美国数学竞赛AMCAB试题及答案

合集下载

2000-2012美国AMC10中文版试题及答案

2000到2012年AMC10美国数学竞赛0 0P 0 A 0 B 0 C 0D 0 全美中学数学分级能力测验(AMC 10)2000年第01届美国AMC10 (2000年2月日时间75分钟)1. 国际数学奥林匹亚将于2001年在美国举办，假设I 、M 、O 分别表示不同的正整数，且满足I ⨯M ⨯O =2001，则试问I +M +O 之最大值为。

(A) 23 (B) 55 (C) 99 (D) 111 (E) 6712. 2000(20002000)为。

(A) 40 (B) 50 (C) 55 (D) 60 (E) 754. Candra 每月要付给网络公司固定的月租费及上网的拨接费，已知她12月的账单为12.48元，而她1月的账单为17.54元，若她1月的上网时间是12月的两倍，试问月租费是元。

(A) 2.53 (B) 5.06 (C) 6.24 (D) 7.42 (E) 8.775. 如图M ，N 分别为PA 与PB 之中点，试问当P 在一条平行AB 的直在线移动时，下列各数值有项会变动。

(a) MN 长 (b) △P AB 之周长 (c) △P AB 之面积 (d) ABNM 之面积(A) 0项 (B) 1项 (C) 2项 (D) 3项 (E) 4项 6. 费氏数列是以两个1开始，接下来各项均为前两项之和，试问在费氏数列各项的个位数字中，最后出现的阿拉伯数字为。

(A) 0 (B) 4 (C) 6 (D) 7 (E) 97. 如图，矩形ABCD 中，AD =1，P 在AB 上，且DP 与DB 三等分∠ADC ，试问△BDP 之周长为。

