2020年美国数学竞赛(AMC10A)的试题与解答

广东省广州市华南师范大学(510631) 李湖南

1.What value ofxsatisfies

译文方程的解x是多少?

解化简可得,故(E)正确.

2.The numbers 3,5,7,a,andbhave an average(arithmetic mean)of 15.What is the average ofaandb?

A.0 B.15 C.30 D.45 E.60

译文数字3,5,7,a和b的平均值(算术平均值)是15.则a和b的平均值是多少?

解依题意有3+5+7+a+b=15×5,则a+b=60,平均值为30,故(C)正确.

3.Assuming34 and5,what is the value in simplest form of the following expression?

A.−1 B.1 C.D.E.

译文设则表达式的最简形式是什么?

解约分即得故(A)正确.

4.A driver travels for 2 hours at 60 miles per hour,during which her car gets 30 miles per gallon of gasoline.She is paid$0.50 per mile,and her only expense is gasoline at$2.00 per gallon.What is her net rate of pay,in dollars per hour,after this expense?

A.20 B.22 C.24 D.25 E.26

译文一位司机以60 英里/小时的速度驾车2 小时,她的车每跑30 英里需要消耗1 加仑汽油.她能获得0.50 美元/英里的报酬,唯一的花费就是2 美元/加仑的汽油.问她每小时除去消耗之后的净收益是多少美元?

解1 个小时她能跑60 英里,获得60×0.50=30 美元,汽油费为60÷30×2 = 4 美元,故净收益为26 美元,(E)正确.

5.What is the sum of all real numbersxfor which

A.12 B.15 C.18 D.21 E.25

译文满足方程的所有实数x之和是多少?

解分别解方程x2−12x+34=2 和x2−12x+34=−2,可得x1,2=4,8 和x3,4=6,故和为18,(C)正确.

6.How many 4-digit positive integers(that is,integers between 1000 and 9999,inclusive)having only even digits are divisible by 5?

A.80 B.100 C.125 D.200 E.500

译文有多少个四位的正整数(也就是在1000 和9999之间的整数)能被5 整除且所有数字均为偶数?

解依题意,符合条件的四位数的个位数只能是0,十位数和百位数可以是0,2,4,6,8,千位数只能是2,4,6,8,共有1×5×5×4=100 种选择,故(B)正确.

7.The 25 integers from−10 to 14,inclusive,can be arranged to form a 5-by-5 square in which the sum of the numbers in each row,the sum of the numbers in each column,the sum of the numbers along each of the main diagonals are all the same.What is the value of this common sum?

A.2 B.5 C.10 D.25 E.50

译文将25 个整数分别是从−10 到14,放入5×5 的格子中,使得格子里的每行、每列和两条对角线的数字和均相等.问这个数字和是多少?

解这是一个5 阶幻方问题,25 个数字之和是(−10)+(−9)+···+13+14 = 50,分别放入5 行,故每行的数字和是10,(C)正确.

8.What is the value of 1+2+3−4+5+6+7−8+···+197+198+199−200?

A.9800 B.9900 C.10000 D.10100 E.10200

译文1+2+3−4+5+6+7−8+···+197+198+199−200的值是多少?

解原式= 1 +(2 +3−4)+5 +(6 +7−8)+···+197+(198+199−200)= 2×(1+5+···+197)=,故(B)正确.

9.A single bench section at a school event can hold either 7 adults or 11 children.WhenNbench sections are connected end to end,an equal number of adults and children seated together will occupy all the bench space.What is the least possible positive integer value ofN?

A.9 B.18 C.27 D.36 E.77

译文在某学校的活动中,一条长凳可以坐7 个成人或者11 个儿童.当N条长凳首尾相接的时候,刚好坐满了相同数量的成人和儿童.问N的最小正整数值是多少?

解设有x条长凳坐了儿童,则有N −x条长凳坐了成人,依题意有11x= 7(N −x),因而故Nmin=18,(B)正确.

