【精品】2020美国数学大联盟)挑战赛三年级真题(附答案+中文翻译+解题思路)
2019年美国数学竞赛8年级(AMC8)真题(附答案)(电脑版)
2019 年美国数学竞赛 8 年级(AMC8) 真题(附答案 )(电脑版 )
2019 年美国数学竞赛 8 年级(AMC8) 真题(附答案 )(电脑版 )
2019 年美国数学竞赛 8 年级(AMC8) 真题(附答案 )(电脑版 )
2019 年美国数学竞赛 8 年级(AMC8) 真题(附答案 )(电脑版 )
2019 年美国数学竞赛 8 年级(AMC8) 真题(附答案 )(电脑版 )
2019 年美国数学竞赛 8 年级(AMC8) 真题(附答案 )(电脑版 )
AMC 系列比赛一共有以下几个比赛：
American Mathematics Competition 8 - C 8
American Mathematics Competition 10/12 - AMC 10/12
American Invitational Mathematics Exam - AIME United States Mathematical Olympiad and Junior Mathematical Olympiad - USA(J)MO
2019 年美国数学竞赛 8 年级(AMC8) 真题(附答案 )(电脑版 )
2019 年美国数学竞赛 8 年级(AMC8) 真题(附答案 )(电脑版 )
2019 年美国数学竞赛 8 年级(AMC8) 真题(附答案 )(电脑版 )
2019 年美国数学竞赛 8 年级(AMC8) 真题(附答案 )(电脑版 )
2019 年美国数学竞赛 8 年级(AMC8) 真题(附答案 )(电脑版 )
美国数学竞赛 8 年级(AMC8)真题（附答案）
AMC 系列全称 American Mathematics Competitions 的（1950 年开始举办） 的和最负盛名初高中生数学竞赛。 日下午结束。
美国“数学大联盟杯赛” 中国赛区 初赛五年级试卷
2017-2018年度美国“数学大联盟杯赛”(中国赛区)初赛(五年级)(初赛时间：2017年11月26日，考试时间90分钟，总分200分)学生诚信协议：考试期间，我确定没有就所涉及的问题或结论，与任何人、用任何方式交流或讨论， 我确定我所填写的答案均为我个人独立完成的成果，否则愿接受本次成绩无效的处罚。
请在装订线内签名表示你同意遵守以上规定。
考前注意事项：1. 本试卷是五年级试卷，请确保和你的参赛年级一致；2. 本试卷共4页(正反面都有试题)，请检查是否有空白页，页数是否齐全；3. 请确保你已经拿到以下材料：本试卷(共4页，正反面都有试题)、答题卡、答题卡使用说明、英文词汇手册、 草稿纸。
考试完毕，请务必将英文词汇手册带回家，上面有如何查询初赛成绩、 及如何参加复赛的说明。
其他材料均不能带走，请留在原地。
选择题：每小题5分，答对加5分，答错不扣分，共200分，答案请填涂在答题卡上。
1. The smallest possible sum of two different prime numbers isA) 3B) 4C) 5D) 62. The greatest common factor of two numbers is3. The product of these two numbers mustbe divisible byA) 6 B) 9 C) 12 D) 18 3. The sum of 5 consecutive one-digit integers is at most A) 15 B) 25 C) 35 D) 45 4. How many two-digit multiples of 10 are also multiples of 12?A) 4B) 3C) 2D) 15. I have read exactly13of the total number of chapters in my 120-page book. If each chapter has the same whole number of pages, then the total number of chapters I have left could beA) 16 B) 24 C) 32 D) 50 6. What is the greatest odd factor of 44 × 55 × 66?A) 36 B) 55 C) 35 × 55 D) 36 × 55 7. What is the sum of the factors of the prime number 2017? A) 2016B) 2017C) 2018D) 20198. Lynn ran in 6 times as many races as the number of racesshe won. How many of her 126 races did Lynn not win?A) 21B) 90C) 96D) 1059. The least common multiple of 8 and 12 is the greatest common factor of 120 andA) 80B) 124C) 144D) 18010. January has the greatest possible number of Saturdays when January 1 occurs on any ofthe following days of the week exceptA) Thursday B) Friday C) Saturday D) Sunday 11. The number that is 10% of 1000 is 10 more than 10% ofA) 90B) 100C) 900D) 99012. The sum of 16 fours has the same value as the product of ? fours.A) 2 B) 3 C) 4 D) 16 13. Of the following, which is the sum of two consecutive integers?A) 111 111B) 222 222C) 444 444D) 888 88814. Abe drove for 2 hours at 30 km/hr. and for 3 hours at 50 km/hr. What was Abe’s averagespeed over the 5 hours?A) 35 km/hr.B) 40 km/hr.C) 42 km/hr.D) 45 km/hr.15. My broken watch runs twice as fast as it should. If my watch first broke at 6:15 P.M.,what time was displayed on my watch 65 minutes later?A) 7:20 P.M. B) 7:25 P.M.C) 8:20 P.M. D) 8:25 P.M.16. (2018 × 2017) + (2018 × 1) =A) 20172 B) 20182 C) 20183D) (2018 + 2017)217. A prized bird lays 2, 3, or 4 eggs each day. If the bird laid 17 eggs in 1 week,on at most how many days that week did the bird lay exactly 2 eggs?A) 2B) 3C) 4D) 518. Of the following, which could be the perimeter of a rectangle whoseside-lengths, in cm, are prime numbers?A) 10 cmB) 22 cmC) 34 cmD) 58 cm19. The average of all possible total values of a 4-coin stack of nickels and dimes (containingat least one of each coin) isA) 20¢B) 30¢C) 40¢D) 60¢20. The diameter of Ann’s drum i s 40 cm more than the radius. What is half the circumference of the drum?A) 120π cmB) 80π cmC) 60π cmD) 40π cm21. Of the following, which expression has the greatest number offactors that are multiples of 2018?A) 2018 × 12B) 20182C) 20192D) 20192019第1页，共4页 第2页，共4页22. When the sum of the factors of a prime number is divided by that prime number, theremainder isA) 0 B) 1 C) 2 D) 3 