2014美国数学竞赛AMC10A、10B试题及答案

2014 AMC 10B Problems Problem 1 Leah has 13 coins^ all of which are pennies and nickels. If 5h已 had one more nickel than she has now, then she would h^ve the same number of pennies and nickels, lr )匚ents, how much are Leah's coins worth?(A) 33 (B) 35 (C) 37 (D) 39 (E) 41Problem 223 4 2s 2-彳 + 2-^ Problem 3Randy drove the first third of his trip on a gravel road^ the next 20 miles on pavement, and ths remaining □ne -fifth on a dirt road . In miles how long was Randy's tripi?(A) 30 (B)等 (C)粤(D) 40 (E)竽Problem 4Susie pays for 4 muffins and 3 bananas. Calvin spends twice as much paying for 2 muffins and 16 bananas. A muffin is. how many times as expensive as a banana?m 吨了(A) | (B)彩 (C) f (D) 2 (E)-Problem 5Doug constructs a square window using 8 equal-size panes of glass, as shown ・ The ratio of the height to width ft3『 each pane is 5 : 2f and the borders around and between the panes are 2 inches wide, in inches, what 谄 the sid& length of th© square wind 口w?(A) 26 (B) 28 (C) 30 (D) 32 (E) 34Problem 6Orvin went to the store with just enough rrnoney to buy 30 balloons. When h@ arrived^ he discovered that the store had a special sale on b a I loons ：: buy 1 balloon at the regular p 『i 匚已 and get a second at g off the regular price. What is the greatest number of balloons Orvin could buy?(A) 33 (B) 34 (C) 36 (D) 38 (E) 39what im(A) 16 (B) 24 (C) 32 (D) 48 (E) 64Problem 7Suppose A> U > 0 and A is 近% greater than B. what is £?⑷100(#) (B) 1 叫字)(C) 100(字)(D) 1 叫乎)(E) 100(和SolutionProblem 8A truck tr^ve>l5 f feet every t seconds. There ars 3 feet in a yard i How many yards dDss the truck travel6in 3 rninute-s?b frn 血m10i 1 口⑷廖(X (①〒⑴)石㈣〒Solutionproblem 9For real numbers w and z,丄+丄予二于=2014,ut ww + zWhat is ----------- ?w ― z、一1 J 1(A) - 2014 (B)艄(C) —(D) 1 (E) 2014Problem 10In the addition shown below A. [3. C.and D are distinct digits. How many different values are po^ible ForD?ABBCB+ BCADADI3DDD(A) 2 (E) 4 (C) 7 (D) 8 (E) 9Problem 1111. For the consume^ a single discount of n%is more adv^ntagsous than any of the following discounts:(1)two ^ucces^ive 15% discounts(2)three successive 10% discounts(3) a 25f/J discount Fellow sd by s 5xu discountWhat is the- smallest possibls positive intsgsr valus uf n?(A) 27 (B) 2S (C) 29 (D) 31 (E) 33Problem 1212. The largest divisor of 2, 014h 000* 000 is rtsslf. What is its fifth largest divisor?(A) 125t875.000 (B) 20L400. 000 (C) 251, 750.000 (D) 402： 800.000 (E) 503.500.000Problem 13Sin regular hexagons surround 召regular hexagon of ^ide length 1 as shown. What is the ar?a of A J4BC?(A) 2V5 {B) 3V3 (C) 1+3辺(D) 2 + 2“$(E) 3+2\/3Problem 14Oariica drove her new car on a trip for a whole number of hour勺averaging 55 miles per hour. At the beginning of the trip」abc miles was displayed on thm Dclannstsr^ where abc is 耳3-digit number with ti > 1 and « + 6 —c < 7. At the end of the trip，the odometer showed eba miles. What is(i~ + fc2+ c~ ?(A) 26 (B) 27 (C) 36 (D) 37 (E) 41Problem 15In rectangle 打<7 = 2GR and points E3nd F lie on AF£□ that ED and FD tnsect Z.ADCas shown, what is the ratio of the area of ADEF to the ares of rectangle ABCD?S)晋(B)^ ©瞬㈣笫Problem 16Four fair six-sided dice are rolled. What is the probability that at least three of the four dice show the sarme value? (A)吉(B)12(C) I {D)島㈣|Problem 17What is thm greatest power of 2 t hat if a fmutor of 101— 4U：，01?Problem 18(A) 21DO2(B) 21003(C) 21004(D) 21C05(E) 21D0C A list of 11 positive! integers has a mean of 10；a median of 9j and a uiniqus> mads af 8- What is the largest passible value of an integer in the list?(A) 24 {B) 30 (C) 31 (D) 33 (E) 35Problem 19Two ccrncentric circles have radii 1 and 2. Two points an the outer circle are chosen independently and uniformly at random. What iw the probabi^it/ that the chord joining the two points in tersects the inner circle?(A) g (B) i (C)匕詳(D) i (E) +Problem 20For how many integers 工is the number —51J:2 + 50 negative?(A) 8 (B) 10 (C) 12 (D) 14(E) 16Problem 21Trapezoid ABCD has parallel sides AB af length 33 and CD of length 21. The other two sides are of lengths 1(' and 14. The angles at A and B are acute. What js the Imn gth of the shorter diagonal of ABCD?{A) 10?6 (B) 25 (C) S\/10 (D) 18“(E) 26SolutiariProblem 22Eight semicircles line the insid目af a ^qu^re with 吕id日length 2 as shown. What is the radium Q F the circle tangent tc ell of these semicircles?A sphere is inscribed in a truncated right circular cons as shown. The volume of thm truncated cone is twice that of the sphere. What is the ratro of the radius af the bottam base of the truncated cone to the radius of the tap base of the truncated cone?Problem 23(A) | ©近(D)2 (E)^^ Problem 24The numbers 1, 2} 3}斗』5 are to be arranged in a circle. An arrangement rs bad if rt is not true that fcr every n from 1 to 15 one can find a subset of the numberm that appear consecutively on the circle that sum to J ?. Arrangements that differ only by a rotation or a reflection mne considered the same. Haw many different bad arrangements are there?(A) 1 (B) 2 (C) 3 (D) 4 (E) 5 Problem 25In a small pond there are eleven lily pads in a row labeled 0 through 10, A frog is sitting on pad 1. When the frog is on pad N } 0 < N < 10, it 呷illju 叩 to pad N-l with probability — and to pad N +1 with probability 1 - Each jump is independent of the previous jumps. If the frog reaches pad 0 it will be eaten by a patiently waiting snake, If the frog reaches pad 10 it will exit the pond> never to return, what is the probability that the frog will escape being eaten by the snake?