﬀerence in the50th row must be252−151=101.Thus,the40th entry in the50th row(that is,the number in the50th row and the40th column)is151+39(101)=151+3939=4090.(b)We follow the same procedure as in(a).First,we determine theﬁrst entry in the R th row.Since theﬁrst column is an arithmetic sequence with common diﬀerence3,then the R th entry in theﬁrst column(that is,theﬁrst entry in the R th row)is4+(R−1)(3)or 4+3R−3=3R+1.Second,we determine the common diﬀerence in the R th row by determining the second entry in the R th row.Since the second column is an arithmetic sequence with common diﬀerence5,then the R th entry in the second column(that is,the second entry in the R th row)is7+(R−1)(5) or7+5R−5=5R+2.Therefore,the common diﬀerence in the R th row must be(5R+2)−(3R+1)=2R+1.Thus,the C th entry in the R th row(that is,the number in the R th row and the C th column)is3R+1+(C−1)(2R+1)=3R+1+2RC+C−2R−1=2RC+R+C(c)Suppose that N is an entry in the table,say in the R th row and C th column.From(b),then N=2RC+R+C and so2N+1=4RC+2R+2C+1.Now4RC+2R+2C+1=2R(2C+1)+2C+1=(2R+1)(2C+1).Since R and C are integers with R≥1and C≥1,then2R+1and2C+1are each integers that are at least3.Therefore,2N+1=(2R+1)(2C+1)must be composite,since it is the product of two integers that are each greater than1.9.(a)If n=2011,then8n−7=16081and so √8n−7≈126.81.Thus,1+√8n−72≈1+126.812≈63.9.Therefore,g(2011)=2(2011)+1+8(2011)−72=4022+ 63.9 =4022+63=4085.(b)To determine a value of n for which f(n)=100,we need to solve the equation2n−1+√8n−72=100(∗)Weﬁrst solve the equation2x−1+√8x−72=100(∗∗)because the left sides of(∗)and(∗∗)do not diﬀer by much and so the solutions are likely close together.We will try integers n in(∗)that are close to the solutions to(∗∗). Manipulating(∗∗),we obtain4x−(1+√8x−7)=2004x−201=√8x−7(4x−201)2=8x−716x2−1608x+40401=8x−716x2−1616x+40408=02x2−202x+5051=0By the quadratic formula,x=202±2022−4(2)(5051)2(2)=202±√3964=101±√992and so x≈55.47or x≈45.53.We try n=55,which is close to55.47:f(55)=2(55)−1+8(55)−72=110−1+√4332Since √433≈20.8,then1+√4332≈10.9,which gives1+√4332=10.Thus,f(55)=110−10=100.Therefore,a value of n for which f(n)=100is n=55.(c)We want to show that each positive integer m is in the range of f or the range of g ,butnot both.To do this,we ﬁrst try to better understand the “complicated”term of each of the func-tions –that is,the term involving the greatest integer function.In particular,we start witha positive integer k ≥1and try to determine the positive integers n that give 1+√8n −72 =k .By deﬁnition of the greatest integer function,the equation 1+√8n −72 =k is equiv-alent to the inequality k ≤1+√8n −72<k +1,from which we obtain the following set of equivalent inequalities 2k ≤1+√8n −7<2k +22k −1≤√8n −7<2k +14k 2−4k +1≤8n −7<4k 2+4k +14k 2−4k +8≤8n <4k 2+4k +812(k 2−k )+1≤n <12(k 2+k )+1If we deﬁne T k =1k (k +1)=1(k 2+k )to be the k th triangular number for k ≥0,thenT k −1=12(k −1)(k )=12(k 2−k ).Therefore, 1+√8n −72 =k for T k −1+1≤n <T k +1.Since n is an integer,then 1+√8n −72=k is true for T k −1+1≤n ≤T k .When k =1,this interval is T 0+1≤n ≤T 1(or 1≤n ≤1).When k =2,this interval is T 1+1≤n ≤T 2(or 2≤n ≤3).When k =3,this interval is T 2+1≤n ≤T 3(or 4≤n ≤6).As k ranges over all positive integers,these intervals include every positive integer n and do not overlap.Therefore,we can determine the range of each of the functions f and g by examining the values f (n )and g (n )when n is in these intervals.For each non-negative integer k ,deﬁne R k to be the set of integers greater than k 2and less than or equal to (k +1)2.Thus,R k ={k 2+1,k 2+2,...,k 2+2k,k 2+2k +1}.For example,R 0={1},R 1={2,3,4},R 2={5,6,7,8,9},and so on.Every positive integer occurs in exactly one of these sets.Also,for each non-negative integer k deﬁne S k ={k 2+2,k 2+4,...,k 2+2k }and deﬁne Q k ={k 2+1,k 2+3,...,k 2+2k +1}.For example,S 0={},S 1={3},S 2={6,8},Q 0={1},Q 1={2,4},Q 2={5,7,9},and so on.Note that R k =Q k ∪S k so every positive integer occurs in exactly one Q k or in exactly one S k ,and that these sets do not overlap since no two S k ’s overlap and no two Q k ’s overlap and no Q k overlaps with an S k .We determine the range of the function g ﬁrst.For T k −1+1≤n ≤T k ,we have 1+√8n −72=k and so 2T k −1+2≤2n ≤2T k 2T k −1+2+k ≤2n + 1+√8n −72 ≤2T k +k k 2−k +2+k ≤g (n )≤k 2+k +k k 2+2≤g (n )≤k 2+2kNote that when n is in this interval and increases by 1,then the 2n term causes the value of g (n )to increase by 2.Therefore,for the values of n in this interval,g (n )takes precisely the values k 2+2,k 2+4,k 2+6,...,k 2+2k .In other words,the range of g over this interval of its domain is precisely the set S k .As k ranges over all positive integers (that is,as these intervals cover the domain of g ),this tells us that the range of g is precisely the integers in the sets S 1,S 2,S 3,....(We could also include S 0in this list since it is the empty set.)We note next that f (1)=2− 1+√8−72 =1,the only element of Q 0.For k ≥1and T k +1≤n ≤T k +1,we have 1+√8n −72=k +1and so 2T k +2≤2n ≤2T k +12T k +2−(k +1)≤2n − 1+√8n −72 ≤2T k +1−(k +1)k 2+k +2−k −1≤f (n )≤(k +1)(k +2)−k −1k 2+1≤f (n )≤k 2+2k +1Note that when n is in this interval and increases by 1,then the 2n term causes the value of f (n )to increase by 2.Therefore,for the values of n in this interval,f (n )takes precisely the values k 2+1,k 2+3,k 2+5,...,k 2+2k +1.In other words,the range of f over this interval of its domain is precisely the set Q k .As k ranges over all positive integers (that is,as these intervals cover the domain of f ),this tells us that the range of f is precisely the integers in the sets Q 0,Q 1,Q 2,....Therefore,the range of f is the set of elements in the sets Q 0,Q 1,Q 2,...and the range of g is the set of elements in the sets S 0,S 1,S 2,....These ranges include every positive integer and do not overlap.10.(a)Suppose that ∠KAB =θ.Since ∠KAC =2∠KAB ,then ∠KAC =2θand ∠BAC =∠KAC +∠KAB =3θ.Since 3∠ABC =2∠BAC ,then ∠ABC =23×3θ=2θ.Since ∠AKC is exterior to AKB ,then ∠AKC =∠KAB +∠ABC =3θ.This gives the following conﬁguration:BNow CAK is similar to CBA since the triangles have a common angle at C and ∠CAK =∠CBA .Therefore,AKBA=CACBordc=baand so d=bca.Also,CKCA=CACBora−xb=baand so a−x=b2aor x=a−b2a=a2−b2a,as required.(b)From(a),bc=ad and a2−b2=ax and so we obtainLS=(a2−b2)(a2−b2+ac)=(ax)(ax+ac)=a2x(x+c) andRS=b2c2=(bc)2=(ad)2=a2d2In order to show that LS=RS,we need to show that x(x+c)=d2(since a>0).Method1:Use the Sine LawFirst,we derive a formula for sin3θwhich we will need in this solution:sin3θ=sin(2θ+θ)=sin2θcosθ+cos2θsinθ=2sinθcos2θ+(1−2sin2θ)sinθ=2sinθ(1−sin2θ)+(1−2sin2θ)sinθ=3sinθ−4sin3θSince∠AKB=180◦−∠KAB−∠KBA=180◦−3θ,then using the Sine Law in AKB givesx sinθ=dsin2θ=csin(180◦−3θ)Since sin(180◦−X)=sin X,then sin(180◦−3θ)=sin3θ,and so x=d sinθsin2θandc=d sin3θsin2θ.This givesx(x+c)=d sinθsin2θd sinθsin2θ+d sin3θsin2θ=d2sinθsin22θ(sinθ+sin3θ)=d2sinθsin22θ(sinθ+3sinθ−4sin3θ)=d2sinθsin22θ(4sinθ−4sin3θ)=4d2sin2θsin22θ(1−sin2θ)=4d2sin2θcos2θsin22θ=4d2sin2θcos2θ(2sinθcosθ)2=4d2sin2θcos2θ4sin2θcos2θ=d2as required.We could have instead used the formula sin A +sin B =2sinA +B 2 cos A −B 2 toshow that sin 3θ+sin θ=2sin 2θcos θ,from which sin θ(sin 3θ+sin θ)=sin θ(2sin 2θcos θ)=2sin θcos θsin 2θ=sin 22θMethod 2:Extend ABExtend AB to E so that BE =BK =x and join KE .ENow KBE is isosceles with ∠BKE =∠KEB .Since ∠KBA is the exterior angle of KBE ,then ∠KBA =2∠KEB =2θ.Thus,∠KEB =∠BKE =θ.But this also tells us that ∠KAE =∠KEA =θ.Thus, KAE is isosceles and so KE =KA =d.ESo KAE is similar to BKE ,since each has two angles equal to θ.Thus,KA BK =AE KE or d x =c +x dand so d 2=x (x +c ),as required.Method 3:Use the Cosine Law and the Sine LawWe apply the Cosine Law in AKB to obtainAK 2=BK 2+BA 2−2(BA )(BK )cos(∠KBA )d 2=x 2+c 2−2cx cos(2θ)d 2=x 2+c 2−2cx (2cos 2θ−1)Using the Sine Law in AKB ,we get x sin θ=d sin 2θor sin 2θsin θ=d x or 2sin θcos θsin θ=d x and so cos θ=d 2x.Combining these two equations,d2=x2+c2−2cx2d24x2−1d2=x2+c2−cd2x+2cxd2+cd2x=x2+2cx+c2d2+cd2x=(x+c)2xd2+cd2=x(x+c)2d2(x+c)=x(x+c)2d2=x(x+c)as required(since x+c=0).(c)Solution1Our goal is toﬁnd a triple of positive integers that satisfy the equation in(b)and are the side lengths of a triangle.First,we note that if(A,B,C)is a triple of real numbers that satisﬁes the equation in(b)and k is another real number,then the triple(kA,kB,kC)also satisﬁes the equationfrom(b),since(k2A2−k2B2)(k2A2−k2B2+kAkC)=k4(A2−B2)(A2−B2+AC)=k4(B2C2)=(kB)2(kC)2 Therefore,we start by trying toﬁnd a triple(a,b,c)of rational numbers that satisﬁes the equation in(b)and forms a triangle,and then“scale up”this triple to form a triple (ka,kb,kc)of integers.To do this,we rewrite the equation from(b)as a quadratic equation in c and solve for c using the quadratic formula.Partially expanding the left side from(b),we obtain(a2−b2)(a2−b2)+ac(a2−b2)=b2c2which we rearrange to obtainb2c2−c(a(a2−b2))−(a2−b2)2=0By the quadratic formula,c=a(a2−b2)±a2(a2−b2)2+4b2(a2−b2)22b2=a(a2−b2)±(a2−b2)2(a2+4b2)2b2Since∠BAC>∠ABC,then a>b and so a2−b2>0,which givesc=a(a2−b2)±(a2−b2)√a2+4b22b2=(a2−b2)2b2(a±√a2+4b2)Since a2+4b2>0,then √a2+4b2>a,so the positive root isc=(a2−b2)2b2(a+a2+(2b)2)We try to ﬁnd integers a and b that give a rational value for c .We will then check to see if this triple (a,b,c )forms the side lengths of a triangle,and then eventually scale these up to get integer values.One way for the value of c to be rational (and in fact the only way)is for a 2+(2b )2to be an integer,or for a and 2b to be the legs of a Pythagorean triple.Since √32+42is an integer,then we try a =3and b =2,which givesc =(32−22)2·22(3+√32+42)=5and so (a,b,c )=(3,2,5).Unfortunately,these lengths do not form a triangle,since 3+2=5.(The Triangle Inequality tells us that three positive real numbers a ,b and c form a triangle if and only if a +b >c and a +c >b and b +c >a .)We can continue to try small Pythagorean triples.Now 152+82=172,but a =15and b =4do not give a value of c that forms a triangle with a and b .However,162+302=342,so we can try a =16and b =15which givesc =(162−152)2·152(16+√162+302)=31450(16+34)=319Now the lengths (a,b,c )=(16,15,319)do form the sides of a triangle since a +b >c and a +c >b and b +c >a .Since these values satisfy the equation from (b),then we can scale them up by a factor of k =9to obtain the triple (144,135,31)which satisﬁes the equation from (b)and are the side lengths of a triangle.(Using other Pythagorean triples,we could obtain other triples of integers that work.)Solution 2We note that the equation in (b)involves only a ,b and c and so appears to depend only on the relationship between the angles ∠CAB and ∠CBA in ABC .Using this premise,we use ABC ,remove the line segment AK and draw the altitude CF .CBA 3θ2θb aa c os 2θbc os 3θF Because we are only looking for one triple that works,we can make a number of assump-tions that may or may not be true in general for such a triangle,but which will help us ﬁnd an example.We assume that 3θand 2θare both acute angles;that is,we assume that θ<30◦.In ABC ,we have AF =b cos 3θ,BF =a cos 2θ,and CF =b sin 3θ=a sin 2θ.Note also that c =b cos 3θ+a cos 2θ.One way to ﬁnd the integers a,b,c that we require is to look for integers a and b and an angle θwith the properties that b cos 3θand a cos 2θare integers and b sin 3θ=a sin 2θ.Using trigonometric formulae,sin 2θ=2sin θcos θcos 2θ=2cos 2θ−1sin 3θ=3sin θ−4sin 3θ(from the calculation in (a),Solution 1,Method 1)cos 3θ=cos(2θ+θ)=cos 2θcos θ−sin 2θsin θ=(2cos 2θ−1)cos θ−2sin 2θcos θ=(2cos 2θ−1)cos θ−2(1−cos 2θ)cos θ=4cos 3θ−3cos θSo we can try to ﬁnd an angle θ<30◦with cos θa rational number and then integers a and b that make b sin 3θ=a sin 2θand ensure that b cos 3θand a cos 2θare integers.Since we are assuming that θ<30◦,then cos θ>√32≈0.866.The rational number with smallest denominator that is larger than √32is 78,so we try the acute angle θwith cos θ=7.In this case,sin θ=√1−cos 2θ=√158,and sosin 2θ=2sin θcos θ=2×78×√158=7√1532cos 2θ=2cos 2θ−1=2×4964−1=1732sin 3θ=3sin θ−4sin 3θ=3×√158−4×15√15512=33√15128cos 3θ=4cos 3θ−3cos θ=4×343512−3×78=7128To have b sin 3θ=a sin 2θ,we need 33√15128b =7√1532a or 33b =28a .To ensure that b cos 3θand a cos 2θare integers,we need 7128b and 1732a to be integers,andso a must be divisible by 32and b must be divisible by 128.The integers a =33and b =28satisfy the equation 33b =28a .Multiplying each by 32gives a =1056and b =896which satisfy the equation 33b =28a and now have the property that b is divisible by 128(with quotient 7)and a is divisible by 32(with quotient 33).With these values of a and b ,we obtain c =b cos 3θ+a cos 2θ=896×7128+1056×1732=610.We can then check that the triple (a,b,c )=(1056,896,610)satisﬁes the equation from(b),as required.As in our discussion in Solution 1,each element of this triple can be divided by 2to obtain the “smaller”triple (a,b,c )=(528,448,305)that satisﬁes the equation too.Using other values for cos θand integers a and b ,we could obtain other triples (a,b,c )of integers that work.。