23. What is the sum of the digits of the greatest integer that has a square root less than 100? A) 18B) 36C) 99D) 10024. My favorite number has 6 different factors. If the product of all 6 factors is 123, what isthe sum of the factors of my favorite number?A) 24B) 28C) 32D) 3625. For how many different pairs of unequal positive integers less than 10 is the least commonmultiple of the numbers less than their product?A) 6B) 7C) 8D) 926. Exactly 12 of the students in my class have at least one brother, and 12 have at least onesister. If 13have no siblings, what fraction of the students in my class have at least onebrother and at least one sister?A) 16 B) 15 C) 14 D) 1327. Each day, Sal swims a lap 1 second faster than on the daybefore. If Sal swims a lap in 60 minutes on the 1st day, on what day does he swim a lap in 10% less time than the 1st day?A) 359th B) 360th C) 361st D) 362nd 28. 20172018 × 20172019 = 2017 ? × 20171009A) 1010B) 2010C) 3028D) 403829. Both arcs AB and AD are quarter circles of radius 5, figure on the right.Arc BCD is a semi-circle of radius 5. What is the area of the region ABCD ?A) 25 B) 10 + 5π C) 50D) 50 + 5π30. For every $5 I earn from my job, I save $2. For every $4 I save from my job, I am givenan additional $1 from my parents to add to my savings. How much must I earn in order to have $40 in savings?A) $160B) $120C) $100D) $8031. In the figure on the right, the side-length of the smaller squareis 4. The four arcs are four semi-circles. Each side of square ABCD is tangent to one of the semi-circles. The area of ABCD isA) 32B) 36C) 48D) 6432. A million is a large number, a “1” follo wed by 6 zeros. A googol is a large number, a “1”followed by one hundred zeros. A googolplex is a large number, a “1” followed by a googol of zeros. A googolplexian is a large number, a “1” fo llowed by a googolplex of zeros. A googolplexian isA) 10100 B) 1001010C) 100101010D) None of the above33. An integral triangle is a triangle with positive integral side-lengths and a positive area.Such a triangle can have a perimeter as small as 3. What is the next smallest possible perimeter of an integral triangle?A) 4B) 5C) 6D) 734. 2 liter of 2% fat milk + 3 liter of 3% fat milk = 5 liter of ? fat milkA) 2.5%B) 2.6%C) 5%D) 6%35. One day, a motorist came to a hill that was ten-mile drive up one side and a ten-mile drivedown the other. He drove up the hill at an average speed of 30 miles per hour. How fast will he have to drive down the other side to average 60 miles per hour for the entire 20-mile distance?A) 30 miles per hour B) 60 miles per hour C) 90 miles per hour D) None of the above 36. What is the weight of a fish if it weighs ten pounds plus half its weight?A) 10B) 15C) 20D) 2537. Without using pennies, how many different combinations of coins (nickels, dimes,quarters) will make 30 cents?A) 3B) 4C) 5D) 638. A man once bought a fine suit for which he paid $30 more than14of its price. How much did he pay for the suit? A) $30B) $35C) $40D) $4539. A father is five times as old as his son. In fifteen years he will be only twice as old. Howold is the father at present?A) 40B) 35C) 30D) 2540. It takes 30 minutes to completely fill a tank. If, however, a hole allows13of the water that is entering the tank to escape, how long will it then take to fill the tank?A) 40 B) 45 C) 60 D) 90第3页，共4页第4页，共4页。
2019年美国数学竞赛(AMC10B)的试题与解答
这时第二个容器里充满了 3 的水. 问小容器与大容器的容积 4
之比是多少?
解: 设第一、二个容器的容积分别为 V1 和 V2, 依题意有
5 6 V1 =
3 4 V2,
从而
V1 V2
=
3 4 5 6
=
9 , 故 (D) 正确. 10
2. Consider the statement, ”If n is not prime, then n − 2 is
√ 由垂线段最短知, m2 + n2
|√am + bn|
=
√ 5.
当且仅当
a2 + b2
OA 与直线 ax + by = 5 垂直即 an = bm 时取最小值.
解法 2 由柯西不等式, (a2 +b2)(m2 +n2) (am+bn)2,
即 5(m2 + n2) 等号成立.
√ 25, m2 + n2
6 ter and the second was empty. She poured all the water from the
first container into the second container, at which point the sec3
ond container was full of water. What is the ratio of the volume 4
解: 反例是指符合命题的条件但不符合命题的结论的例
子, 即满足“若 n 不是素数, 则 n − 2 是素数. ”选项中只有 27
满足, 故 (E) 正确.