| (B)曙(C)磊(D)纟(E) \答案：1. C2. E3. E4. B5.A6. C7. A8. E9. A10. C11. C12. C13. B14. D15. A16. B17. D18. E19. D20. C21. B22. B23. E24. B25. Cwwt 冷10诲+ * +矿叮(A) 3 (B) & (C) y (D)爭 (E) 170Problem 2Roy's 亡吕t m吕t 百 of a can cf eat food ev&ry morning and y of a ean of eat food every evening. Before 怡目ding hiis 匚 a )t □“ Mlonday morn in Rciy ope me d a bck containing 6 cans of eat feotl*. On what day of the wisek dbd the 匚已t finish eating all the cat food in the box? (A) Tuesday (U) Wednesday (C) Thur 吕d 暫(U) Friday (E) Saturday Problem 3 Bridget bakes 48 loaves of bread for her bakery. She sells half of them in the morning for 32,501 each. In the aftemaon she sells two thirds of what she has left 3 and bscauss they are not freshshe charges only half priice. Tn the late aftern 口口n shethe r^mainiiing loaves -at B dollar leach. Each loaf 匚口百上鼻S (J_75 farher to make, In dollarSj what is her profit for the day? (A) 24 (B) 36 (C) 44 (D) 4S (E) 52Problem 4Walking down Jane Street, Ralph passed four houses in 吕 row, each painted a different 匚olcr- He passed th® orange house before the red housSi and the passed the blue house before ths yellow house. Ths blue house was not n&^t to t.h& ysillow house. How many orderings of the colored fhouses 囂「白 possible?(A) 2 (B) 3 (C) 4 (D) 5 (E) 6Problem 5On an 3l^eb rs quiz d 10% of Elis students scored 70 point 勺 35% scored 80 points^ 30% scored 90 pointSj and the rest scored 100 points. What is the differsincc between ths mean and median score cf ths students' scares on this quis?(A) 1 (B) 2 (C) 3 (D) 4 (E) 5Problerri 6SuppoBis that o 匚OWE ; give 6- gallons of milk in n days. At thus rate, how marTpr gallanE 口f nnilk will d cows give in c days?Probfem 7IN 口 r^zSrd 『自 £l nurfib&f£ ir. 些、ci f and b S-Stiisfy J : < LL <L b. Hdw marhy Gf th€i Follow in g id 白口u£ lit i 自弓 mustba true?(I) H + 空 V ci + b(II) Ji — y < a — 6(III) xy < ab(A) 0 (B) 1 (C) 2 (D) 3 (E) 4(C)cibdc (D) bcdeProblem 8 Which of the following niumbers Is a perfect square?Problem 9(A) 14115!~^T~(D)17fl81~2~18?193~2~Tfie two legs of a right trimngl已which are altitudes, ha\/e lengths 2\/3 and 6. How long is the third altitude of the triangte?(A) 1 (B) 2 (C) 3 (D) 4 (E) 5Problem 10Five positive consecutive integers starting with a have average b. what is the average of 5 consecutive integers that start with b?(A) d + 3 (B) ti + 4 (C) a 4^5 (D) u 4- 6 {E) <i + 7Problem 11A cusEomer who intends to purchase art appliance has three coupon5^ only ore of which may be uw总d;Coupon 1: .10/( off the listed price if the listed price rs at least S50Coupon 2. $20 off the listed price if the listed pnce is at Itfast 1100coupon 3： 18*X off the amount by which the listed price e^teeds $100For which of the following listed prices will coupari 1 offer a greater price reduction than either coupon 2 or coupon 3?(A) S179.95 (B) S199.95 (C) S219.95 (D) $239.95 (E) $259.95Problem 12A regular hexagon has side length 6. Ccngruerit arcs with radius 3 are drawn with the center at each of the vertical, creating circular sac tors 启弓shown. The region inside the ha^agon but outside the sectors is shaded as shown Whe t is thm^rea of the shaded regian?(A)幫価—跖(B) 27V3-&r (C) 54?5-18TT(D) &4^3-12^(E) 伽Problem 13Equilateral A>1R(7 hms side length 1』and squmrms AB DE, BOH I, CAFG bs outbids the triangle. What is the area of hexagon DEFGH I?Problem 14Ths y-intencBptSj P目nd Q, of two perp&ndicjldr lines intersecting mt the point j4(6h 8)have m sunn of zero. What is the area of AAPQ?(A) 45 (B) 4S (c) 54 (D) M (E) 72Problem 15Oavidl drives from his home to the airport to cat匚h a flight. He drives 35 milES in th^ first hour, but realizes that he will be 1 hour late if he continues at this speed. He increases his speed by 15 miles per hour for the rest of the way to the airport and arrives 30 minutes early. How many miles is the airport from his home?(A) 140 (B) 175 (C) 210 (D) 245 (E) 280Problem 16Fn rectangle A拜C0 A/J =L OC7 = 2』and points E, F、and <7 are midpoints of f 门』-and respectively・Point H is the midpoint of GE. What is the arsa of the shaded region?12 + 3>/3-4-(C) 3 + v5 (D)(E) 6/ Z?⑷吉㈣兽©害 (D)卷(E) 111 / 12Problem 17Three fair six-sided dn^e are rolled. What 祐 the prob ability that the values shown on two of the dice sum to the value shown an the rematning die?(A) j (B)第(C)籟(D)备(E) | Problem 18A square in the coordinate plane has vertices whose y-co ordinates are 0』1』4, and 5. What is the area of thm square?(A) 16 (B) 17 (C) 25 (D) 26 (E) 27 Problem 19Four cubes with edge lengths 1, 2, 3, and 4 are stacked as shown. What is the length of the portion of XV contdin^d in th© cub@ with edge length 3?(A)警 (B) 2^/3 (C)警 (D) 4 ㈣ 3闪Problem 20The product (8)(888 . .. 8), where ttie second factor has k digits, is an integer whose digits have a sum of 1000. What is fc?(A) 901 (B) 911 (C) 919 (D) 991 (E) 999Problem 21Positive integers a and b are such that the grsplis of = EAX + & and y = 驻+ b intersect this ir-a^is at ths same point. What is the sum erf all possible jr-coondinat&s of ths SB points of intersection?⑷-20 (B) -18 (C) 一苗(D) -12 (E) —8Problem 22[n r直亡tanglm AnC'D r AB= 20 and BG =10. Let £? be a point Ort CD such that Z.CBE= 15°. What is AE?(A) (B) IO A/3(C) 18 (D) 11 辺(E) 20Problem 23A rectmngubr piece of paper whose length is V3 times the width has area A, The paper is divided into three equal sections along the opposite I eng th Sj artd then a dotted fine is drawn from the first divider to the 亡end divider cn the opposite side shown. The paper 左then fbldEd flat along thiim dotted line to cnemtm a. new whmp日with area /?. What is the ratio J? : A?(A) 1 : 2 (B) 3 : &Problem 24A ssquanc^ cf natural nuimber^ i£cori£trut?ted by listing the first 4H then skipping one^ fci sting the newt 5, skipping 2. listing 6』skipping 3d andj on the nth itEration, listing n + 3 and skipping n. The sequence bsgins L 2h 3h 4, 6, 7. 8. 9.10. 13. What is ths 500- QOOth number in the sequence?(A) 996,506 (B) 996507 (C) 996508 (D) 996509 (E) 996510Problem 25The number 5B67is between 22011 and 2Z014r now many pairs or integers ^re there such thmt(C) 2 : 3 (D) 3 : 4 (E) 4 : 5Problem 201 < m < 2012 and(A) 278 (B) 279 (C) 280 (D) 281 (E) 282答案：1.C 2.D 3.D 4.B 5.E 6.B 7.B 8.B 9.D 10.C 11.C 12.C 13.C 14.C 15.B 16.B 17.D 18.E 19.D 20.B 21.C 22.A 23.C 24.B 25.A。

