1、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 : M O=：；0i；L What's the largest possible value of the sum , / 。？(A) 23(B) 55(C) 99(D) 111(E) 671factors are the lowest.Solution

2、The sum is the highest if two，.So, 1 - 3，2 DI and1 + 3 + 667 = 671 =|(E)|2,000,0002000(200(?°)=(A) 200產(chǎn)I(B) 4000aooc (C) 2000400°(D) 4t 000,0002rH"'：) 2000Solution2000 , 2OOO2000 = 20001，2OOO2000 = 2OQO2000 " = 2OOO2001. H3Each day, Jenny ate如闕 of the jellybeans that were in

3、her jar at thebeginning of the day. At the end of the second day,remained. Howmany jellybeans were in the jar originally?(A) 40(B) 50(C) 55(D) 60(E) 75Solutionx(.8)(-8) = 32t(,64) = 324Chandra pays an online service provider a fixed monthly fee plus anhourly charge for connect time. Her December bil

4、l was'?匚'' but inJanuary her bill was S。；：4 because she used twice as much connect time as in December. What is the fixxed monthly fee?(A) $2.53(B) $5.06(C)蚯24(D) $742(E) $8,77SolutionLet c_jbe the fixed fee, and匕be the amount she pays for the minutesshe used in the first month.£ +

5、* = 12.48x += 17.54y = 5.06£=7.42We want the fixed fee, which is 回5and 匚of .As .1 .'1 , how many of the fourPoints 叵and 暨are the midpoints of sides moves along a line that is parallel to side quantities listed below change?(a) the length of the segment MN(b) the perimeter of 肉一 ,“ .I(c) the

6、 area of P.L(d) the area of trapezoid； A M(A) 0(B) 1(C) 2(D) 3(E)4MN = AB“, so 皿更 doesn'tSolution(a) Clearly /i.9does not change, and change either.(b) Obviously, the perimeter changes.(c) The area clearly doesn't change, as both the base 艮二land its corresponding height remain the same.(d) T

7、he bases!4 Band W*do not change, and neither does the height,so the trapezoid remains the same.Only quantity changes, so the correct answer is.6The Fibonacci Sequence11 ； starts with two 1s andeach term afterwards is the sum of its predecessors. Which one of the ten digits is the last to appear in t

8、het units position of a number in the Fibonacci Sequence?(A) 0(B) 4(C) 6(D) 7(E) 9SolutionThe pattern of the units digits are1,1,2,3,5,8,3,1,4,5,9,4,3T 7,0, 一 7,4,1,5,6In order of appearance:2 隊(duì)& 8,4,9,7,0? 6.1 is the last.匚I 7In rectangle HBC'D,.匯。= 1,?3is on LOl, and D8 and1 trisect 一 '

9、;l . What is the perimeter of'？.5/3(A) 3 +?(B)2+緣(C)2 + 2調(diào) (D)(E) 2 +JA£Solution ，.Solution，；二二I,Since _ iDi is trisected,Thus, 口J.P.E爐 一 /三。一 3,DB = 2BP =V3-2 +Adding,73 2瓜下 ,4/3 丁.824At Olympic High School, of the freshmen and of the sophomores took the AMC-10. Given that the number of fr

10、eshmen and sophomore contestants was the same, which of the following must be true?There are five times as many sophomores as freshmen.B'There are twice as many sophomores as freshmen.'There are as many freshmen as sophomores.U-There are twice as many freshmen as sophomores.【E'There are

11、five times as many freshmen as sophomores.SolutionLet 仁be the number of freshman and be the number of sophomores.2f 45/ =守，2/=卻f = 2sThere are twice as many freshmen as sophomores.回9If 卜 二 R where 2, then -_(A) -2(B) 2(C) 2-2P(D) 2p-2(E) |2p-2|Solution|簟一2| p二 Z, so :'P.£ + & = 2.M -P =