滑铁卢竞赛数学题

滑铁卢竞赛数学题概述
滑铁卢竞赛数学题通常比较难，涉及的知识点广泛，包括代数、几何、数论、组合数学等多个领域。

以下是一些滑铁卢竞赛数学题的示例：
1. 有100个球，其中有一个与其他99个重量不同，但外观相同。

用一个天平，最少需要称多少次才能确定这个重量不同的球？
2. 一个正方形的面积为1，将其四边中点连接起来，形成另一个正方形。

如此重复，得到第五、第六个正方形，求第五个正方形的面积。

3. 一个圆被分成n个相等的扇形，其中一个是空心的，其他n-1个是实心的。

求空心扇形的圆心角是多少度？
4. 有100个人站成一排，从第1个人开始报数，每次报到奇数的人离开队伍。

经过若干轮后，只剩下一个人。

求这个人最初站在第几位？
5. 有5个不同质因数的最小正整数是多少？
以上仅是滑铁卢竞赛数学题的一些示例，实际上还有更多难题和技巧题。

如果想要深入了解滑铁卢竞赛数学题的解题技巧和策略，建议参考相关的竞赛书籍和资料，或者参加专业的数学竞赛培训课程。

Pascal滑铁卢数学竞赛(Grade 9)-数学Mathematics-2004-试题 exam

(A) 21 (D) 13
(B) 40 (E) 34
(C) 19
20. At the beginning of the game “Clock 7”, the arrow points to one of the seven numbers. On each turn, the arrow is rotated clockwise by the number of spaces indicated by the arrow at the beginning of the turn. For example, if “Clock 7” starts with the arrow pointing at 4, then on the first turn, the arrow is rotated clockwise 4 spaces so that it now points at 1. The arrow will then move 1 space on the next turn, and so on. If the arrow points at 6 after the 21st turn, at which number did the arrow point after the first turn?
1. To win a skateboard, the skill testing question is 5 × (10 – 6) ÷ 2 . The correct answer is
(A)ቤተ መጻሕፍቲ ባይዱ10
(B) 35
(C) 32
(D) 22
(E) 40
2. The average of 2, x and 12 is 8. What is the value of x?

2010PascalSolution滑铁卢竞赛题答案

1.In cents,theﬁve given choices are50,90,95,101,and115cents.The diﬀerences between each of these and$1.00(or100cents),in cents,are100−50=50100−90=10100−95=5101−100=1115−100=15 The diﬀerence between$1.01and$1.00is the smallest(1cent),so$1.01is closest to$1.00.Answer:(D) ing the correct order of operations,(20−16)×(12+8)4=4×204=804=20Answer:(C)3.We divide the750mL ofﬂour into portions of250mL.We do this by calculating750÷250=3.Therefore,750mL is three portions of250mL.Since50mL of milk is required for each250mL ofﬂour,then3×50=150mL of milk is required in total.Answer:(C) 4.There are8ﬁgures in total.Of these,3are triangles.Therefore,the probability is3.Answer:(A) 5.We simplify the left side and express it as a fraction with numerator1:1 9+118=218+118=318=16Therefore,the number that replaces the is6.Answer:(C) 6.There are16horizontal segments on the perimeter.Each has length1,so the horizontalsegments contribute16to the perimeter.There are10vertical segments on the perimeter.Each has length1,so the vertical segments contribute10to the perimeter.Therefore,the perimeter is10+16=26.(We could arrive at this total instead by starting at aﬁxed point and travelling around the outside of theﬁgure counting the number of segments.)Answer:(E) 7.Since33=3×3×3=3×9=27,then√33+33+33=√27+27+27=√81=9Answer:(B)8.The diﬀerence between the two given numbers is7.62−7.46=0.16.This length of the number line is divided into8equal segments.The length of each of these segments is thus0.16÷8=0.02.Point P is three of these segments to the right of7.46.Thus,the number represented is7.46+3(0.02)=7.46+0.06=7.52.Answer:(E)9.A 12by 12grid of squares will have 11interior vertical lines and 11interior horizontal lines.(In the given 4by 4example,there are 3interior vertical lines and 3interior horizontal lines.)Each of the 11interior vertical lines intersects each of the 11interior horizontal lines and creates an interior intersection point.Thus,each interior vertical line accounts for 11intersection points.Therefore,the number of interior intersection points is 11×11=121.Answer:(B)10.Because the central angle for the interior sector “Less than 1hour”is 90◦,then the fraction of the students who do less than 1hour of homework per day is 90◦360◦=14.In other words,25%of the students do less than 1hour of homework per day.Therefore,100%−25%=75%of the students do at least 1hour of homework per day.Answer:(E)11.Solution 1Since there is more than 1four-legged table,then there are at least 2four-legged tables.Since there are 23legs in total,then there must be fewer than 6four-legged tables,since 6four-legged tables would have 6×4=24legs.Thus,there are between 2and 5four-legged tables.If there are 2four-legged tables,then these tables account for 2×4=8legs,leaving 23−8=15legs for the three-legged tables.Since 15is divisible by 3,then this must be the solution,so there are 15÷3=5three-legged tables.(We can check that if there are 3or 4four-legged tables,then the number of remaining legs is not divisible by 3,and if there are 5four-legged tables,then there is only 1three-legged table,which is not allowed.)Solution 2Since there is more than 1table of each type,then there are at least 2three-legged tables and 2four-legged tables.These tables account for 2(3)+2(4)=14legs.There are 23−14=9more legs that need to be accounted for.These must come from a combination of three-legged and four-legged tables.The only way to make 9from 3s and 4s is to use three 3s.Therefore,there are 2+3=5three-legged tables and 2four-legged tables.Answer:(E)12.Solution 1The total area of the rectangle is 3×4=12.The total area of the shaded regions equals the total area of the rectangle (12)minus the area of the unshaded region.The unshaded region is a triangle with base of length 1and height 4;the area of this region is 12(1)(4)=2.Therefore,the total area of the shaded regions is 12−2=10.Solution 2The shaded triangle on the left has base of length 2and height of length 4,so has an area of 12(2)(4)=4.The shaded triangle on the right has base of length3(at the top)and height of length4,sohas an area of12(3)(4)=6.Therefore,the total area of the shaded regions is4+6=10.Answer:(C)13.Since the ratio of boys to girls at Cayley H.S.is3:2,then33+2=35of the students at CayleyH.S.are boys.Thus,there are35(400)=12005=240boys at Cayley H.S.Since the ratio of boys to girls at Fermat C.I.is2:3,then22+3=25of the students at FermatC.I.are boys.Thus,there are25(600)=12005=240boys at Fermat C.I.There are400+600=1000students in total at the two schools.Of these,240+240=480are boys,and so the remaining1000−480=520students are girls.Therefore,the overall ratio of boys to girls is480:520=48:52=12:13.Answer:(B) 14.When the given net is folded,the face numbered5will be opposite the face numbered1.Therefore,the remaining four faces share an edge with the face numbered1,so the product of the numbers is2×3×4×6=144.Answer:(B)15.The percentage10%is equivalent to the fraction110.Therefore,t=110s,or s=10t.Answer:(D)16.Since the base of the folded box measures5cm by4cm,then the area of the base of the boxis5(4)=20cm2.Since the volume of the box is60cm3and the area of the base is20cm2,then the height of the box is60=3cm.Therefore,each of the four identical squares has side length3cm,because the edges of these squares form the vertical edges of thebox.Therefore,the rectangular sheet measures3+5+3=11cm by3+4+3=10cm,and so has area11(10)=110cm2.Answer:(B) 17.Solution1Since SUR is a straight line,then∠RUV=180◦−∠SUV=180◦−120◦=60◦.Since P W and QX are parallel,then∠RV W=∠V T X=112◦.Since UV W is a straight line,then∠RV U=180◦−∠RV W=180◦−112◦=68◦.Since the measures of the angles in a triangle add to180◦,then∠URV=180◦−∠RUV−∠RV U=180◦−60◦−68◦=52◦Solution2Since SUR is a straight line,then∠RUV=180◦−∠SUV=180◦−120◦=60◦.Since P W and QX are parallel,then∠RST=∠RUV=60◦.Since ST X is a straight line,then∠RT S=180◦−∠V T X=180◦−112◦=68◦.Since the measures of the angles in a triangle add to180◦,then∠URV=∠SRT=180◦−∠RST−∠RT S=180◦−60◦−68◦=52◦Answer:(A) 18.Solution1When Catherine adds30litres of gasoline,the tank goes from18full to34full.Since34−18=68−18=58,then58of the capacity of the tank is30litres.Thus,18of the capacity of the tank is30÷5=6litres.Also,the full capacity of the tank is8×6=48litres.Toﬁll the remaining14of the tank,Catherine must add an additional14×48=12litres of gas.Because each litre costs$1.38,it will cost12×$1.38=$16.56toﬁll the rest of the tank. Solution2Suppose that the capacity of the gas tank is x litres.Starting with1of a tank,30litres of gas makes the tank3full,so1x+30=3x or5x=30or x=48.The remaining capacity of the tank is14x=14(48)=12litres.At$1.38per litre,it will cost Catherine12×$1.38=$16.56toﬁll the rest of the tank.Answer:(C) 19.The area of a semi-circle with radius r is1πr2so the area of a semi-circle with diameter d is1 2π(12d)2=18πd2.The semicircles with diameters UV,V W,W X,XY,and Y Z each have equal diameter andthus equal area.The area of each of these semicircles is18π(52)=258π.The large semicircle has diameter UZ=5(5)=25,so has area18π(252)=6258π.The shaded area equals the area of the large semicircle,minus the area of two small semicircles, plus the area of three small semicircles,which equals the area of the large semicircle plus the area of one small semicircle.Therefore,the shaded area equals6258π+258π=6508π=3254π.Answer:(A)20.The sum of the odd numbers from5to21is5+7+9+11+13+15+17+19+21=117Therefore,the sum of the numbers in any row is one-third of this total,or39.This means as well that the sum of the numbers in any column or diagonal is also39.Since the numbers in the middle row add to39,then the number in the centre square is 39−9−17=13.Since the numbers in the middle column add to39,then the number in the middle square in the bottom row is39−5−13=21.591317x21Since the numbers in the bottom row add to39,then the number in the bottom right square is39−21−x=18−x.Since the numbers in the bottom left to top right diagonal add to39,then the number in the top right square is39−13−x=26−x.Since the numbers in the rightmost column add to39,then(26−x)+17+(18−x)=39or 61−2x=39or2x=22,and so x=11.We can complete the magic square as follows:195159131711217Answer:(B) 21.We label the numbers in the empty boxes as y and z,so the numbers in the boxes are thus8,y,z,26,x.Since the average of z and x is26,then x+z=2(26)=52or z=52−x.We rewrite the list as8,y,52−x,26,x.Since the average of26and y is52−x,then26+y=2(52−x)or y=104−26−2x=78−2x.We rewrite the list as8,78−2x,52−x,26,x.Since the average of8and52−x is78−2x,then8+(52−x)=2(78−2x)60−x=156−4x3x=96x=32Therefore,x=32.Answer:(D) 22.Since JKLM is a rectangle,then the angles at J and K are each90◦,so each of SJP andQKP is right-angled.By the Pythagorean Theorem in SJP,we haveSP2=JS2+JP2=522+392=2704+1521=4225Since SP>0,then SP=√4225=65.Since P QRS is a rhombus,then P Q=P S=65.By the Pythagorean Theorem in QKP,we haveKP2=P Q2−KQ2=652−252=4225−625=3600Since KP>0,then KP=√3600=60.(Instead of using the Pythagorean Theorem,we could note instead that SJP is a scaled-up version of a3-4-5right-angled triangle and that QKP is a scaled-up version of a5-12-13 right-angled triangle.This would allow us to use the known ratios of side lengths to calculate the missing side length.)Since KQ and P Z are parallel and P K and W Q are parallel,then P KQW is a rectangle,andso P W=KQ=25.Similarly,JP ZS is a rectangle and so P Z=JS=52.Thus,W Z=P Z−P W=52−25=27.Also,SY RM is a rectangle.Since JM and KL are parallel(JKLM is a rectangle),JK and ML are parallel,and P Q and SR are parallel(P QRS is a rhombus),then∠MSR=∠KQP and∠SRM=∠QP K.Since SMR and QKP have two equal angles,then their third angles must be equal too.Thus,the triangles have the same proportions.Since the hypotenuses of the triangles are equal, then the triangles must in fact be exactly the same size;that is,the lengths of the corresponding sides must be equal.(We say that SMR is congruent to QKP by“angle-side-angle”.) In particular,MR=KP=60.Thus,ZY=SY−SZ=MR−JP=60−39=21.Therefore,the perimeter of rectangle W XY Z is2(21)+2(27)=96.Answer:(D) 23.First,we note that2010=10(201)=2(5)(3)(67)and so20102=223252672.Consider N consecutive four-digit positive integers.For the product of these N integers to be divisible by20102,it must be the case that two diﬀerent integers are divisible by67(which would mean that there are at least68integers in the list)or one of the integers is divisible by672.Since we want to minimize N(and indeed because none of the answer choices is at least68), we look for a list of integers in which one is divisible by672=4489.Since the integers must all be four-digit integers,then the only multiples of4489the we must consider are4489and8978.First,we consider a list of N consecutive integers including4489.Since the product of these integers must have2factors of5and no single integer within10 of4489has a factor of25,then the list must include two integers that are multiples of5.To minimize the number of integers in the list,we try to include4485and4490.Thus our candidate list is4485,4486,4487,4488,4489,4490.The product of these integers includes2factors of67(in4489),2factors of5(in4485and 4490),2factors of2(in4486and4488),and2factors of3(since each of4485and4488is divisible by3).Thus,the product of these6integers is divisible by20102.Therefore,the shortest possible list including4489has length6.Next,we consider a list of N consecutive integers including8978.Here,there is a nearby integer containing2factors of5,namely8975.So we start with the list8975,8976,8977,8978and check to see if it has the required property.The product of these integers includes2factors of67(in8978),2factors of5(in8975),and2 factors of2(in8976).However,the only integer in this list divisible by3is8976,which has only1factor of3.To include a second factor of3,we must include a second multiple of3in the list.Thus,we extend the list by one number to8979.Therefore,the product of the numbers in the list8975,8976,8977,8978,8979is a multiple of 20102.The length of this list is5.Thus,the smallest possible value of N is5.(Note that a quick way to test if an integer is divisible by3is to add its digit and see if this total is divisible by3.For example,the sum of the digits of8979is33;since33is a multiple of3,then8979is a multiple of3.)Answer:(A)24.We label the terms x1,x2,x3,...,x2009,x2010.Suppose that S is the sum of the odd-numbered terms in the sequence;that is,S=x1+x3+x5+···+x2007+x2009We know that the sum of all of the terms is5307;that is,x1+x2+x3+···+x2009+x2010=5307Next,we pair up the terms:each odd-numbered term with the following even-numbered term.That is,we pair theﬁrst term with the second,the third term with the fourth,and so on,until we pair the2009th term with the2010th term.There are1005such pairs.In each pair,the even-numbered term is one bigger than the odd-numbered term.That is, x2−x1=1,x4−x3=1,and so on.Therefore,the sum of the even-numbered terms is1005greater than the sum of the odd-numbered terms.Thus,the sum of the even-numbered terms is S+1005.Since the sum of all of the terms equals the sum of the odd-numbered terms plus the sum of the even-numbered terms,then S+(S+1005)=5307or2S=4302or S=2151.Thus,the required sum is2151.Answer:(C) 25.Before we answer the given question,we determine the number of ways of choosing3objectsfrom5objects and the number of ways of choosing2objects from5objects.Consider5objects labelled B,C,D,E,F.The possible pairs are:BC,BD,BE,BF,CD,CE,CF,DE,DF,EF.There are10such pairs.The possible triples are:DEF,CEF,CDF,CDE,BEF,BDF,BDE,BCF,BCE,BCD.There are10such triples.(Can you see why there are the same number of pairs and triples?)Label the six teams A,B,C,D,E,F.We start by considering team A.Team A plays3games,so we must choose3of the remaining5teams for A to play.As we saw above,there are10ways to do this.Without loss of generality,we pick one of these sets of3teams for A to play,say A plays B,C and D.We keep track of everything by drawing diagrams,joining the teams that play each other witha line.Thus far,we haveAB C DThere are two possible cases now–either none of B,C and D play each other,or at least one pair of B,C,D plays each other.Case1:None of the teams that play A play each otherIn the conﬁguration above,each of B,C and D play two more games.They already play A and cannot play each other,so they must each play E and F.This givesAE FB C DNo further choices are possible.There are10possible schedules in this type of conﬁguration.These10combinations come from choosing the3teams that play A.Case2:Some of the teams that play A play each otherHere,at least one pair of the teams that play A play each other.Given the teams B,C and D playing A,there are3possible pairs(BC,BD,CD).We pick one of these pairs,say BC.(This gives10×3=30conﬁgurations so far.)AB C DIt is now not possible for B or C to also play D.If it was the case that C,say,played D,then we would have the conﬁgurationAB C DE FHere,A and C have each played3games and B and D have each played2games.Teams E and F are unaccounted for thus far.They cannot both play3games in this conﬁguration as the possible opponents for E are B,D and F,and the possible opponents for F are B,D and E,with the“B”and“D”possibilities only to be used once.A similar argument shows thatB cannot play D.Thus,B or C cannot also play D.So we have the conﬁgurationAB C DHere,A has played3games,B and C have each played2games,and D has played1game.B andC must play1more game and cannot playD or A.They must play E and F in some order.There are2possible ways to assign these games(BE and CF,or BF and CE.)This gives30×2=60conﬁgurations so far.Suppose that B plays E and C plays F.AB C DE FSo far,A,B and C each play3games and E,F and D each play1game.The only way to complete the conﬁguration is to join D,E and F.AB C DE FTherefore,there are60possible schedules in this case.In total,there are10+60=70possible schedules.Answer:(E)。