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 first odd 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 isSolutionLet 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: . Aswe 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 of is . 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.)。

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.)。

AMC 美国数学竞赛试题+详解英文版

2013 AMC8 Problems1.Danica wants to arrange her model cars in rows with exactly 6 cars in each row. She now has 23 model cars. What is the smallest number of additional cars she must buy in order to be able to arrange all her cars this way?2.A sign at the fish market says, "50% off, today only: half-pound packages for just $3 perpackage." What is the regular price for a full pound of fish, in dollars?What is the value of?3.4.Eight friends ate at a restaurant and agreed to share the bill equally. Because Judi forgot her money, each of her seven friends paid an extra $2.50 to cover her portion of the total bill. What was the total bill? 5.Hammie is in thegrade and weighs 106 pounds. His quadruplet sisters are tiny babiesand weigh 5, 5, 6, and 8 pounds. Which is greater, the average (mean) weight of these five children or the median weight, and by how many pounds?6.The number in each box below is the product of the numbers in the two boxes that touch it in the row above. For example, . What is the missing number in the top row?7.Trey and his mom stopped at a railroad crossing to let a train pass. As the train began to pass, Trey counted 6 cars in the first 10 seconds. It took the train 2 minutes and 45 seconds to clear the crossing at a constant speed. Which of the following was the most likely number of cars in the train?8.A fair coin is tossed 3 times. What is the probability of at least two consecutive heads?9.The Incredible Hulk can double the distance he jumps with each succeeding jump. If his first jump is 1 meter, the second jump is 2 meters, the third jump is 4 meters, and so on, then on which jump will he first be able to jump more than 1 kilometer?10.What is the ratio of the least common multiple of 180 and 594 to the greatest common factor of 180 and 594?11.Ted's grandfather used his treadmill on 3 days this week. He went 2 miles each day. On Monday he jogged at a speed of 5 miles per hour. He walked at the rate of 3 miles per hour on Wednesday and at 4 miles per hour on Friday. If Grandfather had always walked at 4 miles per hour, he would have spent less time on the treadmill. How many minutes less?12.At the 2013 Winnebago County Fair a vendor is offering a "fair special" on sandals. If you buy one pair of sandals at the regular price of $50, you get a second pair at a 40% discount, and a third pair at half the regular price. Javier took advantage of the "fair special" to buy three pairs of sandals. What percentage of the $150 regular price did he save?13.When Clara totaled her scores, she inadvertently reversed the units digit and the tens digit of one score. By which of the following might her incorrect sum have differed from the correct one?14.Abe holds 1 green and 1 red jelly bean in his hand. Bea holds 1 green, 1 yellow, and 2 red jelly beans in her hand. Each randomly picks a jelly bean to show the other. What is the probability that the colors match?15.If , , and , what is the product of , , and ?16.A number of students from Fibonacci Middle School are taking part in a community serviceproject. The ratio of -graders to -graders is , and the the ratio of -graders to-graders is . What is the smallest number of students that could be participating in the project?17.The sum of six consecutive positive integers is 2013. What is the largest of these six integers?18.Isabella uses one-foot cubical blocks to build a rectangular fort that is 12 feet long, 10 feet wide, and 5 feet high. The floor and the four walls are all one foot thick. How many blocks does the fort contain?19.Bridget, Cassie, and Hannah are discussing the results of their last math test. Hannah shows Bridget and Cassie her test, but Bridget and Cassie don't show theirs to anyone. Cassie says, 'I didn't get the lowest score in our class,' and Bridget adds, 'I didn't get the highest score.' What is the ranking of the three girls from highest to lowest?20.A rectangle is inscribed in a semicircle with longer side on the diameter. What is thearea of the semicircle?21.Samantha lives 2 blocks west and 1 block south of the southwest corner of City Park. Her school is 2 blocks east and 2 blocks north of the northeast corner of City Park. On school days she bikes on streets to the southwest corner of City Park, then takes a diagonal path through the park to the northeast corner, and then bikes on streets to school. If her route is as short as possible, how many different routes can she take?22.Toothpicks are used to make a grid that is 60 toothpicks long and 32 toothpicks wide. How many toothpicks are used altogether?23.Angle of is a right angle. The sides of are the diameters of semicirclesas shown. The area of the semicircle on equals , and the arc of the semicircle onhas length . What is the radius of the semicircle on ?24.Squares , , and are equal in area. Points and are the midpointsof sides and , respectively. What is the ratio of the area of the shaded pentagonto the sum of the areas of the three squares?25.A ball with diameter 4 inches starts at point A to roll along the track shown. The track iscomprised of 3 semicircular arcs whose radii are inches, inches, andinches, respectively. The ball always remains in contact with the track and does notslip. What is the distance the center of the ball travels over the course from A to B?2013 AMC8 Problems/Solutions1. ProblemDanica wants to arrange her model cars in rows with exactly 6 cars in each row. She now has 23 model cars. What is the smallest number of additional cars she must buy in order to be able to arrange all her cars this way?Solution:In order to have her model cars in perfect, complete rows of 6, Danica must have a number ofcars that is a multiple of 6. The smallest multiple of 6 which is larger than 23 is 24, so she'll need to buy more model car.2.A sign at the fish market says, "50% off, today only: half-pound packages for just $3 per package." What is the regular price for a full pound of fish, in dollars?ProblemSolution: The 50% off price of half a pound of fish is $3, so the 100%, or the regular price, of a half pound of fish is $6. Consequently, if half a pound of fish costs $6, then a whole pound of fish is dollars.What is the value of?3. ProblemNotice that we can pair up every two numbers to make a sum of 1:SolutionTherefore, the answer is .4. ProblemEight friends ate at a restaurant and agreed to share the bill equally. Because Judi forgot her money, each of her seven friends paid an extra $2.50 to cover her portion of the total bill.What was the total bill?Each of her seven friends paidto cover Judi's portion. Therefore, Judi's portion mustbe. Since Judi was supposed to payof the total bill, the total bill must be.Solution5.Hammie is in thegrade and weighs 106 pounds. His quadruplet sisters are tiny babiesand weigh 5, 5, 6, and 8 pounds. Which is greater, the average (mean) weight of these fivechildren or the median weight, and by how many pounds?ProblemLining up the numbers (5, 5, 6, 8, 106), we see that the median weight is 6 pounds. SolutionThe average weight of the five kids is .Therefore, the average weight is bigger, bypounds, making the answer.6. The number in each box below is the product of the numbers in the two boxes that touch it in the row above. For example,. What is the missing number in the top row?ProblemSolutionLet the value in the empty box in the middle row be , and the value in the empty box in the top row be . is the answer we're looking for.Solution 1: Working BackwardsWe see that, making.It follows that, so.Another way to do this problem is to realize what makes up the bottommost number. Thismethod doesn't work quite as well for this problem, but in a larger tree, it might be faster. (In this case, Solution 1 would be faster since there's only two missing numbers.)Solution 2: Jumping Back to the StartAgain, let the value in the empty box in the middle row be , and the value in the empty box in the top row be . is the answer we're looking for.We can write some equations:Now we can substitute into the first equation using the two others:7. Trey and his mom stopped at a railroad crossing to let a train pass. As the train began to pass,Trey counted 6 cars in the first 10 seconds. It took the train 2 minutes and 45 seconds to clearthe crossing at a constant speed. Which of the following was the most likely number of cars inthe train?ProblemIf Trey saw, then he saw.Solution 12 minutes and 45 seconds can also be expressed asseconds.Trey's rate of seeing cars,, can be multiplied byon the top andbottom (and preserve the same rate):. It follows that the most likely number of cars is.2 minutes and 45 seconds is equal to.Solution 2Since Trey probably counts around 6 cars every 10 seconds, there are groups of 6cars that Trey most likely counts. Since, the closest answer choice is.8. A fair coin is tossed 3 times. What is the probability of at least two consecutive heads?ProblemFirst, there areways to flip the coins, in order.Solution The ways to get two consecutive heads are HHT and THH. The way to get three consecutive heads is HHH.Therefore, the probability of flipping at least two consecutive heads is .9. The Incredible Hulk can double the distance he jumps with each succeeding jump. If his first jump is 1 meter, the second jump is 2 meters, the third jump is 4 meters, and so on, then onwhich jump will he first be able to jump more than 1 kilometer?ProblemThis is a geometric sequence in which the common ratio is 2. To find the jump that would be over a 1000 meters, we note that. SolutionHowever, because the first term isand not, the solution to the problem is10. What is the ratio of the least common multiple of 180 and 594 to the greatest common factorof 180 and 594?ProblemTo find either the LCM or the GCF of two