12、 2 2*. fcT|101 , and .J , not a possibleThe sides of a triangle with positive area have lengthsThe sides of a second triangle with positive area have lengths and . What is the smallest positive number that is value of k；"？(A) 2(B) 4(C) 6(D) 8(E) 10SolutionFrom the triangle inequality,M b 'a

13、nd "". The smallestpositive number not possible is 。一 . .', which is 8.回 11Two different prime numbers betweenand ' are chosen. When theirsum is subtracted from their product, which of the following numbers could be obtained?(A) 21(B) 60(C) 119(D) 180(E) 231Two prime numbers betwee

14、n&and 1T are both odd.odd * odd) (odd + odd) = odd - even = oddThus, we can discard the even choices.曲(Q + b) = ab G b = (o -1)(6 1) 1Both 。- Hand a - 1 are even, so one more than:，一 Q ： is a multipleof four.I I :'is the only possible choice.satisfy this, 1. 一】 】二一2 1 I .-：. 口12Figures 也.,_,

15、 and consist of ., , 13, and , nonoverlapping unit squares, respectively. If the pattern were continued, how many(A) 10401(B) 19801(C) 20201(D) 39801(E) 40801Solution Solution 1We have a recursion:五 一 =力士I.E. we add increasing multiples of: each time we go up a figure.So, to go from Figure 0 to 100,

16、 we add4 1+42 + 499 + 4* 100 =4- 5050 = 2020020201Solution 2We can divide up figure too get the sum of the sum of the first心 十-odd numbers and the sum of the first卜網(wǎng) odd numbers. If you do not seethis, here is the example for 門 一 3:5The sum of the first巨 odd numbers is g so for figure m, there are22

17、+-unit squares. We plug in I ' 'to get E L “， which ischoice 回 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

18、 of the same color? * 閨 0 (B)l (C)5!4!.3!.2!*l!(D)(E) 15!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

19、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: *Y Y )After this we can proceed to fill in the whole pegboard, so there is onlyI arrangement of the pegs. The answer is回YR ¥GRY

20、B G R YO BG R Y14Mrs. 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. Th

21、e scores (listed in ascending order) were；。，&J, -, and -： 1. What was the last score Mrs. Walter entered?(A) 71(B) 76(C) 80(D) 82(E) 91Solution71,76,80,82,91The sum of the firstscores must be even, so we must choose .evensor the _odds to be the first two scores.Let us look at the numbers in mod

22、If we choose the two odds, the next number must be a multiple ofII,of which there is none.Similarly, if we choose 76,80 or 里，里,the next number must be a multiple of 區(qū) of which there is none.So we choose 小first.The next number must be 1 in mod 3, of which only'H remains.The sum of the first three

23、 scores is f77 9. This is equivalent to I in mod .Thus, we need to choose one number that is3in mod :. Ujis the onlyone that works.Thus, S：iis the last score entered.回15Two non-zero real numbers,and , satisfy 此一 ".Which of thea bI abfollowing is a possible value of 匯 。？(A) -2(B)(C) 1(D) i (E)2f

24、t0也Solutionaba2+lr4 = wb aab+ u* + ft2一 abSubstituting : 一“，we get-（a - fi）2 + aa + i? _ -a2 + 2ab -J? +a2 + S2 abab2abab2回 16The diagram shows neighbors. Segment segment !，；多.花 lattice points, each one unit from its nearest區(qū)&meets segment*D at 臼 Find the length of手 （B）等 （C）竽（D）2海（E）管SolutionSol

25、ution 1Let be the line containingEland Sand let 同be the line containing 士and D. If we set the bottom left point at（°*。），then 4 = Q3）, 芭 一 ：；），。- ¥；，and 一匚；'；.The line is given by the equation . 一，i； ! . The -intercept is .ii_；.；）, so . We are given two points on O, hence we can0-3 _ -1

26、compute the slope,三to be I： - , so is the line2-0Similarly,西is given by S _. %.The slope in this case is I -so 4=" + g. Plugging in the point 。）gives us 與=-2, sois theline 之 一二一At "the intersection point, both of the equations must be true, so-11y =工-2： y = -jr-x + 3 => H-2 = "-j：