评注 运用柯西不等式关键是“配凑”形式, 充分利用已
知条件. 当出现分式和的形式时, 一般构造分式分母和作为
AMC 美国数学竞赛 2002 AMC 10A 试题及答案解析
2002 AMC 10A1、The ratio is closest to which of the following numbers?SolutionWe factor as . As , ouranswer is .2、For the nonzero numbers , , , define.Find .Solution. Ouranswer is then .Alternate solution for the lazy: Without computing the answer exactly,we see that , , and . The sumis , and as all the options are integers, the correct one is obviously .3、According to the standard convention for exponentiation,.If the order in which the exponentiations are performed is changed, how many other values are possible?SolutionThe best way to solve this problem is by simple brute force.It is convenient to drop the usual way how exponentiation is denoted,and to write the formula as , where denotes exponentiation. We are now examining all ways to add parentheses to this expression. There are 5 ways to do so:1.2.3.4.5.We can note that . Therefore options 1 and 2 are equal, and options 3 and 4 are equal. Option 1 is the one given in the problem statement. Thus we only need to evaluate options 3 and 5.Thus the only other result is , and our answer is .4、For how many positive integers does there exist at least one positive integer such that ?infinitely manySolutionSolution 1For any we can pick , we get , therefore theanswer is .Solution 2Another solution, slightly similar to this first one would be using Simon's Favorite Factoring Trick.Let , thenThis means that there are infinitely many numbers that can satisfythe inequality. So the answer is .5、Each of the small circles in the figure has radius one. The innermost circle is tangent to the six circles that surround it, and each of those circles is tangent to the large circle and to its small-circle neighbors. Find the area of the shaded region.SolutionThe outer circle has radius , and thus area . The littlecircles have area each; since there are 7, their total area is . Thus,our answer is .6、Cindy was asked by her teacher to subtract from a certain numberand then divide the result by . Instead, she subtracted and thendivided the result by , giving an answer of . What would heranswer have been had she worked the problem correctly?SolutionWe work backwards; the number that Cindy started with is. Now, the correct result is . Ouranswer is .7、If an arc of on circle has the same length as an arc of oncircle , then the ratio of the area of circle to the area of circle isSolutionLet and be the radii of circles A and B, respectively.It is well known that in a circle with radius r, a subtended arc oppositean angle of degrees has length .Using that here, the arc of circle A has length . The arcof circle B has length . We know that they are equal,so , so we multiply through and simplify to get . As all circles are similar to one another, the ratio of the areas is just thesquare of the ratios of the radii, so our answer is .8、Betsy designed a flag using blue triangles, small white squares, anda red center square, as shown. Let be the total area of the bluetriangles, the total area of the white squares, and the area of thered square. Which of the following is correct?SolutionThe blue that's touching the center red square makes up 8 triangles, or 4 squares. Each of the corners is 2 squares and each of the edges is 1, totaling 12 squares. There are 12 white squares, thus we have.9、There are 3 numbers A, B, and C, such that ,and . What is the average of A, B, and C?More than 1SolutionNotice that we don't need to find what A, B, and C actually are, just their average. In other words, if we can find A+B+C, we will be done.Adding up the equations gives soand the average is . Our answer is .10、Compute the sum of all the roots of.SolutionSolution 1We expand to get which isafter combining like terms. Using the quadratic partof Vieta's Formulas, we find the sum of the roots is . Solution 2Combine terms to get, hence the rootsare and , thus our answer is .11、Jamal wants to store computer files on floppy disks, each ofwhich has a capacity of megabytes (MB). Three of his files requireMB of memory each, more require MB each, and theremaining require MB each. No file can be split between floppydisks. What is the minimal number of floppy disks that will hold all the files?SolutionA 0.8 MB file can either be on its own disk, or share it with a 0.4 MB. Clearly it is not worse to pick the second possibility. Thus we will have 3 disks, each with one 0.8 MB file and one 0.4 MB file.We are left with 12 files of 0.7 MB each, and 12 files of 0.4 MB each.Their total size is MB. The total capacity of 9 disks is MB, hence we need at least 10 more disks. And wecan easily verify that 10 disks are indeed enough: six of them will carry two 0.7 MB files each, and four will carry three 0.4 MB files each.Thus our answer is .12、Mr. Earl E. Bird leaves his house for work at exactly 8:00 A.M. every morning. When he averages miles per hour, he arrives at hisworkplace three minutes late. When he averages miles per hour, hearrives three minutes early. At what average speed, in miles per hour, should Mr. Bird drive to arrive at his workplace precisely on time?SolutionSolution 1Let the time he needs to get there in be t and the distance he travelsbe d. From the given equations, we know that and. Setting the two equal, we have andwe find of an hour. Substituting t back in, we find . From, we find that r, and our answer, is .Solution 2Since either time he arrives at is 3 minutes from the desired time, the answer is merely the harmonic mean of 40 and 60. The harmonicmean of a and b is . In this case, a and b are 40 and 60,so our answer is , so .Solution 3A more general form of the argument in Solution 2, with proof:Let be the distance to work, and let be the correct average speed.Then the time needed to get to work is .We know that and . Summing these twoequations, we get: .Substituting and dividing both sides by , we get ,hence .(Note that this approach would work even if the time by which he is late was different from the time by which he is early in the other case - we would simply take a weighed sum in step two, and hence obtaina weighed harmonic mean in step three.)13、Give a triangle with side lengths 15, 20, and 25, find the triangle's smallest height.SolutionSolution 1This is a Pythagorean triple (a 3-4-5 actually) with legs 15 and 20. Thearea is then . Now, consider an altitude drawn to anyside. Since the area remains constant, the altitude and side to which it is drawn are inversely proportional. To get the smallest altitude, it must be drawn to the hypotenuse. Let the length be x; we have, so and x is 12. Our answer is then.Solution 2By Heron's formula, the area is , hence the shortest altitude'slength is .14、Both roots of the quadratic equation are prime numbers. The number of possible values of isSolutionConsider a general quadratic with the coefficient of being and theroots being and . It can be factored as which is just. Thus, the sum of the roots is the negative of the coefficient of and the product is the constant term. (In general, this leads to Vieta's Formulas).We now have that the sum of the two roots is while the product is. Since both roots are primes, one must be , otherwise the sumwould be even. That means the other root is and the product mustbe . Hence, our answer is .15、Using the digits 1, 2, 3, 4, 5, 6, 7, and 9, form 4 two-digit prime numbers, using each digit only once. What is the sum of the 4 prime numbers?SolutionOnly odd numbers can finish a two-digit prime number, and a two-digit number ending in 5 is divisible by 5 and thus composite,hence our answer is .