numbers, always prime factorize first. Solution 1The prime factorization of . The prime factorization of .Then, find the greatest power of all the numbers there are; if one number is one but not the other, use it (this is ). Multiply all of these to get 5940.For the GCF of 180 and 594, use the least power of all of the numbers that are in bothfactorizations and multiply. = 18. Thus the answer = =.We start off with a similar approach as the original solution. From the prime factorizations, the GCF is 18.Similar SolutionIt is a well known fact that. So we have,.Dividing by 18 yields .Therefore, .11. Ted's grandfather used his treadmill on 3 days this week. He went 2 miles each day. On Monday he jogged at a speed of 5 miles per hour. He walked at the rate of 3 miles per hour on Wednesday and at 4 miles per hour on Friday. If Grandfather had always walked at 4 miles per hour, he would have spent less time on the treadmill. How many minutes less?ProblemWe use that fact that . Let d= distance, r= rate or speed, and t=time. In this case, letrepresent the time.SolutionOn Monday, he was at a rate of . So,.For Wednesday, he walked at a rate of . Therefore,.On Friday, he walked at a rate of. So,. Adding up the hours yields++=.We now find the amount of time Grandfather would have taken if he walked atperday. Set up the equation,.To find the amount of time saved, subtract the two amounts: -=.To convert this to minutes, we multiply by 60.Thus, the solution to this problem is12. At the 2013 Winnebago County Fair a vendor is offering a "fair special" on sandals. If you buy one pair of sandals at the regular price of $50, you get a second pair at a 40% discount, and a third pair at half the regular price. Javier took advantage of the "fair special" to buy three pairs of sandals. What percentage of the $150 regular price did he save?ProblemFirst, find the amount of money one will pay for three sandals without the discount. We have.SolutionThen, find the amount of money using the discount: .Finding the percentage yields .To find the percent saved, we have13. ProblemWhen Clara totaled her scores, she inadvertently reversed the units digit and the tens digit of one score. By which of the following might her incorrect sum have differed from the correct one?Let the two digits be and. SolutionThe correct score was . Clara misinterpreted it as. The difference between thetwo iswhich factors into. Therefore, since the difference is a multiple of 9,the only answer choice that is a multiple of 9 is.14.Abe holds 1 green and 1 red jelly bean in his hand. Bea holds 1 green, 1 yellow, and 2 red jelly beans in her hand. Each randomly picks a jelly bean to show the other. What is the probability that the colors match?ProblemThe probability that both show a green bean is. The probability that both show ared bean is . Therefore the probability isSolution15. If ,, and , what is the product of, , and ?ProblemSolutionTherefore,.Therefore,.To most people, it would not be immediately evident that , so we can multiply 6'suntil we get the desired number:, so.Therefore the answer is16. A number of students from Fibonacci Middle School are taking part in a community serviceproject. The ratio of-graders to-graders is, and the the ratio of-graders to-graders is . What is the smallest number of students that could be participating inthe project?ProblemSolutionWe multiply the first ratio by 8 on both sides, and the second ratio by 5 to get the same number for 8th graders, in order that we can put the two ratios together:Solution 1: AlgebraTherefore, the ratio of 8th graders to 7th graders to 6th graders is. Since the ratiois in lowest terms, the smallest number of students participating in the project is.The number of 8th graders has to be a multiple of 8 and 5, so assume it is 40 (the smallest possibility). Then there are 6th graders and7th graders. The numbers ofstudents isSolution 2: Fakesolving17. The sum of six consecutive positive integers is 2013. What is the largest of these six integers?ProblemThe mean of these numbers is. Therefore the numbers are, so the answer isSolution 1Let thenumber be . Then our desired number is.Solution 2Our integers are , so we have that.Let the first term be. Our integers are. We have,Solution 318.Isabella uses one-foot cubical blocks to build a rectangular fort that is 12 feet long, 10 feet wide, and 5 feet high. The floor and the four walls are all one foot thick. How many blocks does the fort contain?ProblemThere arecubes on the base of the box. Then, for each of the 4 layers abovethe bottom (as since each cube is 1 foot by 1 foot by 1 foot and the box is 5 feet tall, there are4 feet left), there arecubes. Hence, the answer is.Solution 1 We can just calculate the volume of the prism that was cut out of the originalbox. Each interior side of the fort will be 2 feet shorter than each side of the outside. Since thefloor is 1 foot, the height will be 4 feet. So the volume of the interior box is.Solution 2The volume of the original box is . Therefore, the number of blockscontained in the fort is19. Bridget, Cassie, and Hannah are discussing the results of their last math test. Hannah shows Bridget and Cassie her test, but Bridget and Cassie don't show theirs to anyone. Cassie says, 'I didn't get the lowest score in our class,' and Bridget adds, 'I didn't get the highest score.' What is the ranking of the three girls from highest to lowest?ProblemIf Hannah did better than Cassie, there would be no way she could know for sure that she didn't get the lowest score in the class. Therefore, Hannah did worse than Cassie. Similarly, ifHannah did worse than Bridget, there is no way Bridget could have known that she didn't getthe highest in the class. Therefore, Hannah did better than Bridget, so our order isSolution20. Arectangle is inscribed in a semicircle with longer side on the diameter. What is thearea of the semicircle?ProblemSolutionA semicircle has symmetry, so the center is exactly at the midpoint of the 2 side on the rectangle, making the radius, by the Pythagorean Theorem,. The area is21. ProblemSamantha lives 2 blocks west and 1 block south of the southwest corner of City Park. Her school is 2 blocks east and 2 blocks north of the northeast corner of City Park. On school days she bikes on streets to the southwest corner of City Park, then takes a diagonal path through the park to the northeast corner, and then bikes on streets to school. If her route is as short as possible, how many different routes can she take?SolutionThe number of ways to get from Samantha's house to City Park is, and the number ofways to get from City Park to school is. Since there's one way to go through CityPark (just walking straight through), the number of different ways to go from Samantha's house to City Park to school22.Toothpicks are used to make a grid that is 60 toothpicks long and 32 toothpicks wide. How many toothpicks are used altogether?ProblemThere are 61 vertical columns with a length of 32 toothpicks, and there are 33 horizontal rowswith a length of 60 toothpicks. An effective way to verify this is to try a small case, i.e. a grid of toothpicks. Thus, our answer isSolution23.Angleof is a right angle. The sides ofare the diameters of semicircles as shown. The area of the semicircle on equals, and the arc of the semicircle onhas length . What is the radius of the semicircle on?ProblemIf the semicircle on AB were a full circle, the area would be 16pi. Therefore the diameter of the first circle is 8. The arc of the largest semicircle would normally have a complete diameter of 17. The Pythagorean theorem says that the other side has length 15, so the radius is.Solution 1We go as in Solution 1, finding the diameter of the circle on AC and AB. Then, an extended version of the theorem says that the sum of the semicircles on the left is equal to the biggest one, so the area of the largest is , and the middle one is , so the radius is .Solution 224. Squares, , andare equal in area. Pointsandare the midpointsof sidesand, respectively. What is the ratio of the area of the shaded pentagonto the sum of the areas of the three squares?ProblemSolution 1First let(whereis the side length of the squares) for simplicity. We can extenduntil it hits the extension of. Call this point. The area of trianglethen isThe area of rectangleis. Thus, our desired area is. Now, the ratio of the shaded area to the combined area of the three squares is.Solution 2Let the side length of each square be 1.Let the intersection ofandbe .Since, . Sinceand are vertical angles, theyare congruent. We also haveby definition.So we haveby congruence. Therefore,.Since andare midpoints of sides,. This combined withyields.The area of trapezoidis.The area of triangleis.So the area of the pentagon is .The area of the 3 squares is . Therefore, .Solution 3Let the intersection of andbe .Now we haveand .Because both triangles has a side on congruent squares therefore.Becauseand are vertical angles. Also bothand are right angles so .Therefore by AAS (Angle, Angle, Side) . Then translating/rotating the shadedinto the position ofSo the shaded area now completely covers the squareSet the area of a square asTherefore, .25.A ball with diameter 4 inches starts at point A to roll along the track shown. The track is comprised of 3 semicircular arcs whose radii are inches, inches, andinches, respectively. The ball always remains in contact with the track and does not slip. What is the distance the center of the ball travels over the course from A to B?ProblemThe radius of the ball is 2 inches. If you think about the ball rolling or draw a path for the ball (see figure below), you see that in A and C it loses inches, and it gains inches on B.So, the departurefrom the length of the track means that the answer is .Solution 1The total length of all of the arcs is . Since we want the path fromthe center, the actual distance will be shorter. Therefore, the only answer choice less thanis . This solution may be invalid because the actual distance can be longer if the path the center travels is on the outside of the curve, as it is in the middle bump. Solution 2。