10.Seven cubes,whose volumes are 1,8,27,64,125,216,and 343 cubic units,are stacked vertically to form a tower in which the volumes of the cubes decrease from bottom to top.Except for the bottom cube,the bottom face of each cube lies completely on top of the cube below it.What is the total surface area of the tower(including the bottom)in square units?

A.644 B.658 C.664 D.720 E.749

译文七个立方体,体积分别是1,8,27,64,125,216,343个立方单位,依次按照体积大小由底到顶垂直地堆积成一座塔.除了最底部的立方体,每个立方体的底面都完全被下面的立方体的顶面覆盖.问这座塔的表面积(包括底面)是多少个平方单位?

解这七个数都是立方数,则这七个立方体的棱长分别是1,2,3,4,5,6,7,从而塔的侧面积为4×(12+22+...+72)=560,而上、下底面积之和为2×72=98,共658,故(B)正确.

11.What is the median of the following list of 4040 numbers?

1,2,3,...,2020,12,22,32,...,20202

A.1974.5 B.1975.5 C.1976.5 D.1977.5 E.1978.5

译文下列4040 个数:1,2,3,...,2020,12,22,32,...,20202的中位数是多少?

解由于442=1936,452=2025,从而以上数列按递增排列的话,就成为:

12.Triangle ∆AMCis isosceles withAM=AC.Mediansandare perpendicular to each other,andMV=CU=12.What is the area of ∆AMC?

A.48 B.72 C.96 D.144 E.192

译文等腰∆AMC中,AM=AC,中线和互相垂直,且MV=CU=12.则∆AMC的面积是多少?

解如图示,设MV交UC于点E,过A作AD ⊥ MC于D,交UV于F,则AD是MC的中垂线.依题意,U,V分别是AM,AC的中点,则UV是∆AMC的中位线,且UM=V C=,从而∆UMC∆V CM(SSS),可得∠UCM= ∠V MC= 45°,因 此EM=EC,点E在AD上.

图1

于是∆UV E～∆CME,且均是等腰直角三角形,因而进而∆AMC,得故(C)正确.

13.A frog sitting at the point (1,2)begins a sequence of jumps,where each jump is parallel to one of the coordinate axes and has length 1,and the direction of each jump (up,down,left,right)is chosen independently at random.The sequence ends when the frog reaches a side of the square with vertices(0,0),(0,4),(4,0),and(4,4).What is the probability that the sequence of jumps ends on a vertical side of the square?

译文一只青蛙坐在点(1,2)上,开始一系列的跳跃,每次跳跃都平行于坐标轴且长度为1,方向(上、下、左、右)是随机的且独立,当青蛙到达由点(0,0),(0,4),(4,0),(4,4)构成的正方形的一条边的时候,跳跃终止.问跳跃终止于正方形竖直的两条边上的概率是多少?

解如图示,青蛙在点F1处,它可以向四个方向跳跃,概率均为向左跳跃,立刻达成目标; 向上、向右、向下分别跳跃到点A1,C,A3处,再通过其它跳跃达成目标.根据对称性,青蛙由点A1,A2,A3,A4出发达成目标的概率是一样的,设为a;青蛙由点B1,B2出发达成目标的概率是一样的,设为b;青蛙由点F1,F2出发达成目标的概率是一样的,设为x;青蛙由点C出发达成目标的概率设为c.

图2

因此,P(青蛙由F1出发达成目标)=P(青蛙向左)+P(青蛙向上)×P(青蛙由A1出发达成目标)+P(青蛙向右)×P(青蛙由C出发达成目标)+P(青蛙向下)×P(青蛙由A3出发达成目标),即有同理,可得方程组成立,解得故(B)正确.

14.Real numbersxandysatisfiesx+y=4 andxy=−2.What is the value ofx++y?

A.360 B.400 C.420 D.440 E.480

译文实数x,y满足方程x+y= 4 和xy=−2.则的值是多少?