国际数学竞赛试题及答案解析大全

国际数学竞赛试题及答案解析大全引言：国际数学竞赛是一项旨在测试学生数学能力、逻辑思维和解决问题技巧的全球性竞赛。

它不仅能够激发学生对数学的兴趣，还能培养他们的创新思维和团队合作精神。

本大全收录了近年来国际数学竞赛的典型试题及其详细答案解析，旨在帮助参赛者更好地准备和理解竞赛内容。

试题一：题目：证明对于任意正整数$ n $，$ 1^2 + 2^2 + 3^2 + \ldots + n^2 = \frac{n(n + 1)(2n + 1)}{6} $。

答案解析：我们可以通过数学归纳法来证明这个公式。

首先验证$ n = 1 $时成立：\[ 1^2 = \frac{1 \cdot 2 \cdot 3}{6} \]假设对于某个正整数$ k $，公式成立，即：\[ 1^2 + 2^2 + \ldots + k^2 = \frac{k(k + 1)(2k + 1)}{6} \]我们需要证明当$ n = k + 1 $时，公式仍然成立：\[ 1^2 + 2^2 + \ldots + k^2 + (k + 1)^2 = \frac{k(k + 1)(2k + 1)}{6} + (k + 1)^2 \]\[ = \frac{k(k + 1)(2k + 1) + 6(k + 1)^2}{6} \]\[ = \frac{(k + 1)(2k^2 + k + 6k + 6)}{6} \]\[ = \frac{(k + 1)(2k + 1)(k + 3)}{6} \]这正是我们想要证明的公式，因此通过数学归纳法证明了该公式对所有正整数$ n $都成立。

试题二：题目：在一个圆内接四边形ABCD中，已知AB = 2，BC = 3，CD = 4，DA = 5，求圆的半径。

答案解析：设圆的半径为$ r $。

根据圆内接四边形的性质，对角和为180度，即$ ∠A + ∠C = 180^\circ $。

利用余弦定理，我们可以求出$ AC $的长度：\[ AC^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC \cdot \cos(\angle ABC) \]\[ AC^2 = 2^2 + 3^2 - 2 \cdot 2 \cdot 3 \cdot \cos(180^\circ - \angle ACD) \]\[ AC^2 = 4 + 9 - 12 \cdot \cos(\angle ACD) \]同理，我们可以求出$ BD $的长度：\[ BD^2 = CD^2 + DA^2 - 2 \cdot CD \cdot DA \cdot \cos(\angle CDA) \]\[ BD^2 = 4^2 + 5^2 - 2 \cdot 4 \cdot 5 \cdot \cos(180^\circ - \angle ABC) \]\[ BD^2 = 16 + 25 - 40 \cdot \cos(\angle ABC) \]由于ABCD是圆内接四边形，$ AC + BD = 2r $。

滑铁卢智商测试题(3篇)

第1篇摘要：滑铁卢智商挑战测试题，一套精心设计的智力游戏，旨在挑战你的逻辑思维、观察力、记忆力以及创新能力。

本测试题适合15岁以上成年人，共包含50个小题，时间限制为60分钟。

完成挑战后，你将对自己的智力水平有一个全新的认识。

准备好了吗？让我们一起开始这场智慧的旅程！一、选择题（每题2分，共20分）1. 以下哪个数字不是2的倍数？A. 4B. 5C. 6D. 82. 下面哪个词与其他词不属于同一类别？A. 苹果B. 香蕉C. 轮胎D. 橙子3. 下列哪个城市不在亚洲？A. 东京B. 新德里C. 巴黎D. 北京4. 下列哪个词是反义词？A. 高兴B. 悲伤C. 美丽D. 高大5. 下面哪个物品不是交通工具？A. 自行车B. 汽车C. 手机D. 飞机6. 下列哪个数字是质数？A. 4B. 5C. 6D. 77. 下面哪个成语与其他成语不属于同一类别？A. 鹿死谁手B. 水滴石穿C. 狐假虎威D. 青出于蓝8. 下列哪个城市不在非洲？A. 开罗B. 约翰内斯堡C. 巴黎D. 马达加斯加9. 下面哪个词是名词？A. 吃饭B. 看书C. 学习D. 旅游10. 下列哪个数字是偶数？A. 3B. 5C. 7D. 9二、填空题（每题3分，共30分）11. 填入缺失的字母，使句子完整：“日____月____，年年有余。