2023年AMC10美国数学竞赛A卷附中文翻译和答案

2023AMC10美国数学竞赛A卷1. A cell phone plan costs $20 each month, plus 5¢ per text message sent, plus 10¢ for each minute used over 30 hours. In January Michelle sent 100 text messages and talked for 30.5 hours. How much did she have to pay?(A) $24.00 (B) $24.50 (C) $25.50 (D) $28.00 (E) $30.002. A small bottle of shampoo can hold 35 milliliters of shampoo, Whereas a large bottle can hold 500 milliliters of shampoo. Jasmine wants to buy the minimum number of small bottles necessary to completely fill a large bottle. How many bottles must she buy?(A) 11 (B) 12 (C) 13 (D) 14 (E) 153. Suppose [a b] denotes the average of a and b, and {a b c} denotes the average of a, b, and c. What is {{1 1 0} [0 1] 0}?(A) (B)(C)(D) (E)4. Let X and Y be the following sums of arithmetic sequences:X= 10 + 12 + 14 + …+ 100.Y= 12 + 14 + 16 + …+ 102.What is the value of(A) 92 (B) 98 (C) 100 (D) 102 (E) 1125. At an elementary school, the students in third grade, fourth grade, and fifth grade run an average of 12, 15, and 10 minutes per day, respectively. There are twice as many third graders as fourth graders, and twice as many fourth graders as fifth graders. What is the average number of minutes run per day by these students?(A) 12 (B) (C) (D) 13 (E) 146. Set A has 20 elements, and set B has 15 elements. What is the smallest possible number of elements in A∪B, the union of A and B?(A) 5 (B) 15 (C) 20 (D) 35 (E) 3007. Which of the following equations does NOT have a solution?(A) (B) (C)(D) (E)8. Last summer 30% of the birds living on Town Lake were geese, 25% were swans, 10% were herons, and 35% were ducks. What percent of the birds that were not swans were geese?(A) 20 (B) 30 (C) 40 (D) 50 (E) 609. A rectangular region is bounded by the graphs of the equations y=a, y=-b, x=-c, and x=d, where a, b, c, and d are all positive numbers. Which of the following represents the area of this region?(A) ac + ad + bc + bd (B) ac – ad + bc – bd (C) ac + ad – bc – bd(D) –ac –ad + bc + bd (E) ac – ad – bc + bd10. A majority of the 20 students in Ms. Deameanor’s class bought pencils at the school bookstore. Each of these students bought the same number of pencils, and this number was greater than 1. The cost of a pencil in cents was greater than the number of pencils each student bought, and the total cost of all the pencils was $17.71. What was the cost of a pencil in cents?(A) 7 (B) 11 (C) 17 (D) 23 (E) 7711. Square EFGH has one vertex on each side of square ABCD. Point E is on AB with AE=7·EB. What is the ratio of the area of EFGH to the area of ABCD?(A) (B)(C)(D) (E)12. The players on a basketball team made some three-point shots, some two-point shots, some one-point free throws. They scored as many points with two-point shots as with three-point shots. Their number of successful free throws was one more than their number of successful two-point shots. The team’s total score was 61 points. How many free throws did they make?(A) 13 (B) 14 (C) 15 (D) 16 (E) 1713. How many even integers are there between 200 and 700 whose digits are alldifferent and come from the set {1, 2, 5, 7, 8, 9}?(A) 12 (B)20 (C)72 (D) 120 (E) 20014. A pair of standard 6-sided fair dice is rolled once. The sum of the numbers rolled determines the diameter of a circle. What is the probability that the numerical value of the area of the circle is less than the numerical value of the circle’s circumference?(A) (B)(C)(D) (E)15. Roy bought a new battery-gasoline hybrid car. On a trip the car ran exclusively on its battery for the first 40 miles, then ran exclusively on gasoline for the rest of the trip, using gasoline at a rate of 0.02 gallons per mile. On the whole trip he averaged55 miles per gallon. How long was the trip in miles?(A) 140 (B) 240 (C) 440 (D) 640 (E) 84016. Which of the following in equal to(A) (B) (C) (D) (E)17. In the eight-term sequence A, B, C, D, E, F, G, H, the value of C is 5 and the sum of any three consecutive terms is 30. What is A + H?(A) 17 (B) 18 (C) 25 (D) 26 (E) 4318. Circles A, B, and C each have radius 1. Circles A and B share one point of tangency. Circle C has a point of tangency with the midpoint of AB. What is the area inside Circle C but outside Circle A and Circle B?(B) (C) (D) (E)(A)19. In 1991 the population of a town was a perfect square. Ten years later, after an increase of 150 people, the population was 9 more than a perfect square. Now, in 2023, with an increase of another 150 people, the population is once again a perfect square. Which of the following is closest to the percent growth of the town’s popu lation during this twenty-year period?(A) 42 (B) 47 (C) 52 (D) 57 (E) 6220. Two points on the circumference of a circle of radius r are selected independently and at random. From each point a chord of length r is drawn in a clockwise direction. What is the probability that the two chords intersect?(A) (B) (C) (D) (E)21. Two counterfeit coins of equal weight are mixed with 8 identical genuine coins. The weight of each of the counterfeit coins is different from the weight of each of the genuine coins. A pair of coins is selected at random without replacement from the 10 coins. A second pair is selected at random without replacement from the remaining 8coins. The combined weight of the first pair is equal to the combined weight of the second pair. What is the probability that all 4 selected coins are genuine?(A) (B) (C) (D) (E)22. Each vertex of convex pentagon ABCDE is to be assigned a color. There are 6 colors to choose from, and the ends of each diagonal must have different colors. How many different colorings are possible?(A) 2500 (B) 2880 (C) 3120 (D) 3250 (E) 375023. Seven students count from 1 to 1000 as follows:·Alice says all the numbers, except she skips the middle number in each consecutive group of thre e numbers. That is Alice says 1, 3, 4, 6, 7, 9, …, 997, 999, 1000.·Barbara says all of the numbers that Alice doesn’t say, except she also skips the middle number in each consecutive grope of three numbers.·Candice says all of the numbers that neither Alice nor Barbara says, except she also skips the middle number in each consecutive group of three numbers. ·Debbie, Eliza, and Fatima say all of the numbers that none of the students with the first names beginning before theirs in the alphabet say, except each also skips the middle number in each of her consecutive groups of three numbers.·Finally, George says the only number that no one else says.What number does George say?(A) 37 (B) 242 (C) 365 (D) 728 (E) 99824. Two distinct regular tetrahedra have all their vertices among the vertices of the same unit cube. What is the volume of the region formed by the intersection of the tetrahedra?(A) (B) (C) (D) (E)an integer. A point X in the interior of R is25. Let R be a square region andcalled n-ray partitional if there are n rays emanating from X that divide R into N triangles of equal area. How many points are 100-ray partitional but not 60-ray partitional?(A) 1500 (B) 1560 (C) 2320 (D) 2480 (E) 25002023AMC10美国数学竞赛A卷1. 某通讯公司手机每月基本费为20美元, 每传送一则简讯收 5美分（一美元=100 美分）。

美国数学竞赛AMC8 -- 2019年真题解析(英文解析+中文解析)

美国数学竞赛AMC8 -- 2019年真题解析（英文解析+中文解析）Problem 1Answer: DSolution:We maximize the number of sandwiches Mike and Ike can buy by finding the lowest multiple of 4.5 that is less than 30 This number is 6.Therefore, they can buy 6 sandwiches for 4.5*6=27. They spend the remaining money on soft drinks, so they buy30-27=3 soft drinks. Combining the items, Mike and Ike buy 6+3=9 items.中文解析：4.5*6=27. 因此买6个Sandwiches，剩下的3元钱买3个Drinks.Problem 2Answer: ESolution:Using the diagram we find that the larger side of the small rectangle is 2 times the length of the smaller side. Therefore the longer side is 5*2=10. So the area of the identical rectangles is5*10=50. We have 3 identical rectangles that form the large rectangle. Therefore the area of the large rectangle is 50*3=150.中文解析：长方形的短边的长度是5，则AD=5+5=10. CD=AD=10. 即长方形的长边是10. ABCD的面积是: AB *BC =(10+5)*10=150.答案是E。