解依题意可得x2+y2= (x+y)2−2xy= 20,x3+y3=(x+y)(x2−xy+y2)=88,原式==440.故(D)正确.

15.A positive integer divisor of 12! is chosen at random.The probability that the divisor is a perfect square can be expressed as,wheremandnare relatively prime positive integers.What ism+n?

A.3 B.5 C.12 D.18 E.23

译文随机选取12!的一个正整数因子,该因子是一个完全平方数的概率可以表示为,其中m,n为互素的正整数.则m+n是多少?

解由于12! = 210×35×52×7×11,则12! 有11×6×3×2×2=792 个正因子;

设k是12!的平方因子,则k=2a×3b×5c×7d×11i,其中a≤10,b≤5,c≤2,d≤1,i≤1,且a,b,c,d,i均为非负偶数,即a=0,2,4,6,8,10,b=0,2,4,c=0,2,d=i=0,此时k有6×3×2=36 种选择.从而所求概率为故m+n=23,(E)正确.

16.A point is chosen at random within the square in the coordinate plane whose vertices are(0,0),(2020,0),(2020,2020),and(0,2020).The probability that the point lies withindunits of a lattice point is.(A point(x,y)is a lattice point ifxandyare both integers.)What isdto the nearest tenth?

A.0.3 B.0.4 C.0.5 D.0.6 E.0.7

译文坐标平面上有一个以(0,0),(2020,0),(2020,2020)和(0,2020)为顶点的正方形.在正方形内随机选择一个点,该点位于格点的d个单位内的概率是.(点(x,y)称为格点,若x和y均为整数.)则d精确到十分位是多少?

图3

解如图示,以格点为圆心,d为半径作一些圆,则正方形内的圆内部分就是符合条件的点集.因此,该点落在此区域的概率为求得故(B)正确.

17.DefineP(x)=(x −12)(x −22)···(x −1002).How many integersnare there such thatP(n)≤0?

A.4900 B.4950 C.5000 D.5050 E.5100

译文定义P(x)= (x −12)(x −22)···(x −1002).则有多少个整数n使得P(n)≤0?

解解不等式P(n)=(n−12)(n−22)···(n−1002)≤0,得12≤n≤22,32≤n≤42,··· ,992≤n≤1002,因此符合条件的n有(22−12+1)+(42−32+1)+···+(1002−992+1)=2×(2+4+···+100)=5100 个.故(E)正确.

18.Let(a,b,c,d)be an ordered quadruple of not necessarily distinct integers,each one of them in the set{0,1,2,3}.For how many such quadruples is it true thatad−bcis odd? (For example,(0,3,1,1)is one such quadruple,because 0·1−3·1=−3 is odd.)

A.48 B.64 C.96 D.128 E.192

译文设(a,b,c,d)是一个四元数,其中a,b,c,d ∈{0,1,2,3}且可以相同.则有多少个这样的四元数使得ad −bc是奇数?(例如,(0,3,1,1)就是一个符合条件的四元数,因为0·1−3·1=−3 是奇数.)

解要使得ad −bc是奇数,ad和bc必一奇一偶,分两种情况:

(1)ad是奇数,bc是偶数:此时a,d ∈{1,3},b,c可以是一奇一偶、一偶一奇、两个偶数,共有2×2×(2×2×3)=48种选择;(2)ad是偶数,bc是奇数:同理可得48 种选择.故共有96 个,(C)正确.

19.As shown in the figure below,a regular dodecahedron(the polyhedron consisting of 12 congruent regular pentagonal faces)floats in space with two horizontal faces.Note that there is a ring of five slanted faces adjacent to the top face,and a ring of five slanted faces adjacent to the bottom face.How many ways are there to move from the top face to the bottom face via a sequence of adjacent faces so that each face is visited at most once and moves are not permitted from the bottom ring to the top ring?