”12. 填入缺失的字母，使句子完整：“千____一____，______一____。

”13. 填入缺失的字母，使句子完整：“水________，山________。

”14. 填入缺失的字母，使句子完整：“上________，下________。

”15. 填入缺失的字母，使句子完整：“人________，物________。

”16. 填入缺失的字母，使句子完整：“天________，地________。

”17. 填入缺失的字母，使句子完整：“海________，山________。

”18. 填入缺失的字母，使句子完整：“云________，风________。

滑铁卢数学竞赛

滑铁卢数学竞赛1、21.|x|>3表示的区间是（）[单选题] *A.（-∞，3）B.(-3,3)C. [-3,3]D. （-∞，-3）∪（3，+ ∞）(正确答案)2、15.下列数中，是无理数的为（）[单选题] *A.-3.14B.6/11C.√3(正确答案)D.03、9.一棵树在离地5米处断裂，树顶落在离树根12米处，问树断之前有多高（）[单选题] *A. 17(正确答案)B. 17.5C. 18D. 204、若tan（π-α）>0且cosα>0，则角α的终边在（）[单选题] *A.第一象限B.第二象限C.第三象限D.第四象限(正确答案)5、下列说法正确的是[单选题] *A.带“+”号和带“-”号的数互为相反数B.数轴上原点两侧的两个点表示的数是相反数C.和一个点距离相等的两个点所表示的数一定互为相反数D.一个数前面添上“-”号即为原数的相反数(正确答案)6、16、在中,则( ). [单选题] *A. AB<2AC (正确答案)B. AB=2ACC. AB>2ACD. AB与2AC关系不确定7、260°是第（）象限角？[单选题] *第一象限第二象限第三象限(正确答案)第四象限8、4.已知两圆的半径分别为3㎝和4㎝，两个圆的圆心距为10㎝,则两圆的位置关系是（）[单选题] *Ａ.内切Ｂ.相交Ｃ.外切Ｄ.外离(正确答案)9、下列各角中，是界限角的是（）[单选题] *A. 1200°B. -1140°C. -1350°(正确答案)D. 1850°10、下列各对象可以组成集合的是（）[单选题] *A、与1非常接近的全体实数B、与2非常接近的全体实数(正确答案)C、高一年级视力比较好的同学D、与无理数相差很小的全体实数11、若2?＝a2＝4 ?，则a?等于( ) [单选题] *A. 43B. 82C. 83(正确答案)D. 4?12、47、若△ABC≌△DEF，AB＝2，AC＝4，且△DEF的周长为奇数，则EF的值为（）[单选题] *A．3B．4C．1或3D．3或5(正确答案)13、8. 下列事件中，不可能发生的事件是(? ? )．[单选题] *A．明天气温为30℃B．学校新调进一位女教师C．大伟身长丈八(正确答案)D．打开电视机，就看到广告14、3．中国是最早采用正负数表示相反意义的量，并进行负数运算的国家．若零上10℃记作+10℃，则零下10℃可记作（）[单选题] *A．10℃B．0℃C.-10 ℃(正确答案)D．-20℃15、19．对于实数a、b、c,“a>b”是“ac2(c平方)>bc2(c平方) ; ”的（）[单选题] * A．充分不必要条件B．必要不充分条件(正确答案)C．充要条件D．既不充分也不必要条件16、22．如图棋盘上有黑、白两色棋子若干，找出所有使三颗颜色相同的棋在同一直线上的直线，满足这种条件的直线共有（）[单选题] *A．5条(正确答案)B．4条C．3条D．2条17、f（x）=-2x+5在x=1处的函数值为（）[单选题] *A、-3B、-4C、5D、3(正确答案)18、10. 如图所示，小明周末到外婆家，走到十字路口处，记不清哪条路通往外婆家，那么他一次选对路的概率是(? ? ?)．[单选题] *A．1/2B．1/3(正确答案)C．1/4D．119、15．如图所示，下列数轴的画法正确的是（）[单选题] *A．B．C．(正确答案)D．20、-950°是（）[单选题] *A. 第一象限角B. 第二象限角(正确答案)C. 第三象限角D. 第四象限角21、1．如图，∠AOB＝120°，∠AOC＝∠BOC，OM平分∠BOC，则∠AOM的度数为（）[单选题] *A．45°B．65°C．75°(正确答案)D．80°22、27．下列计算正确的是（）[单选题] *A．（﹣a3）2＝a6(正确答案)B．3a+2b＝5abC．a6÷a3＝a2D．（a+b）2＝a2+b223、16．“x2（x平方）－4x－5=0”是“x=5”的( ) [单选题] *A．充分不必要条件B．必要不充分条件(正确答案)C．充要条件D．既不充分也不必要条件24、19．下列两个数互为相反数的是（）[单选题] *A．（﹣）和﹣（﹣）B．﹣5和(正确答案)C．π和﹣14D．+20和﹣（﹣20）25、13．在数轴上，下列四个数中离原点最近的数是（）[单选题] *A．﹣4(正确答案)B．3C．﹣2D．626、14．命题“?x∈R，?n∈N*，使得n≥x2（x平方）”的否定形式是（）[单选题] * A．?x∈R，?n∈N*，使得n<x2B．?x∈R，?x∈N*，使得n<x2C．?x∈R，?n∈N*，使得n<x2D．?x∈R，?n∈N*，使得n<x2(正确答案)27、8．(2020·课标Ⅱ)已知集合U={－2，－1,0,1,2,3}，A={－1,0,1}，B={1,2}，则?U(A∪B)=( ) [单选题] *A．{－2,3}(正确答案)B．{－2,2,3}C．{－2，－1,0,3}D．{－2，－1,0,2,3}28、7人小组选出2名同学作正副组长，共有选法（）种。

欧几里得滑铁卢数学竞赛_2006EuclidSolution

y = x2−2(21)x+212−212+440 = (x−21)2−212+440 = (x−21)2−441+440 = (x−21)2−1
From this form, we immediately see that the coordinates of the vertex are (21, −1).
(c) Solution 1 To determine B, the point of intersection of the lines y = x and x + 2y = 12, we set y = x in the second equation to obtain x + 2x = 12 or 3x = 12 or x = 4.
3. (a) Answer: (21, −1)
Solution 1
The x-intercepts of the given parabola are x = 20 and x = 22.
The x-coordinate of the vertex of the parabola is the average of the x-intercepts, or
10A + B
resulting number is
.
10
So we want to determine A and B so that
10A + B
A+B
=
10
2
10A + B = 5(A + B)
5A = 4B
Since A and B are digits such that 5A = 4B, then A = 4 and B = 5 is the only possibility. Therefore, n = 45. (We can quickly check that the average of the digits of n is 4.5, the number obtained by putting a decimal point between the digits of n.)

2007 AMC 10A 试题及答案解析

2007 AMC 10A1、One ticket to a show costs at full price. Susan buys 4 ticketsusing a coupon that gives her a 25% discount. Pam buys 5 tickets using a coupon that gives her a 30% discount. How many more dollars does Pam pay than Susan?Solution= the amount Pam spent = the amount Susan spent▪▪Pam pays 10 more dollars than Susan2、Define and . What is ?Solution3、An aquarium has a rectangular base that measures 100 cm by 40cm and has a height of 50 cm. It is filled with water to a height of 40 cm. A brick with a rectangular base that measures 40 cm by 20 cm and a height of 10 cm is placed in the aquarium. By how many centimeters does the water rise?SolutionThe volume of the brick is . Thus the water volume rose .4、The larger of two consecutive odd integers is three times thesmaller. What is their sum?SolutionLet the two consecutive odd integers be , . Then , so and their sum is .5、A school store sells 7 pencils and 8 notebooks for . It also sells 5 pencils and 3 notebooks for . How much do 16 pencils and 10 notebooks cost?SolutionWe let cost of pencils in cents, number of notebooks in cents. ThenSubtracting these equations yields . Backwards solving gives . Thus the answer is .6、At Euclid High School, the number of students taking the AMC10 was 60 in 2002, 66 in 2003, 70 in 2004, 76 in 2005, 78 and 2006, and is 85 in 2007. Between what two consecutive years was there the largest percentage increase?SolutionWe compute the percentage increases:1.2.3.4.5.The answer is .In fact, the answer follows directly from examining the differences between each year. The largest differences are , and it is easy to see that due to the decreased starting number of students in 2002 that that will be our answer.7、Last year Mr. John Q. Public received an inheritance. He paid 20% in federal taxes on the inheritance, and paid 10% of what he has left in state taxes. He paid a total of for both taxes. How many dollars was the inheritance?SolutionAfter paying his taxes, he has of the inheritance left. Since is of the inheritance, the whole inheritance is.8、Triangles and are isosceles with and . Point D is inside . , and . What is the degree measure of ?SolutionWe angle chase, and find out that:▪▪▪9、Real numbers and satisfy the equations and. What is ?SolutionAndSubstitution gives , and solving for yields . Thus .10、The Dunbar family consists of a mother, a father, and somechildren. The average age of the members of the family is , the father is years old, and the average age of the mother and children is . How many children are in the family?SolutionLet be the number of children. Then the total ages of the family is , and the total number of people in the family is . So11、The numbers from to are placed at the vertices of a cube in sucha manner that the sum of the four numbers on each face is the same.What is this common sum?SolutionThe sum of the numbers on the top face of a cube is equal to the sum of the numbers on the bottom face of the cube; these numbersrepresent all of the vertices of the cube. Thus the answer is.12、Two tour guides are leading six tourists. The guides decide to split up. Each tourist must choose one of the guides, but with the stipulation that each guide must take at least one tourist. How many different groupings of guides and tourists are possible?SolutionEach tourist has to pick in between the guides, so for tourists there are possible groupings. However, since each guide must take at least one tourist, we subtract the cases where a guide has no tourist. Thus the answer is .13、Yan is somewhere between his home and the stadium. To get to the stadium he can walk directly to the stadium, or else he can walk home and then ride his bicycle to the stadium. He rides times as fast as he walks, and both choices require the same amount of time. What is the ratio of Yan's distance from his home to his distance from the stadium?SolutionLet the distance from Yan's initial position to the stadium be and the distance from Yan's initial position to home be . We are trying to find , and we have the following identity given by the problem:Thus and the answer is14、A triangle with side lengths in the ratio is inscribed in a circle of radius . What is the area of the triangle?SolutionSince 3-4-5 is a Pythagorean triple, the triangle is a right triangle. Since the hypotenuse is a diameter of the circumcircle, the hypotenuse is . Then the other legs are and . The area is15、Four circles of radius are each tangent to two sides of a square and externally tangent to a circle of radius , as shown. What is the area of the square?SolutionThe diagonal has length . Therefore the sides have length , and the area isAlternate SolutionExtend two radii from the larger circle to the centers of the two smaller circles above. This forms a right triangle of sides . The length of the hypotenuse of the right triangle plus twice the radius of the smaller circle is equal to the side of the square. It follows, then16、Integers , , , and , not necessarily distinct, are chosen independently and at random from to , inclusive. What is the probability that is even?SolutionThe only times when is even is when and are of the same parity. The chance of being odd is , so it has a probability of being even. Therefore, the probability that will be even is.17、Suppose that and are positive integers such that . What is the minimum possible value of ?Solutionmust be a perfect cube, so each power of a prime in the factorization for must be divisible by . Thus the minimum value of is , which makes . These sum to .18、Consider the -sided polygon , as shown. Each of its sides has length , and each two consecutive sides form a right angle. Suppose that and meet at . What is the area of quadrilateral ?SolutionSolution 1We can obtain the solution by calculating the area of rectangleminus the combined area of triangles and .We know that triangles and are similar because. Also, since , the ratio of the distance from toto the distance from to is also . Solving with the fact that the distance from to is 4, we see that the distance from tois .The area of is simply , the area of is, and the area of rectangle is .Taking the area of rectangle and subtracting the combined area of and yields .Solution 2Extend and and call their intersection .The triangles and are clearly similar with ratio , hence and thus . The area of the triangle is . The triangles and are similar as well, and we now know that the ratio of their dimensions is .Draw altitudes from onto and , let their feet be and . We get that . Hence .Then the area of is , and the area ofcan be obtained by subtracting the area of , which is . Hence the answer is .19、A paint brush is swept along both diagonals of a square to produce the symmetric painted area, as shown. Half the area of the square is painted. What is the ratio of the side length of the square to the brush width?SolutionSolution 1Without loss of generality, let the side length of the square be unit. The area of the painted area is of the area of the larger square, sothe total unpainted area is also . Each of the unpainted triangle hasarea . It is easy to tell that these triangles are isosceles right triangles, so let be the side length of one of the smaller triangles:The diagonal of the triangle is . The corners of the painted areasare also isosceles right triangles with side length . Its diagonal is equal to the width of the paint, and is . The answer we are looking for is thus . Multiply the numerator and the denominator by to simplify, and you get orwhich is .Solution 2Again, have the length of the square equal to and let the width of each individual stripe be . Note that you can split each stripe into two rectangles and two isosceles right triangles at the corners. Thenthe area of each stripe is . The area covered by the two total stripes is twice the area of one stripe, minus the area in the intersection of the stripes, which is a square with side length . This area is equal to So:.By the quadratic formula,It's easy to tell that is too large, so . We want to find , and . Multiply the numerator and the denominator by ,20、Suppose that the number satisfies the equation . What is the value of ?SolutionSolution 1Notice that . Thus.Solution 2. We apply the quadratic formula to get.Thus (so it doesn't matter which root of we use). Using the binomial theorem we can expand this out and collect terms to get .Solution 3We know that . We can square both sides to get, so . Squaring both sides again gives, so . Q. E. D.Solution 4We let and be roots of a certain quadratic. Specifically. We use Newton's Sums given the coefficients to find .21、A sphere is inscribed in a cube that has a surface area of square meters. A second cube is then inscribed within the sphere. What is the surface area in square meters of the inner cube?SolutionSolution 1We rotate the smaller cube around the sphere such that two opposite vertices of the cube are on opposite faces of the larger cube. Thus the main diagonal of the smaller cube is the side length of the outer square.Let be the surface area of the inner square. The ratio of the areas of two similar figures is equal to the square of the ratio of their sides. As the diagonal of a cube has length where is a side of the cube, the ratio of a side of the inner square to that of the outer square (and the side of the outer square = the diagonal of the inner square), we have . Thus .Solution 2 (computation)The area of each face of the outer cube is , and the edge length of the outer cube is . This is also the diameter of the sphere, and thus the length of a long diagonal of the inner cube.A long diagonal of a cube is the hypotenuse of a right triangle with a side of the cube and a face diagonal of the cube as legs. If a side of the cube is , we see that .Thus the surface area of the inner cube is .22、A finite sequence of three-digit integers has the property that the tens and units digits of each terms are, respectively, the hundreds and tens digits of the next term, and the tens and units digits of the last term are, respectively, the hundreds and tens digits of the first term. For example, such a sequence might begin with terms , , and and end with the term . Let be the sum of all the terms in the sequence. What is the largest prime number that always divides ?SolutionA given digit appears as the hundreds digit, the tens digit, and the units digit of a term the same number of times. Let be the sum of theunits digits in all the terms. Then , so must be divisible by . To see that it need not be divisible by any larger prime, the sequence gives .23、How many ordered pairs of positive integers, with , have the property that their squares differ by ?SolutionFor every two factors , we have. Since ,, from which it follows that the number of ordered pairs is given by the number of ordered pairs. There are factors of , which give us six pairs . However, since are positive integers, we also need that are positive integers, so and must have the same parity. Thus we exclude the factors , and we are left with four pairs .24、Circles centered at and each have radius , as shown. Point is the midpoint of , and . Segments and are tangent to the circles centered at and , respectively, and is a common tangent. What is the area of the shaded region ?SolutionThe area we are trying to find is simply. Obviously, . Thus, is a rectangle, and so its area is.Since is tangent to circle , is a right . We knowand , so is isosceles, a -right , andhas with length . The area of . For obvious reasons, , and so the area of is also .(or , for that matter) is the area of its circle. Thusand both have an area of .Plugging all of these areas back into the original equation yields.25、For each positive integer , let denote the sum of the digits of For how many values of isSolutionSolution 1For the sake of notation let . Obviously . Then the maximum value of is when , and the sum becomes . So the minimum bound is . We do casework upon the tens digit:Case 1: . Easy to directly disprove.Case 2: . , and if and otherwise.Subcase a:.This exceeds our bounds, so no solution here.Subcase b:.First solution.Case 3: . , and if andotherwise.Subcase a:. Second solution.Subcase b: . Third solution.Case 4: . But , and the these clearly sum to .Case 5: . So and (recall that ), and . Fourth solution.In total we have solutions, which are and . Solution 2Clearly, . We can break this up into three cases:Case 1:Inspection gives .Case 2: , ,If you set up an equation, it reduces towhich has as its only solution satisfying the constraints , .Case 3: , ,This reduces to. The only two solutions satisfying the constraints for this equation are , and , .The solutions are thus and the answer is . Solution 3As in Solution 1, we note that and . Obviously, .As , this means that , or equivalently that .Thus . For each possible we get three possible .(E. g., if , then is a number such that and, therefore .)For each of these nine possibilities we compute asand check whether .We'll find out that out of the 9 cases, in 4 the value has the correct sum of digits.This happens for .。