(Note that we did not need to actually construct the primes. If we had to, one way to match the tens and ones digits to form four primes is , , , and .)16、Let . What is?SolutionLet . Since one ofthe sums involves a, b, c, and d, it makes sense to consider 4x. We have. Rearranging, we have , so .Thus, our answer is .17、Sarah pours four ounces of coffee into an eight-ounce cup and fourounces of cream into a second cup of the same size. She then transfers half the coffee from the first cup to the second and, after stirring thoroughly, transfers half the liquid in the second cup back to the first. What fraction of the liquid in the first cup is now cream?SolutionWe will simulate the process in steps.In the beginning, we have:▪ounces of coffee in cup▪ounces of cream in cupIn the first step we pour ounces of coffee from cup to cup ,getting:▪ounces of coffee in cup▪ounces of coffee and ounces of cream in cupIn the second step we pour ounce of coffee and ounces of cream from cup to cup , getting:▪ounces of coffee and ounces of cream in cup▪the rest in cupHence at the end we have ounces of liquid in cup , and outof these ounces is cream. Thus the answer is .18、A cube is formed by gluing together 27 standard cubicaldice. (On a standard die, the sum of the numbers on any pair of opposite faces is 7.) The smallest possible sum of all the numbers showing on the surface of the cube isSolutionIn a 3x3x3 cube, there are 8 cubes with three faces showing, 12 with two faces showing and 6 with one face showing. The smallest sum with three faces showing is 1+2+3=6, with two faces showing is 1+2=3, and with one face showing is 1. Hence, the smallest possiblesum is . Our answer is thus.19、Spot's doghouse has a regular hexagonal base that measures oneyard on each side. He is tethered to a vertex with a two-yard rope.What is the area, in square yards, of the region outside of the doghouse that Spot can reach?SolutionPart of what Spot can reach is of a circle with radius 2, whichgives him . He can also reach two parts of a unit circle, whichcombines to give . The total area is then , which gives .20、Points and lie, in that order, on , dividing it intofive segments, each of length 1. Point is not on line . Point lieson , and point lies on . The line segments andare parallel. Find .SolutionAs is parallel to , angles FHD and FGA are congruent. Also,angle F is clearly congruent to itself. From SSS similarity,; hence . Similarly, . Thus,.21、The mean, median, unique mode, and range of a collection of eight integers are all equal to 8. The largest integer that can be an element of this collection isSolutionAs the unique mode is , there are at least two s.As the range is and one of the numbers is , the largest one can beat most .If the largest one is , then the smallest one is , and thus the meanis strictly larger than , which is a contradiction.If the largest one is , then the smallest one is . This means that wealready know four of the values: , , , . Since the mean of all thenumbers is , their sum must be . Thus the sum of the missing fournumbers is . But if is the smallest number,then the sum of the missing numbers must be at least ,which is again a contradiction.If the largest number is , we can easily find the solution. Hence, our answer is .NoteThe solution for is, in fact, unique. As the median must be , thismeans that both the and the number, when ordered by size,must be s. This gives the partial solution . For themean to be each missing variable must be replaced by the smallestallowed value.22、A sit of tiles numbered 1 through 100 is modified repeatedly by the following operation: remove all tiles numbered with a perfect square, and renumber the remaining tiles consecutively starting with 1. How many times must the operation be performed to reduce the number of tiles in the set to one?SolutionSolution 1The pattern is quite simple to see after listing a couple of terms.Solution 2Given tiles, a step removes tiles, leaving tiles behind. Now,, so in the next step tilesare removed. This gives , another perfect square.Thus each two steps we cycle down a perfect square, and insteps, we are left with tile, hence our answer is.23、Points and lie on a line, in that order, with and. Point is not on the line, and . The perimeterof is twice the perimeter of . Find .SolutionFirst, we draw an altitude to BC from E.Let it intersect at M. As triangle BEC is isosceles, we immediately get MB=MC=6, so the altitude is 8. Now, let . Using the Pythagorean Theorem on triangleEMA, we find . From symmetry,as well. Now, we use the fact that the perimeter of is twice the perimeter of .We have so. Squaring both sides, we havewhich nicely rearranges into. Hence, AB is 9 so our answer is .24、Tina randomly selects two distinct numbers from the setand Sergio randomly selects a number from the set. The probability that Sergio's number is larger than the sum of the two numbers chosen by Tina isSolutionThis is not too bad using casework.Tina gets a sum of 3: This happens in only one way (1,2) and Sergio can choose a number from 4 to 10, inclusive. There are 7 ways that Sergio gets a desirable number here.Tina gets a sum of 4: This once again happens in only one way (1,3). Sergio can choose a number from 5 to 10, so 6 ways here.Tina gets a sum of 5: This can happen in two ways (1,4) and (2,3). Sergio can choose a number from 6 to 10, so 2*5=10 ways here.Tina gets a sum of 6: Two ways here (1,5) and (2,4). Sergio can choose a number from 7 to 10, so 2*4=8 here.Tina gets a sum of 7: Two ways here (2,5) and (3,4). Sergio can choose from 8 to 10, so 2*3=6 ways here.Tina gets a sum of 8: Only one way possible (3,5). Sergio chooses 9 or 10, so 2 ways here.Tina gets a sum of 9: Only one way (4,5). Sergio must choose 10, so 1 way.In all, there are ways. Tina chooses twodistinct numbers in ways while Sergio chooses a number inways, so there are ways in all. Since , ouranswer is .25、In trapezoid with bases and , we have ,, , and . The area of isSolutionSolution 1It shouldn't be hard to use trigonometry to bash this and find the height, but there is a much easier way. Extend and to meet atpoint :Since we have , with the ratio ofproportionality being . Thus So the sides of are , which we recognize to be aright triangle. Therefore (we could simplify some of the calculation using that the ratio of areas is equal to the ratio of the sides squared),Solution 2Draw altitudes from points and :Translate the triangle so that coincides with . We getthe following triangle:The length of in this triangle is equal to the length of the original, minus the length of . Thus .Therefore is a well-known right triangle. Its area is, and therefore its altitude is.Now the area of the original trapezoid is.。
2019-2020年度 美国大联盟六年级 初赛试题
C) 5
D) 9
23. What is the ones' digit of the smallest whole number larger than 1000 whose digits are all
different?