美国数学竞赛美国高中数学竞赛 AMC 10 试题及答案 2000年-2015年

2000 AMC 10 ProblemsProblem 1In the year 2001, the United States will host the International Mathematical Olympiad. Let , , and be distinct positive integers such that the product 2001=••O M I . What is the largest possible value of the sum O M I ++ ?（A ）23 （B ）55 （C ）99 （D ）111 （E ）671Problem 22000 （20002000）＝（A ）20012000 （B ）20004000 （C ）40002000 （D ）2000000,000,4 （E ）000,000,42000Problem 3Each day, Jenny ate 20% of the jellybeans that were in her jar at the beginning of that day. At the end of the second day, 32 remained. How many jellybeans were in the jar originally?（A ）40 （B ）50 （C ）55 （D ）60 （E ）75Problem 4Chandra pays an on-line 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 fixed monthly fee?（A ）2.53 （B ）5.06 （C ）6.24 （D ）7.42 （E ）8.77Problem 5Points M and N are the midpoints of sides PA and PB of △PAB. As P moves along a line that is parallel to side AB, how many of the four quantities listed below change? (a) the length of the segment MN (b) the perimeter of △PAB(c) the area of △PAB (d) the area of trapezoid ABNM（A ）0 （B ）1 （C ）2 （D ）3 （E ）4Problem 6The Fibonacci sequence 1，1，2，3，5，8，13，21，…… starts with two s, and each term afterwards is the sum of its two predecessors. Which one of the ten digits is the last to appear in the units position of a number in the Fibonacci sequence?（A ）0 （B ）4 （C ）6 （D ）7 （E ）9Problem 7 In rectangle , , is on , and and trisect .What is the perimeter of ?（A ）333+ （B ）3342+ （C ）222+ （D ）2533+ （E ）3352+ Problem 8At Olympic High School, 52 of the freshmen and 54 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?（A ）There are five times as many sophomores as freshmen.（B ）There are twice as many sophomores as freshmen.（C ）There are as many freshmen as sophomores.（D ）There are twice as many freshmen as sophomores.（E ）There are five times as many freshmen as sophomores.Problem 9If , where , then（A ）－2 （B ）2 （C ）2－2p （D ）2p-2 （E ）｜2p-2｜Problem 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 ｜x-y ｜?（A ）2 （B ）4 （C ）6 （D ）8 （E ）10Problem 11Two different prime numbers between 4 and 18 are chosen. When their sum is subtracted from their product, which of the following numbers could be obtained?（A ）21 （B ）60 （C ）119 （D ）180 （E ）231Problem 12Figures 0, 1, 2 and 3 consist of 1, 5, 13, and 25 nonoverlapping unit squares, respectively. If the pattern were continued, how many nonoverlapping unit squares would there be infigure 100?Figure 0 Figure 1 Figure 2 Figure 3（A ）10401 （B ）19801 （C ）20201 （D ）39801 （E ）40801 Problem 13There 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?Problem 14Mrs. 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 71, 76, 80, 82, and 91. What was the last score Mrs. Walter entered?（A ）71 （B ）76 （C ）80 （D ）82 （E ）91Problem 15Two non-zero real numbers, and , satisfy. Find a possible value of ab ab b a -+ . （A ）-2 （B ）-21 （C ）31 （D ）21 （E ）2 Problem 16The diagram shows 28 lattice points, each one unit from its nearest neighbors. Segment AB meets segment CD at E. Find the length of segment AE.（A ）354 （B ）355 （C ）7512 （D ）52 （E ）9565Boris 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?（A ）<dollar/>3.63 （B ）<dollar/>5.13 （C ）<dollar/>6.30 （D ）<dollar/>7.45 （E ）<dollar/>9.07Problem 18Charlyn walks completely around the boundary of a square whose sides are each 5 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?（A ）24 （B ）27 （C ）39 （D ）40 （E ）42Problem 19Through a point on the hypotenuse of a right triangle, lines are drawn parallel to the legs of the triangle so that the triangle 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（A ）121 m （B ）m （C ）1-m （D ）m 41 （E ）281mProblem 20Let A, M, and C be nonnegative integers such that A+M+C=10. What is the maximum value of A ·M ·C + A ·M + M ·C +C ·A?（A ）49 （B ）59 （C ）69 （D ）79 （E ）89Problem 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.（A ）I only （B ）II only （C ）III only （D ）II and III only （E ）None must be true Problem 22One morning each member of Angela's family drank an 8-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?（A ）3 （B ）4 （C ）5 （D ）6 （E ）7When the mean, median, and mode of the list 10, 2, 5, 2, 4, 2, x are arranged in increasing order, they form a non-constant arithmetic progression. What is the sum of all possible real values of ? （A ）3 （B ）6 （C ）9 （D ）17 （E ）20Problem 24Let be a function for which. Find the sum of all values of for which. （A ）-31 （B ）-91 （C ）0 （D ）95 （E ）35 Problem 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 day of year occur?2000 AMC 10 SolutionProblem 1 The following problem is from both the 2000 AMC 12 #1 and 2000 AMC 10 #1, so both problems redirect to this page.The sum is the highest if two factors are the lowest.So, 1·3·776=2001 and 1+3+667=671 (E)Problem 2 The following problem is from both the 2000 AMC 12 #2 and 2000 AMC 10 #2, so both problems redirect to this page.2000 （20002000）＝12000（20002000）=20012000 (A)Problem 3 The following problem is from both the 2000 AMC 12 #3 and 2000 AMC 10 #3, so both problems redirect to this pageSince Jenny eats 20% of her jelly beans per day, 80%=4/5 of her jelly beans remain after one day. Let be the number of jelly beans in the jar originally.325454=••x x=50 (B) Problem 4Let be the fixed fee, and be the amount she pays for the minutes she used in the first month.x+y=12.48 x+2y=17.54 y=5.06 x=7.42 We want the fixed fee, which is (D) Problem 5(a) Clearly AB does not change, and MN=0.5AB, so MN doesn't change either. (b) Obviously, the perimeter changes.(c) The area clearly doesn't change, as both the base AB and its corresponding height remain the same.(d) The bases AB and MN do not change, and neither does the height, so the area of the trapezoid remains the same.Only quantity changes, so the correct answer is .Problem 6 The following problem is from both the 2000 AMC 12 #4 and 2000 AMC 10 #6, so both problems redirect to this page.Note that any digits other than the units digit will not affect the answer. So to make computation quicker, we can just look at the Fibonacci sequence in :The last digit to appear in the units position of a number in the Fibonacci sequence is 6 (C).Problem 7AD=1 Since ∠ADC is trisected, ∠ADP=∠PDB=∠BDC=30º Thus, PD=332 BD=2 BP=332333=- Adding, 3342+Problem 8Let be the number of freshman and be the number of sophomores.s f 5452= f=2s There are twice as many freshmen as sophomores. Problem 9 The following problem is from both the 2000 AMC 12 #5 and 2000 AMC 10 #9, so both problems redirect to this page.When x<2, x-2 is negative so ∣x -2∣=2-x =p and x=2-pThus x-p=(2-p)-p=2-2p (C)Problem10From the triangle inequality, 2<x <10 and 2<y <10. The smallest positive number not possible is 10-2, which is . (D)Problem 11 The following problem is from both the 2000 AMC 12 #6 and 2000 AMC 10 #11, so both problems redirect to this page.All prime numbers between 4 and 18 have an odd product and an even sum. Any odd number minus an even number is an odd number, so we can eliminate B andD. Since the highest two prime numbers we can pick are 13 and 17, the highest number we can make is (13) (17) – (13+17) = 221-30=191. Thus, we can eliminateE. Similarly, the two lowest prime numbers we can pick are 5 and 7, so the lowest number we can make is (5) (7) – (5+7) =23. Therefore, A cannot be an answer. So, the answer must be .Problem 12 The following problem is from both the 2000 AMC 12 #8 and 2000 AMC 10 #12, so both problems redirect to this page.Solution 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 addWe then add to the number of squares in Figure 0 to get, which ischoice Solution 2We can divide up figure to get the sum of the sum of the first odd 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 2n , so for figure , thereare 22)1(n n ++ unit squares. We plug in n=100 to get 20201, which is choice Solution 3Using the recursion from solution 1, we see that the first differences of 4,8,12, … form an arithmetic progression, and consequently that the second differences are constant and all equal to . Thus, the original sequence can be generated from a quadratic function.If c bn an n f ++=2)(, and f(0)=1, f(1)=5, and f(2)=13, we get a system of three equations in three variables:f(0)=1 gives c=1; f(1)=5 gives a+b+c=5; f(2)=13 gives 4a+2b+c=13 Plugging in into the last two equations gives a+b=4 4a+2b=12 Dividing the second equation by 2 gives the system: a+b=4 2a+b=6Subtracting the first equation from the second gives , and hence . Thus, our quadratic function is: 122)(2++=n n n fCalculating the answer to our problem, f(100)=20201Problem 13In 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 forthe 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 is (B)Problem 14 The following problem is from both the 2000 AMC 12 #9 and 2000 AMC 10 #14, so both problems redirect to this page.Solution 1The first number is divisible by 1.The sum of the first two numbers is even.The sum of the first three numbers is divisible by 3.The sum of the first four numbers is divisible by 4.The sum of the first five numbers is 400.Since 400 is divisible by 4, the last score must also be divisible by 4. Therefore, the last score is either 76 or 80.Case 1: 76 is the last number entered.Since 400≡76≡1 (mod 3), the fourth number must be divisible by 3, but none of the scores are divisible by 3. Case 2: 80 is the last number entered. Since 80≡2 (mod 3), the fourth number must be 2 (mod 3). Thatnumber is 71 and only 71. The next number must be 91, since the sum of the first two numbers is even.So the only arrangement of the scores 76, 82, 91,71,80Solution 2We know the first sum of the first three numbers must be divisible by 3, so we write out all 5 numbers (mod 3), which gives 2,1,2,1,1, respectively. Clearly the only way to get a number divisible by 3 by adding three of these is by adding the three ones. So those must go first. Now we have an odd sum, and since the next average must be divisible by 4, 71 must be next. That leaves 80 for last, so the answer is .Problem 15 The following problem is from both the 2000 AMC 12 #11 and 2000 AMC 10 #15, so both problems redirect to this page.22)()()(22222=-=----+=--+=-+ba ab b a b a b a b a b a ab b a ab a b b a (E)Alternatively, we could test simple values, like )21,1(),(=b a , which would yield 2=-+ab ab b a Another way is to solve the equation for giving 1+=a a b , then substituting this into the expression and simplifying gives the answer ofProblem 16Solution 1Let be the line containing A and B and let be the line containing C and D. If we set the bottom left point at (0,0), then A=(0,3), B=(6,0), C=(4,2), and D=(2,0) . The line is given by the equation 11b x m y +=. The -intercept is A=(0,3), so 1b =3. We are given two points on , hence we can compute the slope, to be 210630-=-- , so is the line 321+-=x y Similarly, is given by 22b x m y +=. The slope in this case is 12402=--, so 2b x y +=. Plugging in the point (2,0) gives us 2b =-2, so is the line 2-=x y . At E, the intersection point, both of the equations must be true,2-=x y , 321+-=x y so 3212+-=-x x SO 310=x 34=y We have the coordinates of and , so we can use the distance formula here: 355)334()0310(22=-+- which is answer choiceSolution 2Draw the perpendiculars from andto , respectively. As it turns out, . Let be the point onfor which . , and, so by AA similarity,By the Pythagorean Theorem, wehave, ,。