2009PascalSolution滑铁卢竞赛题答案

1.Calculating,2×9−√36+1=18−6+1=13.Answer:(D)2.On Saturday,Deepit worked6hours.On Sunday,he worked4hours.Therefore,he worked6+4=10hours in total on Saturday and Sunday.Answer:(E) 3.Since1piece of gum costs1cent,then1000pieces of gum cost1000cents.Since there are100cents in a dollar,the total cost is$10.00.Answer:(D) 4.Since each of the18classes has28students,then there are18×28=504students who attendthe school.On Monday,there were496students present,so504−496=8students were absent.Answer:(A) 5.The sum of the angles around any point is360◦.Therefore,5x◦+4x◦+x◦+2x◦=360◦or12x=360or x=30.Answer:(D) 6.When−1is raised to an even exponent,the result is1.When−1is raised to an odd exponent,the result is−1.Thus,(−1)5−(−1)4=−1−1=−2.Answer:(A) 7.Since P Q is horizontal and the y-coordinate of P is1,then the y-coordinate of Q is1.Since QR is vertical and the x-coordinate of R is5,then the x-coordinate of Q is5.Therefore,the coordinates of Q are(5,1).Answer:(C)8.When y=3,we have y3+yy2−y=33+332−3=27+39−3=306=5.Answer:(D)9.Since there are4♣’s in each of theﬁrst two columns,then at least1♣must be moved out ofeach of these columns to make sure that each column contains exactly three♣’s.Therefore,we need to move at least2♣’s in total.If we move the♣from the top left corner to the bottom right corner♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣and the♣from the second row,second column to the third row,ﬁfth column♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣then we have exactly three♣’s in each row and each column.Therefore,since we must move at least2♣’s and we can achieve the conﬁguration that we want by moving2♣’s,then2is the smallest number.(There are also other combinations of moves that will give the required result.)Answer:(B) 10.Solution1Since z=4and x+y=7,then x+y+z=(x+y)+z=7+4=11.Solution2Since z=4and x+z=8,then x+4=8or x=4.Since x=4and x+y=7,then4+y=7or y=3.Therefore,x+y+z=4+3+4=11.Answer:(C) 11.We write out theﬁve numbers to5decimal places each,without doing any rounding:5.076=5.07666...5.076=5.07676...5.07=5.070005.076=5.076005.076=5.07607...We can use these representations to order the numbers as5.07000,5.07600,5.07607...,5.07666...,5.07676...so the number in the middle is5.076.Answer:(E) 12.Solution1Since there are24hours in a day,Francis spends13×24=8hours sleeping.Also,he spends14×24=6hours studying,and18×24=3hours eating.The number of hours that he has left is24−8−6−3=7hours. Solution2Francis spends13+14+18=8+6+324=1724of a day either sleeping,studying or eating.This leaves him1−1724=724of his day.Since there are24hours in a full day,then he has7hours left.Answer:(D) 13.Solution1Since the sum of the angles in a triangle is180◦,then∠QP S=180◦−∠P QS−∠P SQ=180◦−48◦−38◦=94◦Therefore,∠RP S=∠QP S−∠QP R=94◦−67◦=27◦.Solution2Since the sum of the angles in a triangle is180◦,then∠QRP=180◦−∠P QR−∠QP R=180◦−48◦−67◦=65◦Therefore,∠P RS=180◦−∠P RQ=180◦−65◦=115◦.Using P RS,∠RP S=180◦−∠P RS−∠P SR=180◦−115◦−38◦=27◦Answer:(A)14.The perimeter of the shaded region equals the sum of the lengths of OP and OQ plus the lengthof arc P Q.Each of OP and OQ has length5.Arc P Q forms34of the circle with centre O and radius5,because the missing portion corre-sponds to a central angle of90◦,and so is14of the total circle.Thus,the length of arc P Q is34of the circumference of this circle,or34(2π(5))=152π.Therefore,the perimeter is5+5+152π≈33.56which,of the given answers,is closest to34.Answer:(A)15.After some trial and error,we obtain the two lists{4,5,7,8,9}and{3,6,7,8,9}.Why are these the only two?If the largest number of theﬁve integers was8,then the largest that the sum could be would be8+7+6+5+4=30,which is too small.This tells us that we must include one9in the list.(We cannot include any number larger than9,since each number must be a single-digit number.)Therefore,the sum of the remaining four numbers is33−9=24.If the largest of the four remaining numbers is7,then their largest possible sum would be 7+6+5+4=22,which is too small.Therefore,we also need to include an8in the list.Thus,the sum of the remaining three numbers is24−8=16.If the largest of the three remaining numbers is6,then their largest possible sum would be 6+5+4=15,which is too small.Therefore,we also need to include an7in the list.Thus,the sum of the remaining two numbers is16−7=9.This tells us that we need two diﬀerent positive integers,each less than7,that add to9.These must be3and6or4and5.This gives us the two lists above,and shows that they are the only two such lists.Answer:(B) 16.The area of the entire grid is4×9=36.The area of P QR is12(QR)(P Q)=12(3)(4)=6.The area of ST U is12(ST)(UT)=12(4)(3)=6.The area of the rectangle with base RS is2×4=8.Therefore,the total shaded area is6+6+8=20and so the unshaded area is36−20=16.The ratio of the shaded area to the unshaded area is20:16=5:4.Answer:(E) 17.We can suppose that each test is worth100marks.Since the average of herﬁve test marks is73%,then the total number of marks that she received is5×73=365.Once her teacher removes a mark,her new average is76%so the sum of the remaining four marks is4×76=304.Since365−304=61,then the mark removed was61%.Answer:(B) 18.Solution1From December31,1988to December31,2008,a total of20years have elapsed.A time period of20years is the same asﬁve4year periods.Thus,the population of Arloe has doubled 5times over this period to its total of 3456.Doubling 5times is equivalent to multiplying by 25=32.Therefore,the population of Arloe on December 31,1988was 345632=108.Solution 2The population doubles every 4years going forward,so is halved every 4years going backwards in time.The population on December 31,2008was 3456.The population on December 31,2004was 3456÷2=1728.The population on December 31,2000was 1728÷2=864.The population on December 31,1996was 864÷2=432.The population on December 31,1992was 432÷2=216.The population on December 31,1988was 216÷2=108.Answer:(D)19.Since Pat drives 60km at 80km/h,this takes him 60km 80km/h =34h.Since Pat has 2hours in total to complete the trip,then he has 2−34=54hours left to complete the remaining 150−60=90km.Therefore,he must travel at 90km 54h =360km/h =72km/h.Answer:(C)20.Since the three numbers in each straight line must have a product of 3240and must include 45,then the other two numbers in each line must have a product of 3240=72.The possible pairs of positive integers are 1and 72,2and 36,3and 24,4and 18,6and 12,and 8and 9.The sums of the numbers in these pairs are 73,38,27,22,18,and 17.To maximize the sum of the eight numbers,we want to choose the pairs with the largest possible sums,so we choose the ﬁrst four pairs.Thus,the largest possible sum of the eight numbers is 73+38+27+22=160.Answer:(E)21.Since each of Alice and Bob rolls one 6-sided die,then there are 6×6=36possible combinationsof rolls.Each of these 36possibilities is equally likely.Alice wins when the two values rolled diﬀer by 1.The possible combinations that diﬀer by 1are (1,2),(2,3),(3,4),(4,5),(5,6),(2,1),(3,2),(4,3),(5,4),and (6,5).Therefore,there are 10combinations when Alice wins.Thus her probability of winning is 1036=518.Answer:(C)22.Diameters P Q and RS cross at the centre of the circle,which we call O .The area of the shaded region is the sum of the areas of P OS and ROQ plus the sum of the areas of sectors P OR and SOQ .Each of P OS and ROQ is right-angled and has its two perpendicular sides of length 4(the radius of the circle).Therefore,the area of each of these triangles is 12(4)(4)=8.Each of sector P OR and sector SOQ has area 14of the total area of the circle,as each hascentral angle 90◦(that is,∠P OR =∠SOQ =90◦)and 90◦is one-quarter of the total central angle.Therefore,each sector has area 14(π(42))=14(16π)=4π.Thus,the total shaded area is 2(8)+2(4π)=16+8π.Answer:(E)23.The maximum possible mass of a given coin is 7×(1+0.0214)=7×1.0214=7.1498g.The minimum possible mass of a given coin is 7×(1−0.0214)=7×0.9786=6.8502g.What are the possible numbers of coins that could make up 1000g?To ﬁnd the largest number of coins,we want the coins to be as light as possible.If all of the coins were as light as possible,we would have 10006.8502≈145.98coins.Now,we cannot have a non-integer number of coins.This means that we must have at most 145coins.(If we had 146coins,the total mass would have to be at least 146×6.8502=1000.1292g,which is too heavy.)Practically,we can get 145coins to have a total mass of 1000g by taking 145coins at the minimum possible mass and making each slightly heavier.To ﬁnd the smallest number of coins,we want the coins to be as heavy as possible.If all of the coins were as heavy as possible,we would have 10007.1498≈139.86coins.Again,we cannot have a non-integer number of coins.This means that we must have at least 140coins.(If we had 139coins,the total mass would be at most 139×7.1498=993.8222g,which is too light.)Therefore,the diﬀerence between the largest possible number and smallest possible number of coins is 145−140=5.Answer:(B)24.Divide the large cube of side length 40into 8smaller cubes of side length 20,by making threecuts of the large cube through its centre using planes parallel to the pairs of faces.Each of these small cubes has the centre of the large cube as its vertex.Each of these small cubes also just encloses one of the large spheres,in the sense that the sphere just touches each of the faces of the small cube.We call the sphere that ﬁts in the central space the inner sphere.To make this sphere as large possible,its centre will be at the centre of the large cube.(If this was not the case,the centre be outside one of the small cubes,and so would be farther away from one of the large spheres than from another.)To ﬁnd the radius of the inner sphere,we must ﬁnd the shortest distance from the centre of the large cube (that is,the centre of the inner sphere)to one of the large spheres.(Think of starting the inner sphere as a point at the centre of the cube and inﬂating it until it just touches the large spheres.)Consider one of these small cubes and the sphere inside it.Join the centre of the small cube to one of its vertices.Since the small cube has side length 20,then this new segment has length √102+102+102or √300,since to get from the centre to a vertex,we must go over 10,down 10and across 10.(See below for an explanation of why this distance is thus √300.)The inner sphere will touch the large sphere along this segment.Thus,the radius of the inner sphere will be this distance (√300)minus the radius of the large sphere (10),and so is √300−10≈7.32.Of the given answers,this is closest to 7.3.(We need to justify why the distance from the centre of the small cube to its vertex is √102+102+102.Divide the small cube into 8tiny cubes of side length 10each.The distance from the centre of the small cube to its vertex is equal to the length of a diagonal of one of the tiny cubes.Consider a rectangular prism with edge lengths a ,b and c .What is the length,d ,of the diag-onal inside the prism?By the Pythagorean Theorem,a face with side lengths a and b has a diagonal of length √a 2+b 2.Consider the triangle formed by this diagonal,the diagonal of the prism and one of the vertical edges of the prism,of length c .cThis triangle is right-angled since the vertical edge is perpendicular to the top face.By the Pythagorean Theorem again,d 2=(√a 2+b 2)2+c 2,so d 2=a 2+b 2+c 2or d =√a 2+b 2+c 2.)Answer:(B)25.The three machines operate in a way such that if the two numbers in the output have a commonfactor larger than 1,then the two numbers in the input would have to have a common factor larger than 1.To see this,let us look at each machine separately.We use the fact that if two numbers are each multiples of d ,then their sum and diﬀerence are also multiples of d .Suppose that (m,n )is input into Machine A.The output is (n,m ).If n and m have a common factor larger than 1,then m and n do as well.Suppose that (m,n )is input into Machine B.The output is (m +3n,n ).If m +3n and n have a common factor d ,then (m +3n )−n −n −n =m has a factor of d as each part of the subtraction is a multiple of d .Therefore,m and n have a common factor of d .Suppose that (m,n )is input into Machine C.The output is (m −2n,n ).If m −2n and n have a common factor d ,then (m −2n )+n +n =m has a factor of d as each part of the addition is a multiple of d .Therefore,m and n have a common factor of d .In each case,any common factor that exists in the output is present in the input.Let us look at the numbers in the ﬁve candidates.After some work,we can ﬁnd the prime factorizations of the six integers:2009=7(287)=7(7)(41)1016=8(127)=2(2)(2)(127)1004=4(251)=2(2)(251)1002=2(501)=2(3)(167)1008=8(126)=8(3)(42)=16(3)(3)(7)=2(2)(2)(2)(3)(3)(7)1032=8(129)=8(3)(43)=2(2)(2)(3)(43)Therefore,the only one of 1002,1004,1008,1016,1032that has a common factor larger than 1with 2009is 1008,which has a common factor of 7with 2009.How does this help?Since2009and1008have a common factor of7,then whatever pair was input to produce(2009,1008)must have also had a common factor of7.Also,the pair that was input to create this pair also had a common factor of7.This can be traced back through every step to say that the initial pair that produces the eventual output of(2009,1008)must have a common factor of7.Thus,(2009,1008)cannot have come from(0,1).Notes:•This does not tell us that the other four pairs necessarily work.It does tell us,though, that(2009,1008)cannot work.•We can trace the other four outputs back to(0,1)with some eﬀort.(This process is easier to do than it is to describe!)To do this,we notice that if the output of Machine A was(a,b),then its input was(b,a), since Machine A switches the two entries.Also,if the output of Machine B was(a,b),then its input was(a−3b,b),since MachineB adds three times the second number to theﬁrst.Lastly,if the output of Machine C was(a,b),then its input was(a+2b,b),since MachineC subtracts two times the second number from theﬁrst.Consider(2009,1016)for example.We try toﬁnd a way from(2009,1016)back to(0,1).We only need toﬁnd one way that works,rather than looking for a speciﬁc way.We note before doing this that starting with an input of(m,n)and then applying MachineB then MachineC gives an output of((m+3n)−2n,n)=(m+n,n).Thus,if applyingMachine B then Machine C(we call this combination“Machine BC”)gives an output of (a,b),then its input must have been(a−b,b).We can use this combined machine to try to work backwards and arrive at(0,1).This will simplify the process and help us avoid negative numbers.We do this by making a chart and by attempting to make the larger number smaller wher-ever possible:Output Machine Input(2009,1016)BC(993,1016)(993,1016)A(1016,993)(1016,993)BC(23,993)(23,993)A(993,23)(993,23)BC,43times(4,23)(4,23)A(23,4)(23,4)BC,5times(3,4)(3,4)A(4,3)(4,3)BC(1,3)(1,3)A(3,1)(3,1)B(0,1)Therefore,by going up through this table,we can see a way to get from an initial input of(0,1)to aﬁnal output of(2009,1016).In a similar way,we can show that we can obtainﬁnal outputs of each of(2009,1004), (2009,1002),and(2009,1032).Answer:(D)。