A) 0
B) 1
C) 3
D) 4
C) 0.99
D) 101
24. The reciprocal of 1.01 is ？
12. The measure of the complement of ∠A is 30R. What is the measure of the supplement of
∠A?
A) 30R
B) 60R
C) 120R
D) 150R
C) 7
D) 9
27
13. Find the units' digit of 27 .
A) 1
B) 3
14. An equilateral triangle shares a side with an isosceles right triangle as shown. What is
m∠ABC?
第3页
A) 105°
B) 120°
C) 135°
D) 150°
15. If 14 whosits equals 35 whatsits, then 50 whatsits equals ？
A) 125 whosits
B) 70 whosits
C) 20 whosits
D) 7 whosits
16. The scale of a certain map is inch = 12 miles. A park is represented on this map by a
2020年秋期高中三年级期终质量评估 英语+ 数学(理+文)试题答案
A. It doesn't work. B. It makes her tired. C. It makes her have no appetite.
3. How many cookies did John eat?
A. Three.
B. Four.
C. Seven.
4. Where are the speakers?
C. Listen to a song
第二部分 阅读理解（共两节，满分 40 分）
第一节（共 15 小题；每小题 2 分，满分 30 分）
阅读下列短文，从每题所给的 A、B、C 和 D 四个选项中，选出最佳选项。
After the election day in the United States, there is a period of uncertainty. Both candidates（候 选人）
permanently that posts misleading information if necessary.
TikTok
TikTok says it is working with independent fact checkers during the election period.
It says it will remove misinformation related to the 2020 election including the vote itself. It is also
questions about the vote, and what people can expect in the coming days.
YouTube
YouTube says it will not allow "misleading claims about voting or content that encourages interference
AMC 美国数学竞赛 2000 AMC 10 试题及答案解析
USA AMC 10 20001In the year , the United States will host the International Mathematical Olympiad. Let , , and be distinct positive integers such that the product . What's the largest possible value of the sum ?SolutionThe sum is the highest if two factors are the lowest.So, and .2Solution.3Each day, Jenny ate of the jellybeans that were in her jar at the beginning of the day. At the end of the second day, remained. How many jellybeans were in the jar originally?Solution4Chandra pays an online service provider a fixed monthly fee plus an hourly charge for connect time. Her December bill was , but in January her bill was because she used twice as much connect time as in December. What is the fixxed monthly fee?SolutionLet be the fixed fee, and be the amount she pays for the minutes she used in the first month.We want the fixed fee, which is5Points and are the midpoints of sides and of . As moves along a line that is parallel to side , how many of the four quantities listed below change?(a) the length of the segment(b) the perimeter of(c) the area of(d) the area of trapezoidSolution(a) Clearly does not change, and , so doesn't change either.(b) Obviously, the perimeter changes.(c) The area clearly doesn't change, as both the base and its corresponding height remain the same.(d) The bases and do not change, and neither does the height, so the trapezoid remains the same.Only quantity changes, so the correct answer is .6The Fibonacci Sequence starts with two 1s and each term afterwards is the sum of its predecessors. Which one of the ten digits is the last to appear in thet units position of a number in the Fibonacci Sequence?SolutionThe pattern of the units digits areIn order of appearance:.is the last.7In rectangle , , is on , and and trisect . What is the perimeter of ?Solution.Since is trisected, .Thus,.Adding, .8At Olympic High School, of the freshmen and of the sophomores took the AMC-10. Given that the number of freshmen and sophomore contestants was the same, which of the following must be true?There are five times as many sophomores as freshmen.There are twice as many sophomores as freshmen.There are as many freshmen as sophomores.There are twice as many freshmen as sophomores.There are five times as many freshmen as sophomores.SolutionLet be the number of freshman and be the number of sophomores.There are twice as many freshmen as sophomores.9If , where , thenSolution, so ...10The sides of a triangle with positive area have lengths , , and . The sides of a second triangle with positive area have lengths , , and . What is the smallest positive number that is not a possible value of ?SolutionFrom the triangle inequality, and . The smallest positive number not possible is , which is .11Two different prime numbers between and are chosen. When their sum is subtracted from their product, which of the following numbers could be obtained?SolutionTwo prime numbers between and are both odd.Thus, we can discard the even choices.Both and are even, so one more than is a multiple of four.is the only possible choice.satisfy this, .12Figures , , , and consist of , , , and nonoverlapping unit squares, respectively. If the pattern were continued, how many nonoverlapping unit squares would there be in figure 100?SolutionSolution 1We have a recursion:.I.E. we add increasing multiples of each time we go up a