A.125 B.250 C.405 D.640 E.810

译文如下图左所示,一个正十二面体(由12 个完全相同的正五边形面组成的多面体)放置在两个水平面的空间中.注意到顶面附近有一个由五个斜面组成的圆环,底面附近有一个由五个斜面组成的圆环.问有多少种方法可以通过一系列相邻面从顶面移动到底面,使得每个面最多经过一次,并且不允许从底环移动到顶环?

图5

解将该图简化成平面图:顶面为点T,底面为点B,顶环依次为点T1,T2,T3,T4,T5,底环依次为点B1,B2,B3,B4,B5,相邻的两个面用线段连接,如上图右所示.则原问题相当于:有多少条从点T出发,经过顶环,再经过底环,最后到达点B的不重复路径?

(1)点T到顶环,有5 种选择,即T1,T2,T3,T4,T5; (2)不妨设先到T1,又有9 种选择到达底环,分别是:T1→,T1→T2→,T1→T2→T3→,T1→T2→T3→T4→,T1→T2→T3→T4→T5→,T1→T5→,T1→T5→T4→,T1→T5→T4→T3→,T1→T5→T4→T3→T2→; (3)每一个顶环上的点有2 种方式到达底环,如T1→B1或T1→B5; (4)以到达B1为例,有以下9 条路径到达点B,分别是:B1→B,B1→B2→B,B1→B2→B3→B,B1→B2→B3→B4→B,B1→B2→B3→B4→B5→B,B1→B5→B,B1→B5→B4→B,B1→B5→B4→B3→B,B1→B5→B4→B3→B2→B.

综上分析可得,共有5×9×2×9=810 条路径,故(E)正确.

20.QuadrilateralABCDsatisfies ∠ABC= ∠ACD=90°,AC= 20 andCD= 30.DiagonalsACandBDintersects at pointEandAE= 5.What is the area of QuadrilateralABCD?

A.330 B.340 C.350 D.360 E.370

译文四边形ABCD满足∠ABC= ∠ACD= 90°,AC= 20,CD= 30.对角线和交于点E,且AE=5.求四边形ABCD的面积是多少?

图6

解如图示,以AC为直径作一个圆,交BD与点F,依题意可得EC=15,设BE=x,依据相交弦定理AE ·EC=BE ·EF,则得再由切割线定理DC2=DF ·DB,得解得或x=而可得S∆ABC=60,故SABCD=360,(D)正确.

21.There exists a unique strictly increasing sequence of nonnegative integersa1

A.117 B.136 C.137 D.273 E.306

译文存在唯一严格递增的非负整数列a1< a2<··· < ak使得则k是多少?

解令217=x,则

22.For how many positive integersn≤1000 isnot divisible by 3? (recall thatis the greatest integer less than or equal tox.)

A.22 B.23 C.24 D.25 E.26

译文有多少个正整数n≤1000 使得不被3 整除?(注意表示小于等于x的最大整数)

解当n不是998,999 或1000 的因子时,易得

(1)公因子:n=1 时,N=2997 能被3 整除;n=2 时,N=1498 不能被3 整除;

(3)999 的非公因子:N=不能被3 整除;

(4)1000 的非公因子:N=不能被3 整除.

综上可得,符合条件的n有25−3=22 个,故(A)正确.

23.LetTbe the triangle in the coordinate plane with vertices(0,0),(4,0),and(0,3).Consider the following five isometries(rigid transformations)of the plane:rotation of 90°,180°,and 270°counterclockwise around the origin,reflection across thex-axis,and reflection across they-axis.How many of the 125 sequences of three of these transformations(not necessarily distinct)will returnTto its original position? (For example,a 180°rotation,followed by a reflection across thex-axis,followed by a reflection across they-axis will returnTto its original position,but a 90°rotation,followed by a reflection across thex-axis,followed by another reflection across thex-axis will not returnTto its original position.)