NOIP2007年提高组（Pascal语言）初赛试题及答案

NOIP2007年提高组（Pascal语言）初赛试题及答案一、单项选择题题目:1. 在以下各项中, ( D ) 不是CPU的组成部分A. 控制器B. 运算器C. 寄存器D. 主板E. 算术逻辑单元(ALU)2. 在关系数据库中, 存放在数据库中的数据的逻辑结构以( E )为主A. 二叉树B. 多叉树C. 哈希表D. C+树E. 二维表3. 在下列各项中, 只有( D )不是计算机的存储容量常用单位A. ByteB. KBC. MBD. UBE. TB4. ASCII码的含义是 ( B )A. 二—十进制转换码B. 美国信息交换标准代码C. 数字的二进制数码D. 计算机可处理字符的唯一编码E. 常用字符的二进制编码5. 在Pascal语言中, 表达式(23 or 2 xor 5)的值是( A )A. 18B. 1C. 23D. 32E. 246. 在Pascal语言中, 判断整数a等于0或b等于0或c等于0的正确的条件表达式是( B )A. not ((a<>0) or (b<>0) or (c<>0))B. not ((a<>0) and (b<>0) and (c<>0))C. not ((a=0) and (b=0) and (c=0))D. (a=0) and (b=0) and (c=0)E. not ((a=0) or (b=0) or (c=0))7. 地面上有标号为A、B、C的3根细柱, 在A柱上方有10个直径相同中间有孔的圆盘, 从上到下次编号为1, 2, 3, ……，将A柱上的部分盘子经过B柱移入C柱, 也可以在B柱上暂存。

如果B柱上的操作记录为：“进，进，出，进，进，出，出，进，进，出，进，出，出”。

那么, 在C柱上, 从下到上的盘子的编号为( D ).A. 2 4 3 6 5 7B. 2 4 1 2 5 7C. 2 4 3 1 7 6D. 2 4 3 6 7 5E. 2 1 4 3 7 58. 与十进制数17.5625相对应的8进制数是( B )A. 21.5625B. 21.44C. 21.73D. 21.731E. 前4个答案都不对9. ……在以下各个描述中, 不一定是欧拉图的是:DA. 图G中没有度为奇数的顶点B. 包括欧拉环游的图(欧拉环游是指通过图中每边恰好一次的闭路径)C. 包括欧拉闭迹的图(欧拉迹是指通过途中每边恰好一次的路径)D. 存在一条回路, 通过每个顶点恰好一次10. ……, 关于死循环的说法中, 只有( A )是正确的.A. 不存在一种算法, 对任何一个程序及相应输入数据, 都可以判断是否会出现死循环, 因而, 任何编译系统都不作死循环检查.B. 有些编译系统可以检测出死循环.C. 死循环属于语法错误, 既然编译系统能检查各种语法错误, 当然也可以检查出死循环.D. 死循环与多进程中出现的"死锁"差不多, 而死锁是可以检查的, 因而, 死循环也是可以检测的E. 对于死循环, 只能等待发生时作现场处理, 没有什么更积极的手段.二、不定项选择题目11. 设A=B=true, C=D=false, 以下逻辑表达是值为真的是( ABC ) ......那3个符号不会打12. 命题“P->Q”可读做P蕴含Q, 其中P、Q是两个独立的命题. 只有命题P 成立而命题Q不成立时, 命题"P->Q"的值为False, 其它情况均为true. 与命题"P->Q"等角的逻辑关系式是( AD )A.(非P)并Q .D.非((非Q)交P)13. (2070)16+(34)8的结果是 ABDA. (8332)10B. (208C)16C. (100000000110)2D. (20214)814. 已知7个节点的二叉树的先根遍历是1 2 4 5 6 3 7(……), 后根遍历是 4 65 2 7 3 1, 则该二叉树的可能的中根遍历是(ABD )A. 4 2 6 5 1 7 3B. 4 2 5 6 1 3 7C. 4 2 3 1 5 4 7D. 4 2 5 6 1 7 315. ……下面关于冗余数据的说法中, 正确的是( BC )A. 应该在数据库中清除一切冗余数据.B. 与高级语言编写的数据处理系统相比, 用关系数据库编写的系统更容易消除冗余数据.C. 为高查询效率, 在数据库中可以适当保留一些冗余数据, 但更新时要做相容性检查.D. 作相容性检查会降低效率, 可以不理睬数据库中的冗余数据.16. 下列各软件中, 属于NOIP竞赛(复赛)推荐使用的语言环境有(ABD )A. gccB. g++C. Turbo CD. free pascal17. 以下断电后仍能保存数据的有( AB )A. 硬盘B. ROMC. 显存D. RAM18. 在下列关于计算机语言的说法中, 正确的有( CD )A. 高级语言比汇编语言更高级, 是因为他的程序的运行效率更高.B. 随着Pascal、C等高级语言的出现, 机器语言和汇编语言已经退出了历史舞台.C. 高级语言程序比汇编语言程序更容易从一种计算机移植到另一种计算机上.D. C是一种面向过程的高级计算机语言.19. 在下列关于算法复杂度的说法中, 正确的有( BC )A. 算法的时间复杂度, 是指它在某台计算机上具体是现实的运行时间.B. 算法的时间复杂度, 是指对于该算法的一种或几种主要的运算, 运算的次数与问题的规模之间的函数关系.C. 一个问题如果是NPC 类问题, 就意味着在解决该问题时, 不存在一个具有多项式时间复杂度的算法, 但这一点还没有得到理论上证实, 也没有被否定.D. 一个问题如果是NP 类, 与C 有相同的结论.20. ……在下列关于递归的说法中, 正确的是( AC )A. 在1977年前后形成编成的计算机高级语言"FORTRAN77"禁止程序使用递归, 原因之一是该方法可能会占用更多的内存空间.B. 和非递归算法相比, 解决同一个问题, 递归算法一般运行得更快一些.C. 对于较复杂的问题, 用递归方式变成往往比非递归方式更容易一些.D. 对于已定义好的标准数学函数sin(x), 应用程序中的语句"y=sin(sin(x));"就是一种递归调用.二、问题求解（共2题，每题5分，共计10分）。

欧几里得滑铁卢数学竞赛_2009EuclidSolution

Solution 2
We label the point (k, k) as K.
Since K lies on the line segment AB, then the slope of AK equals the slope of AB.
4 − (−4) 8
Line segment AB joins (0, 4) to (8, −4), so has slope
Since this equation is true for all values of x, then the coeﬃcients on the left side and right side must be equal, so a + ab = 2 and 3ab = 12. From the second equation, ab = 4 so the ﬁrst equation becomes a + 4 = 2 or a = −2. Since ab = 4, then −2b = 4 and so b = −2. Thus, a = b = −2.
Canadian Mathematics Competition
An activity of the Centre for Education in Mathematics and Computing, University of Waterloo, Waterloo, Ontario
2009 Euclid Contest
Solution 2
Since x3 − 6x2 + 5x = 0, then x(x2 − 6x + 5) = 0 or x(x − 5)(x − 1) = 0.
The three roots of this equation are x = 0, x = 1 and x = 5.

滑铁卢数学竞赛高中试题

滑铁卢数学竞赛高中试题一、选择题1. 已知函数$ f(x) = ax^2 + bx + c $，其中$ a, b, c $为实数，且$ f(1) = 2 $，$ f(-1) = 0 $，$ f(2) = 6 $。

求$ a $的值。

2. 一个圆的半径为5，圆心位于原点，求圆上点$ P(3,4) $到圆心的距离。

3. 若$ \sin(\alpha + \beta) = \frac{1}{2} $，$ \cos(\alpha + \beta) = \frac{\sqrt{3}}{2} $，且$ \alpha $在第二象限，$ \beta $在第一象限，求$ \sin(\alpha) $的值。

二、填空题1. 计算$ \int_{0}^{1} x^2 dx $。

2. 若$ \log_{2}8 = n $，则$ n $的值为______。

3. 一个等差数列的前三项分别为2，5，8，求该数列的第10项。

三、解答题1. 证明：对于任意正整数$ n $，$ 1^3 + 2^3 + ... + n^3 =\frac{n^2(n+1)^2}{4} $。

2. 一个矩形的长是宽的两倍，若矩形的周长为24，求矩形的面积。

3. 已知一个等比数列的前三项分别为3，9，27，求该数列的第5项。

四、应用题1. 一个工厂每天生产相同数量的零件，如果每天生产100个零件，工厂可以在30天内完成订单。

如果每天生产150个零件，工厂可以在20天内完成订单。

求工厂每天实际生产的零件数量。

2. 一个圆环的外圆半径是内圆半径的两倍，且圆环的面积为π。

求外圆的半径。

五、证明题1. 证明：对于任意实数$ x $，$ \cos(x) + \cos(2x) + \cos(3x) $可以表示为一个单一的余弦函数。

六、开放性问题1. 考虑一个无限大的棋盘，每个格子可以放置一个硬币。

2007滑铁卢竞赛试题答案

12 − 0
The line through A(9, 12) and B(13, 0) has slope
= −3 so, using the point-slope
9 − 13
form, has equation y − 0 = −3(x − 13) or y = −3x + 39.
2.
(a)
Since

4. (a) Solution 1 Rearranging and then squaring both sides, √ 2+ x−2 = x−2 √ x−2 = x−4 x − 2 = (x − 4)2 x − 2 = x2 − 8x + 16 0 = x2 − 9x + 18
the parabola.
Therefore, −18 = a(0 − 3)(0 + 3) or −18 = −9a or a = 2.
Thus, the equation of the parabola is y = 2(x − 3)(x + 3) = 2x2 − 18.