figure. So, to go from Figure 0 to 100, we add.Solution 2We can divide up figure to get the sum of the sum of the firstodd numbers and the sum of the first odd numbers. If you do not see this, here is the example for :The sum of the first odd numbers is , so for figure , there are unit squares. We plug in to get , which is choice13There are 5 yellow pegs, 4 red pegs, 3 green pegs, 2 blue pegs, and 1 orange peg to be placed on a triangular peg board. In how many ways can the pegs be placed so that no (horizontal) row or (vertical) column contains two pegs of the same color?SolutionIn each column there must be one yellow peg. In particular, in the rightmost column, there is only one peg spot, therefore a yellow peg must go there.In the second column from the right, there are two spaces for pegs. One of them is in the same row as the corner peg, so there is only one remaining choice left for the yellow peg in this column.By similar logic, we can fill in the yellow pegs as shown:After this we can proceed to fill in the whole pegboard, so there is only arrangement of the pegs. The answer is14Mrs. Walter gave an exam in a mathematics class of five students. She entered the scores in random order into a spreadsheet, which recalculated the class average after each score was entered. Mrs. Walter noticed that after each score was entered, the average was always an integer. The scores (listed in ascending order) were , , , , and . What was the last score Mrs. Walter entered? SolutionThe sum of the first scores must be even, so we must choose evens or the odds to be the first two scores.Let us look at the numbers in mod .If we choose the two odds, the next number must be a multiple of , of which there is none.Similarly, if we choose or , the next number must be a multiple of , of which there is none.So we choose first.The next number must be 1 in mod 3, of which only remains.The sum of the first three scores is . This is equivalent to in mod .Thus, we need to choose one number that is in mod . is the only one that works.Thus, is the last score entered.15Two non-zero real numbers, and , satisfy . Which of the following is a possible value of ?SolutionSubstituting , we get16The diagram shows lattice points, each one unit from its nearest neighbors. Segment meets segment at . Find the length of segment .SolutionSolution 1Let be the line containing and and let be the line containing and . If we set the bottom left point at , then , , , and .The line is given by the equation . The -intercept is , so . We are given two points on , hence we cancompute the slope, to be , so is the lineSimilarly, is given by . The slope in this case is , so . Plugging in the point gives us , so is the line .At , the intersection point, both of the equations must be true, soWe have the coordinates of and , so we can use the distance formula here:which is answer choiceSolution 2Draw the perpendiculars from and to , respectively. As it turns out, . Let be the point on for which ., and , so by AA similarity,By the Pythagorean Theorem, we have ,, and . Let , so , thenThis is answer choiceAlso, you could extend CD to the end of the box and create two similar triangles. Then use ratios and find that the distance is 5/9 of the diagonal AB. Thus, the answer is B.17Boris has an incredible coin changing machine. When he puts in a quarter, it returns five nickels; when he puts in a nickel, it returns five pennies; and when he puts in a penny, it returns five quarters. Boris starts with just one penny. Which of the following amounts could Boris have after using the machine repeatedly?SolutionConsider what happens each time he puts a coin in. If he puts in a quarter, he gets five nickels back, so the amount of money he has doesn't change. Similarly, if he puts a nickel in the machine, he gets five pennies back and the money value doesn't change. However, if he puts a penny in, he gets five quarters back, increasing the amount of money he has by cents.This implies that the only possible values, in cents, he can have are the ones one more than a multiple of . Of the choices given, the only one is18Charlyn walks completely around the boundary of a square whose sides are each km long. From any point on her path she can see exactly km horizontally in all directions. What is the area of the region consisting of all points Charlyn can see during her walk, expressed in square kilometers and rounded to the nearest whole number?SolutionThe area she sees looks at follows:The part inside the walk has area . The part outside the walk consists of four rectangles, and four arcs. Each of the rectangles has area . The four arcs together form a circle with radius . Therefore the total area she can see is, which rounded to the nearest integer is .19Through a point on the hypotenuse of a right triangle, lines are drawn