精选美国amc中文版试题及答案

2000到2012年A M C10美国数学竞赛P 0 A 0 B 0 CD 0 全美中学数学分级能力测验(AMC 10)2000年第01届美国AMC10 (2000年2月日时间75分钟)1. 国际数学奥林匹亚将于2001年在美国举办，假设I 、M 、O 分别表示不同的正整数，且满足I ?M ?O =2001，则试问I ?M ?O 之最大值为。

(A) 23 (B) 55 (C) 99 (D) 111 (E) 6712. 2000(20002000)为。

5. 如图M ，N 分别为PA 与PB 之中点，试问当P 在一条平行AB 的直在线移动时，下列各数值有项会变动。

(a) MN 长 (b) △PAB 之周长 (c) △PAB 之面积 (d) ABNM 之面积 (A) 0项 (B) 1项 (C) 2项 (D) 3项 (E) 4项 6. 费氏数列是以两个1开始，接下来各项均为前两项之和，试问在费氏数列各项的个位数字中，最后出现的阿拉伯数字为。

(A) 0 (B) 4 (C) 6 (D) 7 (E) 97. 如图，矩形ABCD 中，AD =1，P 在AB 上，且DP 与DB 三等分图3 图2 图1 图0 ?ADC ，试问△BDP 之周长为。

(A) 3?33 (B) 2?334 (C) 2?2 (D) 2533+ (E) 2?335 8. 在奥林匹克高中，有52的新生与54的高二生参加AMC 10年级测验。

美国数学竞赛AMC8 -- 2008年真题解析(英文解析+中文解析)

美国数学竞赛AMC8 – 2008年真题解析(英文解析+中文解析)Problem 1Answer: BSolution:50-12-24=14中文解析：总共花的钱是：12+12*2=36元。

剩余50-36=14元。

答案是BProblem 2Answer: ASolution:We can derive that c=8,L=6, U=7,and E=1. Therefore, the answer is 8671.中文解析：这10个字母的对应关系是： B -0；E-1; S-2; ......K -9. 按照这个对应关系：C-8，L-6，U-7，E-1. 即8671. 答案是A。

Problem 3Answer: ASolution:We can go backwards by days, but we can also backwards by weeks. If we go backwards by weeks, we see that February 6 is a Friday. If we now go backwards by days, February 1 is a Sunday.中文解析：13日是周五，则13-7=6，即6日也是周五，则倒推2月1日是周日。

答案是A。

Problem 4Answer: CSolution:The area outside the small triangle but inside the large triangle is 16-1=15. This is equally distributed between the three trapezoids. Each trapezoid has an area of 15/3=5.中文解析：大三角形的面积等于小的等边三角形的面积加上3个梯形的面积。

据此，三个梯形的面积是16-1=15. 每个梯形的面积是15/3=5. 答案是C。

1、下载文档前请自行甄别文档内容的完整性，平台不提供额外的编辑、内容补充、找答案等附加服务。
2、"仅部分预览"的文档,不可在线预览部分如存在完整性等问题,可反馈申请退款(可完整预览的文档不适用该条件!)。
3、如文档侵犯您的权益，请联系客服反馈,我们会尽快为您处理(人工客服工作时间：9:00-18:30)。

答案：
1. C

2. E
3. E
4. B
5. A
6. C
7. A
8. E
9. A
10. C
11. C
12. C
13. B
14. D
15. A
16. B
17. D
18. E
19. D
20. C
21. B
22. B
23. E
24. B
25. C
答案：