√ But we cannot have x − 2 = −1 (as square roots are always non-negative).
√ Therefore, x − 2 = 2 or x − 2 = 4 or x = 6. We can check as in Solution 1 that x = 6 is indeed a solution.
Overall, the perimeter of ABCDE is
√√
√
√√
AB + BC + CD + DE + EA = 2 2 + 2 2 + 2 + 2 + 2 3 = 4 + 4 2 + 2 3

2008滑铁卢竞赛试题答案

Canadian Mathematics Competition
An activity of the Centre for Education in Mathematics and Computing, University of Waterloo, Waterloo, Ontario
ADB ,
AB 2 = BD2 + DA2 = 92 + 122 = 81 + 144 = 225 √ so AB = 225 = 15 since AB > 0. By the Pythagorean Theorem in ADC , DC 2 = CA2 − AD2 = 202 − 122 = 400 − 144 = 256 √ so DC = 256 = 16 since AD > 0. The perimeter of ABC is AB + BC + CA = AB + (BD + DC ) + CA = 15 + (9 + 16) + 20 = 60 Solution 2 Since BD : DA = 9 : 12 = 3 : 4 and BDA is right-angled at B , then 5 BD = 15. to a 3-4-5 triangle. Thus, AB = 3 Since CA : DA = 20 : 12 = 5 : 3 and ADC is right-angled at D, then to a 3-4-5 triangle. Thus, DC = 4 CA = 16. 5 Therefore, the perimeter of ABC is
ADB is similar ADC is similar

2008PascalSolution滑铁卢竞赛题答案

1.Calculating,2+3+42×3×4=924=38.Answer:(E)2.Since3x−9=12,then3x=12+9=21.Since3x=21,then6x=2(3x)=2(21)=42.(Note that we did not need to determine the value of x.)Answer:(A)3.Calculating,√52−42=√25−16=√9=3.Answer:(B)4.Solution1Since JLMR is a rectangle and JR=2,then LM=2.Similarly,since JL=8,then RM=8.Since RM=8and RQ=3,then QM=8−3=5.Since KLMQ is a rectangle with QM=5and LM=2,its area is5(2)=10.Solution2Since JL=8and JR=2,then the area of rectangle JLMR is2(8)=16.Since RQ=3and JR=2,then the area of rectangle JKQR is2(3)=6.The area of rectangle KLMQ is the diﬀerence between these areas,or16−6=10.Answer:(C) 5.Since x=12and y=−6,then3x+y x−y =3(12)+(−6)12−(−6)=3018=53Answer:(C)6.Solution1Since∠P QS is an exterior angle of QRS,then∠P QS=∠QRS+∠QSR,so136◦=x◦+64◦or x=136−64=72.Solution2Since∠P QS=136◦,then∠RQS=180◦−∠P QS=180◦−136◦=44◦.Since the sum of the angles in QRS is180◦,then44◦+64◦+x◦=180◦or x=180−44−64=72.Answer:(A) 7.In total,there are5+6+7+8=26jelly beans in the bag.Since there are8blue jelly beans,the probability of selecting a blue jelly bean is826=413.Answer:(D)8.Since Olave sold108apples in6hours,then she sold108÷6=18apples in one hour.A time period of1hour and30minutes is equivalent to1.5hours.Therefore,Olave will sell1.5×18=27apples in1hour and30minutes.Answer:(A)9.Since the length of the rectangular grid is 10and the grid is 5squares wide,then the side length of each square in the grid is 10÷5=2.There are 4horizontal wires,each of length 10,which thus have a total length of 4×10=40.Since the side length of each square is 2and the rectangular grid is 3squares high,then the length of each vertical wire is 3×2=6.Since there are 6vertical wires,the total length of the vertical wires is 6×6=36.Therefore,the total length of wire is 40+36=76.Answer:(E)10.Solution 1Since Q is at 46and P is at −14,then the distance along the number line from P to Q is 46−(−14)=60.Since S is three-quarters of the way from P to Q ,then S is at −14+34(60)=−14+45=31.Since T is one-third of the way from P to Q ,then T is at −14+13(60)=−14+20=6.Thus,the distance along the number line from T to S is 31−6=25.Solution 2Since Q is at 46and P is at −14,then the distance along the number line from P to Q is 46−(−14)=60.Since S is three-quarters of the way from P to Q and T is one-third of the way from P to Q ,then the distance from T to S is 60 34−13 =60 912−412 =60 512 =25.Answer:(D)11.In total,30+20=50students wrote the Pascal Contest at Mathville Junior High.Since 30%(or 310)of the boys won certiﬁcates and 40%(or 410)of the girls won certiﬁcates,then the total number of certiﬁcates awarded was 310(30)+410(20)=9+8=17.Therefore,17of 50participating students won certiﬁcates.In other words,1750×100%=34%of the participating students won certiﬁcates.Answer:(A)12.Since the perimeter of the rectangle is 56,then2(x +4)+2(x −2)=562x +8+2x −4=564x +4=564x=52x =13Therefore,the rectangle is x +4=17by x −2=11,so has area 17(11)=187.Answer:(B)ing exponent rules,23×22×33×32=23+2×33+2=25×35=(2×3)5=65.Answer:(A)14.Solution 1The wording of the problem tells us that a +b +c +d +e +f must be the same no matter what numbers abc and def are chosen that satisfy the conditions.An example that works is 889+111=1000.In this case,a +b +c +d +e +f =8+8+9+1+1+1=28,so this must always be the value.Solution 2Consider performing this “long addition”by hand.Consider ﬁrst the units column.Since c +f ends in a 0,then c +f =0or c +f =10.The value of c +f cannot be 20or more,as c and f are digits.Since none of the digits is 0,we cannot have c +f =0+0so c +f =10.(This means that we “carry”a 1to the tens column.)Since the result in the tens column is 0and there is a 1carried into this column,then b +e ends in a 9,so we must have b +e =9.(Since b and e are digits,b +e cannot be 19or more.)In the tens column,we thus have b +e =9plus the carry of 1,so the resulting digit in the tens column is 0,with a 1carried to the hundreds column.Using a similar analysis in the hundreds column to that in the tens column,we must have a +d =9.Therefore,a +b +c +d +e +f =(a +d )+(b +e )+(c +f )=9+9+10=28.Answer:(D)15.Each of P SQ and RSQ is right-angled at S ,so we can use the Pythagorean Theorem inboth triangles.In RSQ ,we have QS 2=QR 2−SR 2=252−202=625−400=225,so QS =√225=15since QS >0.In P SQ ,we have P Q 2=P S 2+QS 2=82+225=64+225=289,so P Q =√289=17since P Q >0.Therefore,the perimeter of P QR is P Q +QR +RP =17+25+(20+8)=70.Answer:(E)16.Suppose the radius of the circle is r cm.Then the area M is πr 2cm 2and the circumference N is 2πr cm.Thus,πr 22πr =20or r 2=20or r =40.Answer:(C)17.Solution 1The large cube has a total surface area of 5400cm 2and its surface is made up of 6identical square faces.Thus,the area of each face,in square centimetres,is 5400÷6=900.Because each face is square,the side length of each face is √900=30cm.Therefore,each edge of the cube has length 30cm and so the large cube has a volume of 303=27000cm 3.Because the large cube is cut into small cubes each having volume 216cm 3,then the number of small cubes equals 27000÷216=125.Solution 2Since the large cube has 6square faces of equal area and the total surface area of the cube is 5400cm 2,then the surface area of each face is 5400÷6=900cm 2.Since each face is square,then the side length of each square face of the cube is √900=30cm,and so the edge length of the cube is 30cm.Since each smaller cube has a volume of 216cm 3,then the side length of each smaller cube is 3√216=6cm.Since the side length of the large cube is 30cm and the side length of each smaller cube is 6cm,then 30÷6=5smaller cubes ﬁt along each edge of the large cube.Thus,the large cube is made up of 53=125smaller cubes.Answer:(B)18.Solution1Alex has265cents in total.Since265is not divisible by10,Alex cannot have only dimes,so must have at least1quarter.If Alex has1quarter,then he has265−25=240cents in dimes,so24dimes.Alex cannot have2quarters,since265−2(25)=215is not divisible by10.If Alex has3quarters,then he has265−3(25)=190cents in dimes,so19dimes.Continuing this argument,we can see that Alex cannot have an even number of quarters,since the total value in cents of these quarters would end in a0,making the total value of the dimes end in a5,which is not possible.If Alex has5quarters,then he has265−5(25)=140cents in dimes,so14dimes.If Alex has7quarters,then he has265−7(25)=90cents in dimes,so9dimes.If Alex has9quarters,then he has265−9(25)=40cents in dimes,so4dimes.If Alex has more than9quarters,then he will have even fewer than4dimes,so we do not need to investigate any more possibilities since we are told that Alex has more dimes than quarters.So the possibilities for the total number of coins that Alex has are1+24=25,3+19=22, 5+14=19,and7+9=16.Therefore,the smallest number of coins that Alex could have is16.(Notice that each time we increase the number of quarters above,we are in eﬀect exchanging 2quarters(worth50cents)for5dimes(also worth50cents).)Solution2Suppose that Alex has d dimes and q quarters,where d and q are non-negative integers.Since Alex has$2.65,then10d+25q=265or2d+5q=53.Since the right side is odd,then the left side must be odd,so5q must be odd,so q must be odd.If q≥11,then5q≥55,which is too large.Therefore,q<11,leaving q=1,3,5,7,9which give d=24,19,14,9,4.The solution with d>q and d+q smallest is q=7and d=9,giving16coins in total.Answer:(B) 19.From the deﬁnition,theﬁrst and second digits of an upright integer automatically determinethe third digit,since it is the sum of theﬁrst two digits.Considerﬁrst those upright integers beginning with1.These are101,112,123,134,145,156,167,178,and189,since1+0=1,1+1=2,and so on.(The second digit cannot be9,otherwise the last“digit”would be1+9=10,which is impossible.)There are9such numbers.Beginning with2,the upright integers are202,213,224,235,246,257,268,and279.There are8of them.We can continue the pattern and determine the numbers of the upright integers beginning with 3,4,5,6,7,8,and9to be7,6,5,4,3,2,and1.Therefore,there are9+8+7+6+5+4+3+2+1=45positive3-digit upright integers.Answer:(D) 20.The sum of the six given integers is1867+1993+2019+2025+2109+2121=12134.The four of these integers that have a mean of2008must have a sum of4(2008)=8032.(We do not know which integers they are,but we do not actually need to know.)Thus,the sum of the remaining two integers must be12134−8032=4102.=2051.Therefore,the mean of the remaining two integers is41022(We can verify that1867,2019,2025and2121do actually have a mean of2008,and that1993 and2109have a mean of2051.)Answer:(D)21.The maximum possible value of pqis when p is as large as possible(that is,10)and q is as smallas possible(that is,12).Thus,the maximum possible value of pqis1012=56.The minimum possible value of pqis when p is as small as possible(that is,3)and q is as largeas possible(that is,21).Thus,the maximum possible value of pqis321=17.The diﬀerence between these two values is 56−17=3542−642=2942.Answer:(A)22.Suppose that the distance from Ginger’s home to her school is d km.Since there are60minutes in an hour,then33minutes(or15minutes)is15×1=1of anhour.Since Ginger walks at4km/h,then it takes her d4hours to walk to school.Since Ginger runs at6km/h,then it takes her d6hours to run to school.Since she saves116of an hour by running,then the diﬀerence between these times is116of anhour,sod 4−d6=1163d 12−2d12=116d12=116d=1216=34Therefore,the distance from Ginger’s home to her school is34km.Answer:(E)23.Suppose that the distance from line M to line L is d m.Therefore,the total length of piece W to the left of the cut is d m.Since piece X is3m from line M,then the length of piece X to the left of L is(d−3)m, because3of the d m to the left of L are empty.Similarly,the lengths of pieces Y and Z to the left of line L are(d−2)m and(d−1.5)m.Therefore,the total length of lumber to the left of line L isd+(d−3)+(d−2)+(d−1.5)=4d−6.5mSince the total length of lumber on each side of the cut is equal,then this total length is1 2(5+3+5+4)=8.5m.(We could insteadﬁnd the lengths of lumber to the right of line L to be5−d,6−d,7−d,and 5.5−d and equate the sum of these lengths to the sum of the lengths on the left side.) Therefore,4d−6.5=8.5or4d=15or d=3.75,so the length of the part of piece W to the left of L is3.75m.Answer:(D)24.We label the ﬁve circles as shown in the diagram.P QRS TWe note that there are 3possible colours and that no two adjacent circles can be coloured the same.Consider circle R .There are three possible colours for this circle.For each of these colours,there are 2possible colours for T (either of the two colours that R is not),since it cannot be the same colour as R .Circles Q and S are then either the same colour as each other,or are diﬀerent colours.Case 1:Q and S are the same colourIn this case,there are 2possible colours for Q (either of the colours that R is not)and 1pos-sibility for S (the same colour as Q ).For each of these possible colours for Q /S ,there are two possible colours for P (either of the colours that Q and S are not).P Q R S T3 choices2 choices2 choices1 choice2 choices In this case,there are thus 3×2×2×1×2=24possible ways of colouring the circles.Case 2:Q and S are diﬀerent coloursIn this case,there are 2possible colours for Q (either of the colours that R is not)and 1possibility for S (since it must be diﬀerent from R and diﬀerent from Q ).For each of these possible colourings of Q and S ,there is 1possible colour for P (since Q and S are diﬀerent colours,P is diﬀerent from these,and there are only 3colours in total).P Q R S T3 choices2 choices1 choice1 choice In this case,there are thus 3×2×2×1×1=12possible ways of colouring the circles.In total,there are thus 24+12=36possible ways to colour the circles.Answer:(D)25.Since P Q =2and M is the midpoint of P Q ,then P M =MQ =12(2)=1.Since P QR is right-angled at P ,then by the Pythagorean Theorem,RQ = P Q 2+P R 2= 22+(2√3)2=√4+12=√16=4(Note that we could say that P QR is a 30◦-60◦-90◦triangle,but we do not actually need this fact.)Since P L is an altitude,then ∠P LR =90◦,so RLP is similar to RP Q (these triangles have right angles at L and P respectively,and a common angle at R ).Therefore,P L QP =RP RQ or P L =(QP )(RP )RQ =2(2√3)4=√3.Similarly,RL RP =RP RQ so RL =(RP )(RP )RQ =(2√3)(2√3)4=3.Therefore,LQ =RQ −RL =4−3=1and P F =P L −F L =√3−F L .So we need to determine the length of F L .Drop a perpendicular from M to X on RQ .RP QM LFX Then MXQ is similar to P LQ ,since these triangles are each right-angled and they share a common angle at Q .Since MQ =12P Q ,then the corresponding sides of MXQ are half as long as those of P LQ .Therefore,QX =12QL =12(1)=12and MX =12P L =12(√3)=√32.Since QX =12,then RX =RQ −QX =4−12=72.Now RLF is similar to RXM(they are each right-angled and share a common angle at R).Therefore,F LMX=RLRXso F L=(MX)(RL)RX=√32(3)72=3√3.Thus,P F=√3−3√37=4√37.Answer:(C)。