A.12 B.15 C.17 D.20 E.25

图7

译文设T是坐标平面上以(0,0),(4,0)和(0,3)为顶点的三角形.考虑以下五种平面上的等距变换(刚体变换):绕原点作90°,180°和270°的逆时针旋转,关于x轴或y轴的反射.任选三种变换(不必不同)可以组成125 种组合,有多少种组合将使得T变回起始位置?(例如,一个关于y轴的反射,接着一个关于x轴的反射,再接着一个180°的旋转,将会使得T变回起始位置;但一个关于x轴的反射,接着另一个关于x轴的反射,再接着一个90°的旋转,将不会使得T变回起始位置.)

解分两种情况:(1)全部由旋转组成:只要三次旋转的角度和为360°或720°即可满足要求,因此有90°+90°+180°,90°+180°+90°,180°+90°+90°,270°+270°+180°,270°+180°+270°,180°+270°+270°共6 种组合;(2)由旋转和反射组合而成:有y轴+x轴+180°,y轴+180°+x轴,180°+x轴+y轴,180°+y轴+x轴,x轴+180°+y轴,x轴+y轴+180°,也是6 种组合.故(A)正确.

24.Letnbe the least positive integer greater than 1000 for which gcd(63,n+120)= 21 and gcd(n+63,120)= 60.What is the sum of digits ofn?

A.12 B.15 C.18 D.21 E.24

译文设n是大于1000 的使gcd(63,n+120)= 21,gcd(n+63,120)=60 成立的最小正整数.则n的数字和是多少?

解由gcd(63,n+120)= 21,可得n ≡6(mod 21);由gcd(n+63,120)= 60,可得n ≡57(mod 60).联立解得n ≡237(mod 420),于是n= 1077,1497,1917,···.当n= 1077 时,gcd(63,n+120)= 63,不符;当n= 1497 时,gcd(n+63,120)= 120,也不符;当n= 1917 时,验证后符合条件,此时数字和为18,故(C)正确.

25.Jason rolls three fair standard six-sided dice.Then he looks at the rolls and chooses a subset of the dice(possibly empty,possibly all three dice)to reroll.After rerolling,he wins if and only if the sum of the numbers faces up on the three dice is exactly 7.Jason always plays to optimize his chances of winning.What is the probability that he chooses to reroll exactly two of the dice?

译文詹森掷3 颗标准、均匀的骰子,他看了结果之后会选择若干(可能是0,也可能是3)颗重掷.当3 颗骰子正面朝上的数字和为7 点的时候,他就赢了.詹森总是按照朝着他赢的最优策略去掷.问他刚好选择2 颗骰子重掷的概率是多少?

解掷1 颗骰子得1,2,3,4,5,6 点的概率均为;掷2颗骰子得3 点只有两种情况:12 和21,概率为,···;掷3颗骰子得7 点有15 种情况:115,151,511,124,142,214,241,412,421,133,313,331,223,232,322,概率为,···.经过计算,所有结果如下表所示:

分类/概率/结果1 2 3 4 5 6 7掷1 颗1 1 1 1 1 1 6 6 6 6 6 6掷2 颗1 2 3 4 5 36 36 36 36 36 1 3 3 10 15掷3 颗216 216 216 216 216

因此,詹森要选择2 颗骰子重掷,则上次掷的结果中,任意两颗骰子的数字和不能小于7 点,否则他将选择重掷1 颗骰子;且不能3 颗骰子都是4 点或者以上,要不然他将选择重掷3 颗骰子.根据以上分析,满足条件的情况有:(1)掷出1 点、6 点、6 点,3 种情况;(2)掷出2 点、5 点、5 点,3 种情况;(3)掷出2 点、5 点、6 点,6 种情况; (4)掷出2 点、6 点、6 点,3 种情况;(5)掷出3 点、4 点、4 点,3 种情况;(6)掷出3 点、4点、5 点,6 种情况;(7)掷出3 点、4 点、6 点,6 种情况;(8)掷出3 点、5 点、5 点,3 种情况;(9)掷出3 点、5 点、6 点,6 种情况;(10)掷出3 点、6 点、6 点,3 种情况.