parallel to the legs of the triangle so that the trangle is divided into a square and two smaller right triangles. The area of one of the two small right triangles is times the area of the square. The ratio of the area of the other small right triangle to the area of the square is SolutionLet the square have area , then it follows that the altitude of one of the triangles is . The area of the other triangle is .By similar triangles, we haveThis is choice(Note that this approach is enough to get the correct answer in the contest. However, if we wanted a completely correct solution, we should also note that scaling the given triangle times changes each of the areas times, and therefore it does not influence the ratio of any two areas. This is why we can pick the side of the square.)20Let , , and be nonnegative integers such that . What is the maximum value of ? SolutionThe trick is to realize that the sum is similar to the product .If we multiply , we get.We know that , therefore.Therefore the maximum value of is equal to the maximum value of . Now we will find this maximum.Suppose that some two of , , and differ by at least . Then this triple is surely not optimal.Proof: WLOG let . We can then increase the value ofby changing and .Therefore the maximum is achieved in the cases where is a rotation of . The value of in this case is . And thus the maximum of is.21If all alligators are ferocious creatures and some creepy crawlers are alligators, which statement(s) must be true?I. All alligators are creepy crawlers.II. Some ferocious creatures are creepy crawlers.III. Some alligators are not creepy crawlers.SolutionWe interpret the problem statement as a query about three abstract concepts denoted as "alligators", "creepy crawlers" and "ferocious creatures". In answering the question, we may NOT refer to reality -- for example to the fact that alligators do exist.To make more clear that we are not using anything outside the problem statement, let's rename the three concepts as , , and .We got the following information:▪If is an , then is an .▪There is some that is a and at the same time an .We CAN NOT conclude that the first statement is true. For example, the situation "Johnny and Freddy are s, but only Johnny is a "meets both conditions, but the first statement is false.We CAN conclude that the second statement is true. We know that there is some that is a and at the same time an . Pick one such and call it Bobby. Additionally, we know that if is an , then is an. Bobby is an , therefore Bobby is an . And this is enough to prove the second statement -- Bobby is an that is also a .We CAN NOT conclude that the third statement is true. For example, consider the situation when , and are equivalent (represent the same set of objects). In such case both conditions are satisfied, but the third statement is false.Therefore the answer is .22One morning each member of Angela's family drank an -ounce mixture of coffee with milk. The amounts of coffee and milk varied from cup to cup, but were never zero. Angela drank a quarter of the total amount of milk and a sixth of the total amount of coffee. How many people are in the family?SolutionThe exact value "8 ounces" is not important. We will only use the fact that each member of the family drank the same amount.Let be the total number of ounces of milk drank by the family and the total number of ounces of coffee. Thus the whole family drank a total of ounces of fluids.Let be the number of family members. Then each family member drank ounces of fluids.We know that Angela drank ounces of fluids.As Angela is a family member, we have .Multiply both sides by to get .If , we have .If , we have .Therefore the only remaining option is .23When the mean, median, and mode of the list are arranged in increasing order, they form a non-constant arithmetic progression. What is the sum of all possible real values of ? SolutionAs occurs three times and each of the three other values just once, regardless of what we choose the mode will always be .The sum of all numbers is , therefore the mean is .The six known values, in sorted order, are . From this sequence we conclude: If , the median will be . If , the median will be . Finally, if , the median will be .We will now examine each of these three cases separately.In the case , both the median and the mode are 2, therefore we can not get any non-constant arithmetic progression.In the case we have , because. Therefore our three values inorder are . We want this to be an arithmetic progression. From the first two terms the difference must be . Therefore thethird term must be .Solving we get the only solution for this case: . The case remains. Once again, we have ,therefore the order is . The only solution is when , i. e., .The sum of all solutions is therefore .24Let be a function for which . Find the sum of all values of for which .SolutionIn the definition of , let . We get: . As we have , we must have , in other words .One can now either explicitly compute the roots, or use Vieta's formulas. According to them, the sum of the roots ofis . In our case this is .