2007年俄罗斯数学奥林匹克试题2

第33届俄罗斯数学奥林匹克
九年级试题
（第二天 2007年4月25日 9：00~14：00）
9.5 在凸100边形的每个顶点上都写有两个不同的数。

证明：可以从每个顶点上划去一个数，使得任意两个相邻顶点上剩下的数都互不相同。

9.6 在锐角△ABC 中，点M 与N 分别是边AB 及BC 的中点，点H 是由顶点B 所作高的垂足。

△AHN 与△CHM 的外接圆相交于点P （P ≠H ）。

证明：直线PH 经过线段MN 的中点。

9.7 在10×10的方格表中写着整数1到100：第1行从左到右依次写着1到10，第2行从左到右依次写着11到20，如此下去。

安德烈试图把方格表全部分成1×2的矩形，计算每个矩形中两数之积，再把得到的50个乘积相加。

他希望所得到的和数尽可能小，那么他应该怎样分割？
9.8 季玛算出了整数80到99中每一个数的阶乘的倒数。

他把所得的十进制小数分别打印在20张无限长的纸上（譬如在最后一张纸上所打印的数是1550
10.0001071599!="" 个）。

萨沙从其中的一张纸条上剪下一段，上面恰好有N 个数字且不带小数点。

如果萨沙不想让季玛猜出他是从哪张纸条上剪下的这N 个数字，那么N 的最大值是多少？。

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1.Calculating,3×(7−5)−5=3×2−5=6−5=1.Answer:(B) 2.Since x is less than−1and greater than−2,then the best estimate of the given choices is−1.3.Answer:(B) 3.The shaded square has side length1so has area12=1.The rectangle has dimensions3by5so has area3×5=15.Thus,the fraction of the rectangle that is shaded is115.Answer:(A)4.Calculating,25−52=32−25=7.Answer:(E) 5.In3hours,Leona earns$24.75,so she makes$24.75÷3=$8.25per hour.Therefore,in a5hour shift,Leona earns5×$8.25=$41.25.Answer:(E)6.Calculating,√64+√36√64+36=8+6√100=1410=75.Answer:(A)7.Solution1In total Megan and Dan inherit$1010000.Since each donates10%,then the total donated is10%of$1010000,or$101000.Solution2Megan donates10%of$1000000,or$100000.Dan donates10%of$10000,or$1000.In total,they donate$100000+$1000=$101000.Answer:(A) 8.We think of BC as the base of ABC.Its length is12.Since the y-coordinate of A is9,then the height of ABC from base BC is9.Therefore,the area of ABC is12(12)(9)=54.Answer:(B)9.Calculating the given diﬀerence using a common denominator,we obtain58−116=916.Since916is larger than each of12=816and716,then neither(D)nor(E)is correct.Since916is less than each of34=1216,then(A)is not correct.As a decimal,9=0.5625.Since35=0.6and59=0.5,then916>59,so(C)is the correct answer.Answer:(C)10.Since M=2007÷3,then M=669.Since N=M÷3,then N=669÷3=223.Since X=M−N,then X=669−223=446.Answer:(E)11.The mean of6,9and18is 6+9+183=333=11.Thus the mean of12and y is11,so the sum of12and y is2(11)=22,so y=10.Answer:(C) 12.In P QR,since P R=RQ,then∠RP Q=∠P QR=48◦.Since∠MP N and∠RP Q are opposite angles,then∠MP N=∠RP Q=48◦.In P MN,P M=P N,so∠P MN=∠P NM.Therefore,∠P MN=12(180◦−∠MP N)=12(180◦−48◦)=12(132◦)=66◦.Answer:(D)13.The prime numbers smaller than10are2,3,5,and7.The two of these numbers which are diﬀerent and add to10are3and7.The product of3and7is3×7=21.Answer:(B) 14.Since there were21males writing and the ratio of males to females writing is3:7,then thereare73×21=49females writing.Therefore,the total number of students writing is49+21=70.Answer:(D) 15.Solution1Theﬁrst stack is made up of1+2+3+4+5=15blocks.The second stack is made up of1+2+3+4+5+6=21blocks.There are36blocks in total.We start building the new stack from the top.Since there are more than21blocks,we need at least6rows.For7rows,1+2+3+4+5+6+7=28blocks are needed.For8rows,1+2+3+4+5+6+7+8=36blocks are needed.Therefore,Clara can build a stack with0blocks leftover.Solution2Since the new stack will be larger than the second stack shown,let us think about adding new rows to this second stack using the blocks from theﬁrst stack.Theﬁrst stack contains1+2+3+4+5=15blocks in total.Theﬁrst two rows that we would add to the bottom of the second stack would have7and8 blocks in them,for a total of15blocks.This uses all of the blocks from theﬁrst stack,with none left over,and creates a similar stack.Therefore,there are0blocks left over.Answer:(A) 16.Solution1The sum of the numbers in the second row is10+16+22=48,so the sum of the numbers in any row,column or diagonal is48.In theﬁrst row,P+4+Q=48so P+Q=44.In the third row,R+28+S=48so R+S=20.Therefore,P+Q+R+S=44+20=64.Solution 2The sum of the numbers in the second row is 10+16+22=48,so the sum of the numbers in any row,column or diagonal is 48.From the ﬁrst row,P +4+Q =48so P +Q =44.From the ﬁrst column,P +10+R =48so P +R =38.Subtracting these two equations gives (P +Q )−(P +R )=44−38or Q −R =6.From one of the diagonals,R +16+Q =48or Q +R =32.Adding these last two equations,2Q =38or Q =19,so R =32−Q =13.Also,P =44−Q =25.From the last row,13+28+S =48,or S =7.Thus,P +Q +R +S =25+19+13+7=64.Solution 3The sum of all of the numbers in the grid isP +Q +R +S +10+16+22+28+4=P +Q +R +S +80But the sum of the three numbers in the second column is 4+16+28=48,so the sum of the three numbers in each column is 48.Thus,the total of the nine numbers in the grid is 3(48)=144,so P +Q +R +S +80=144or P +Q +R +S =64.Answer:(C)17.At present,the sum of Norine’s age and the number of years that she has worked is 50+19=69.This total must increase by 85−69=16before she can retire.As every year passes,this total increases by 2(as her age increases by 1and the number of years that she has worked increases by 1).Thus,it takes 8years for her total to increase from 69to 85,so she will be 50+8=58when she can retire.Answer:(C)18.By the Pythagorean Theorem in P QR ,P Q 2=P R 2−QR 2=132−52=144,so P Q =√144=12.By the Pythagorean Theorem in P QS ,QS 2=P S 2−P Q 2=372−122=1225,so QS =√1225=35.Therefore,the perimeter of P QS is 12+35+37=84.Answer:(D)19.Since the reciprocal of 310is 1x+1 ,then 1x +1=1031x =73x =37so x =37.Answer:(C)20.Draw a line from F to BC ,parallel to AB ,meeting BC at P .AD CB E F PSince EB is parallel to F P and ∠F EB =90◦,then EBP F is a rectangle.Since EB =40,then F P =40;since EF =30,then BP =30.Since AD =80,then BC =80,so P C =80−30=50.Therefore,the area of EBCF is sum of the areas of rectangle EBP F (which is 30×40=1200)and F P C (which is 12(40)(50)=1000),or 1200+1000=2200.Since the areas of AEF CD and EBCF are equal,then each is 2200,so the total area ofrectangle ABCD is 4400.Since AD =80,then AB =4400÷80=55.Therefore,AE =AB −EB =55−40=15.Answer:(D)21.Let us ﬁrst consider the possibilities for each integer separately:•The two-digit prime numbers are 11,13,17,19.The only one whose digits add up to a prime number is 11.Therefore,P =11.•Since Q is a multiple of 5between 2and 19,then the possible values of Q are 5,10,15.•The odd numbers between 2and 19that are not prime are 9and 15,so the possible values of R are 9and 15.•The squares between 2and 19are 4,9and 16.Only 4and 9are squares of prime numbers,so the possible values of S are 4and 9.•Since P =11,the possible value of Q are 5,10and 15,and T is the average of P and Q ,then T could be 8,10.5or 13.Since T is also a prime number,then T must be 13,so Q =15.We now know that P =11,Q =15and T =13.Since the ﬁve numbers are all diﬀerent,then R cannot be 15,so R =9.Since R =9,S cannot be 9,so S =4.Therefore,the largest of the ﬁve integers is Q =15.Answer:(B)22.By the Pythagorean Theorem,P R =QR 2+P Q 2=√152+82=√289=17km.Asafa runs a total distance of 8+15+7=30km at 21km/h in the same time that Florence runs a total distance of 17+7=24km.Therefore,Asafa’s speed is 30=5of Florence’s speed,so Florence’s speed is 4×21=84km/h.Asafa runs the last 7km in 721=13hour,or 20minutes.Florence runs the last7km in 7845=3584=512hour,or25minutes.Since Asafa and Florence arrive at S together,then Florence arrived at R5minutes before Asafa.Answer:(E) 23.The total area of the larger circle isπ(22)=4π,so the total area of the shaded regions mustbe512(4π)=53π.Suppose that∠ADC=x◦.The area of the unshaded portion of the inner circle is thusx360of the total area of the innercircle,orx360(π(12))=x360π(since∠ADC isx360of the largest possible central angle(360◦)).The area of the shaded portion of the inner circle is thusπ−x360π=360−x360π.The total area of the outer ring is the diﬀerence of the areas of the outer and inner circles,orπ(22)−π(12)=3π.The shaded area in the outer ring will bex360of this total area,since∠ADC isx360of thelargest possible central angle(360◦). So the shaded area in the outer ring isx360(3π)=3x360π.So the total shaded area(which must equal 53π)is,in term of x,3x360π+360−x360π=360+2x360π.Therefore,360+2x360=53=600360,so360+2x=600or x=120.Thus,∠ADC=120◦.Answer:(B) 24.First,we complete the next several spaces in the spiral to try to get a better sense of thepattern:1716151413185431219612112078910212223242526We notice from this extended spiral that the odd perfect squares lie on a diagonal extending down and to the right from the1,since1,9,25and so on will complete a square of numbers when they are written.(Try blocking out the numbers larger than each of these to see this.) This pattern does continue since when each of these odd perfect squares is reached,the number of spaces up to that point in the sequence actually does form a square.Theﬁrst odd perfect square larger than2007is452=2025.2025will lie18spaces to the left of2007in this row.(The row with2025will actually be long enough to be able to move18spaces to the left from2025.)The odd perfect square before2025is432=1849,so1850will be the number directly above 2025,as the row containing1849will continue one more space to the right before turning up.Since1850is directly above2025,then1832is directly above2007.The odd perfect square after2025is472=2209,so2208will be the number directly below 2025,since2209will be one space to the right and one down.Since2208is directly below2025,then2190is directly below2007.Therefore,the sum of the numbers directly above and below2007is1832+2190=4022.Answer:(E) 25.For x and3x to each have even digits only,x must be in one of the following forms.(Here,a,b,c represent digits that can each be0,2or8,and n is a digit that can only be2or8.)•nabc(2×3×3×3=54possibilities)•na68(2×3=6possibilities)•n68a(2×3=6possibilities)•68ab(3×3=9possibilities)•n668(2possibilities)•668a(3possibilities)•6668(1possibility)•6868(1possibility)In total,there are82possibilities for x.In general terms,these are the only forms that work,since digits of0,2and8in x pro-duce even digits with an even“carry”(0or2)thus keeping all digits in3x even,while a6may be used,but must be followed by8or68or668in order to give a carry of2.More precisely,why do these forms work,and why are they the only forms that work?First,we note that3×0=0,3×2=6,3×4=12,3×6=18and3×8=24.Thus,each even digit of x will produce an even digit in the corresponding position of3x,but may aﬀect the next digit to the left in3x through its“carry”.Note that a digit of0or2in x produces no carry,while a digit of8in x produces an even carry.Therefore,none of these three digits can possibly create an odd digit in3x(either directly or through carrying),as they each create an even digit in the corresponding position of3x and do not aﬀect whether the next digit to the left is even or odd.(We should note that the carry into any digit in3x can never be more than2,so we do not have to worry about creating a carry of1from a digit in x of0or2,or a carry of3from a digit of8in x through multiple carries.) So a digit of2or8can appear in any position of x and a digit of0can appear in any position of x except for theﬁrst position.A digit of4can never appear in x,as it will always produce a carry of1,and so will al-ways create an odd digit in3x.A digit of6can appear in x,as long as the carry from the previous digit is2to make thecarry forward from the6equal to2.(The carry into the6cannot be larger than2.)When this happens,we have3×6+Carry=20,and so a2is carried forward,which does not aﬀect whether next digit is even or odd.A carry of2can occur if the digit before the6is an8,or if the digit before the6is a6whichis preceded by8or by68.Combining the possible uses of0,2,6,and8gives us the list of possible forms above,and hence82possible values for x.Answer:(A)。