(Note that for the above approach to be completely correct, we should additionally verify that there actually are two distinct real roots. This is, for example, obvious from the facts that and .)25In year , the day of the year is a Tuesday. In year , the day is also a Tuesday. On what day of the week did the of year occur?SolutionClearly, identifying what of these years may/must/may not be a leap year will be key in solving the problem.Let be the day of year , the day of year and the day of year .If year is not a leap year, the day will bedays after . As , that would be a Monday.Therefore year must be a leap year. (Then is days after .) As there can not be two leap years after each other, is not a leap year. Therefore day is days after . We have . Therefore is weekdays before , i.e., is a.(Note that the situation described by the problem statement indeed occurs in our calendar. For example, for we have=Tuesday, October 26th 2004, =Tuesday, July 19th, 2005 and =Thursday, April 10th 2003.)。
五年级美国大联盟第一阶段-数论专题(含题目翻译解析)完整版
五年级美国大联盟第一阶段-数论专题（教师版）学生/课程年级学科授课教师日期时段核心内容熟悉美国大联盟常考数论题课型一对一/一对N教学目标1、掌握各类数的概念与特点；2、根据数的特点求解相应的量；重、难点教学目标1/2知识导图（一）单词dollar integer product penny factor one－digit nickle multiple plusdime even minus quarter odd multipleprime number dividecomposite consecutive（二）词组square root at least a millionpositive integers greatest common factor least common multiple two －digit multiples be divisible by the sum of【参考答案】square root 平方根 at least 至少 a million一百万 positive integers 正整数 greatest common factor 最大公因数 least common multiple 最小公倍数 two －digit multiples 两位数的倍数 be divisible by 被……整除 the sum of总和dollar 美元 integer 整数 product 积 penny 1美分 factor 因数 one －digit 一位数 nickle 5美分 multiple 倍数 plus 加 dime 10美分 even 偶数 minus 减 quarter25美分odd奇数 multiple 乘 prime number 质数 divide 除composite合数consecutive连续的导学一：组合种类知识点讲解1、简单列举有些题目，因其所求的答案有多种，用算式不容易表示，需要采用一一列举的方法解决。
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2017-2018年度美国“数学大联盟杯赛”（中国赛区）题目翻译及解题tips
【翻译】：2018与以下哪个数字相加的总和是偶数？The sum of…总和…；the even number偶数
【翻译】：约翰和吉尔一共有92美元。
约翰的钱是吉尔的三倍。
问约翰有多少钱？
①…has three times（倍数）as many(修饰可数名词)/much(修饰不可数名词)as…A的…是B的几倍
②As···as···和什么一样多
【翻译】：汤姆是一个篮球热爱者！在他的书中，他写了100次“ILOVENBA”（我爱NBA）。
问他写的第500个字母是什么。
（提示：本题考查周期循环规律题）
【翻译】：一个长*宽为8*25的长方形和以下哪个长方形有相同的面积。
【翻译】：前100个正整数（1-100）的和与后50个正整数（51-100）的和之间的差是多少？
①Positive difference···与···的差；②positive integers正整数
【翻译】：你有一根10英尺长的杆子需要被切成10等份。
若每一份需要10秒去切，完成这份工作一共需要多少秒。
【翻译】：Amy将2018四舍五入约至十位（rounded···to the nearest tens）得到的数字与Ben将2018四舍五入约至百位得到的数字，这两个数字之和是多少？
【翻译】：下列哪组数有最小公倍数？
【翻译】：Dan每买2支铅笔的同时也会5支钢笔。
如果他买了10支铅笔，那他一共买了几支钢笔？
【翻译】：星期四的20天后是星期几？
【翻译】：下列哪个角的度数最小？
①an obtuse钝角②an acute锐角③a right直角④a stright平角
【翻译】：我们班的每位学生都要轮流喊一个整数。
第一个人喊的是1。
后面每人喊的数字都比前者多3,（即第二个人喊的是数字4，1+3=4）。
问下面哪个选项的数字是我们班的某一个学生可能喊到的数字？（提示：本题考查等差数列）
①A whole number整数②in turn轮流③shout out大声喊
【翻译】：一个男孩买了一个篮球和一个棒球，一共花了1.25美元。
如果这个篮球比这个棒球贵25美分，那篮球多少钱？（注意：1美元=100美分）
【翻译】：2小时+？分钟+40秒=7600秒
【翻译】：如右图，把数字1-7放入其中，使得每条直线的数字相加为12，请问中间的圆圈填数字几？
【翻译】：若每5个大人吃20个苹果，而3个小孩一共吃了12个苹果，请问平均每人吃多少个苹果？
【翻译】：右边的图形周长是多少？（注意在右边的图形中内角是直角或270°）
【翻译】：30人正在排队等待买披萨。
Andy前面有10个人。
Susan是队伍的最后一个人。
请问Andy和Susan之间有几个人？
【翻译】：早上9点再过39个小时后的时间是多少？
【翻译】：11···11（由2016个1组成的数字）不是下列哪个选项的倍数？
【翻译】：两个一千和两个一百万的平均数是多少？
【翻译】：一个三角形和一个正方形面积相同。
如果这个三角形的底和这个正方形的边长相同，三角形底边上的高是4，请问这个正方形的面积是多少？①the length of…的长度；②a base of…的底边
【翻译】：当Volta发现了一个形状是等腰三角形（isosceles triangle）的草地，她非常兴奋以至于绕着这个草地跑了一圈（lap）。
两边的腰每边长度是505米，而第三边的长度是一个整数。
Volta绕着草地跑一圈的长度最可能是以下哪个？（提示：运用两边之和大于第三边，两边之差小于第三边）
【翻译】：整数除数为奇数的数是？
【翻译】：在一个由8个数字组成的序列中，这8个数字的平均数是15,。
若前三个数的平均数是16，后两个数的平均数是15，那最后三个数字的平均数是多少？
①sequence序列②term项
【翻译】：在2000与2050年之间凡是能被4整除的就是闰年（leap years）.然而在2000和2050之间不是所有年都是闰年。
请问在2010到2026这17年中一共有几天？
提示：闰年一年366天，平年则365天。
【翻译】：2357这个数的百位和十位的数字之和是多少？
【翻译】：下列哪个算式的（商*余数）有最大值？
【翻译】：我有一些dimes(1dime=10美分)和nickels(也是一种硬币)，这些硬币加起来一共价值3美元。
若每一个nickel可以换25美分（a quarter=25美分=1nickel）,那我将有5美元。
请问我一共有几个dimes?
【翻译】：我是一只可爱的猫。
当我将一个整数的每个数字相乘就会得到9（比如有一个整数是19，那么1乘以9就得到9），那么我就把这个整数放在“我最喜欢的数字”清单中，在1000到9999这些整数中，属于“我最喜欢的数字”有多少？
提示：9111和1119都是属于我最喜欢的数，抓住3*3=9和9*1=9这两个算式去做
【翻译】：算式1+12+123+1234+…+123456789所得到的数字中的十位数字和个位数字相加起来得多少？
【翻译】：1至10之间的所有质数之积是多少
选项Dnone of the above以上答案都不是
【翻译】：12、14、16和18四个数的平均数是多少？
【翻译】：当Jon喊出一个整数，Al就会喊出这个数中所有数字的积，Brab就会喊出Al所喊整数的所有数字的积，Cy则喊Brab喊出的整数的所有数字的积。
当Cy喊的数字是18，那Jon可能喊的是什么数字提示：本题最好用代入排除法做。
假设选C，即Jon喊的是899，那么AL喊的是8*9*9=648，那Brab喊6*4*8=192，则Cy喊1*9*2=18符合题意说明就是C。
【翻译】：每个大盒子包含3个中盒子，每个中盒子又包含2个小盒子，每个小盒子有5个苹果。
问1200个苹果至少需要多少个大盒子去装？
【翻译】：从现在起18年后，我的年龄将会是我现在的两倍还多4岁。
Current age现在的年龄；twice两倍
【翻译】：Wanda每次一挥手，就会在她裙子上多出现4颗星星（假设一开始裙子没有任何星星）。
在多次挥手后，Wanda将裙子上星星的总数和挥手的次数相乘，得到的积不可能是以下哪个？
21。