27、+ 3 上fX10至4學(xué)期=3We have the coordinates of I land a I, so we can use the distance回Solution 2Draw the perpendiculars from回and 3to C D, respectively. As it turnsout, _D. Let 臣 be the point on 二'二 for which .-IF .1. CD./ y :" ri】./ 二；：二' ：9；r, and_ 1_：，：；, so by AA similarity,AFE AF BCABCE

28、=> = 一AE BEBy the PythagoreanTheorem, we have " ' -、丁 ： - -:,- 1 - -1'、-. Let 、工，so2s 一啟 and回This is answer choicewhich 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

29、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

30、could Boris have after using the machine repeatedly?(A) $3,63(B) $5.13(C)蚯30(D) $745(E) $9,07SolutionConsider 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,

31、 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 ofOf the choices

32、given, theonly one is 回 18Charlyn 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 squ

33、are kilometers and rounded to the nearest whole number?(A) 24(B) 27(C) 39(D) 40(E) 42SolutionThe area she sees looks at follows:The part inside the walk has area 5 U- 3 - 3 = 16. The part outside the walk consists of four rectangles, and four arcs. Each of the rectangles has area £ ；5. The four

34、 arcs together form a circle with radiusI.Therefore the total area she can see is J I ：li, which rounded to the nearest integeris 口 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 small

35、er right triangles. The area of one of the two small right triangles is尊.times the area of the square. The ratio of thearea of the other small right triangle to the area of the square is i m (C) 1 - m (D)白 (E)白ySolutionJ.-'j .Let the square have area ., then it follows that the altitude of one o

36、fIZIthe triangles is 薪匕.The area of the other triangle is 國x _ 1 x _ 1By similar triangles, we have 12鵬 “宏This is choice 回 (Note that this approach is enough to get the correct answer in thecontest. However, if we wanted a completely correct solution, we should also note that scaling the given trian

37、gleJ times changes eachof the areas I: times, and therefore it does not influence the ratio ofany two areas. This is why we can pick the side of the square.)20Let 4, &；, and be nonnegative integers such that一1 l ：，.What is the maximum value ofA; C .-：一二； 一匚 二？(A) 49(B) 59(C) 69(D) 79(E) 89Soluti

38、onThe trick is to realize that the sum-lM.C，= .* .MC = G/'is similarto the product ：月'，晨 j:.If we multiply5+i)(時(shí)干i)(C + i),we getWe know that ： J - 。, therefore(A + 1)(M + 1)(C + 1) = (AMC + AM + 4C + MC) +11.Therefore the maximum value of.j A.-； 一廣.M+1片-is equal tothe maximumvalue of1 ” &qu

39、ot; ； 乂 ， ：1-. Now we will find this maximum.-.Then thisSuppose that some two of 士 衛(wèi) and ddiffer by at least triple I多:3-； is surely not optimal.，.Proof： WLOG let 一£1一 1 We can then increase the value of- 1 ''by changing； Land C C + .Therefore the maximum is achieved in the cases where1

40、1 is arotation ofThe value ofNC' _ 1)inthis case is44-5 = &L And thus the maximum of islC + AM + AC+MC is 80-11=畫叵|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 cre

41、atures are creepy crawlers.III. Some alligators are not creepy crawlers.(A) 1 only (B) Il only (C) III only (D) H and III only (El None must be trueSolutionWe interpret the problem statement as a query about three abstract concepts denoted as "alligators", "creepy crawlers" and &

42、quot;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 asI I, C and 因.We got the following informatio

43、n:If 匚is an . , then is an / .There is some that is a and at the same time an 一 1.We CAN NOT conclude that the first statement is true. For example, the situation "Johnny and Freddy are E3s, but only Johnny is a k ' meets both conditions, but the first statement is false.We CAN conclude tha

44、t the second statement is true. We know thatthere is someRhat is aCand at the same time an H. Pick one suchand call it Bobby. Additionally, we know that ifis an - , then is anto.Bobby is an *, therefore Bobby is an 田.And this is enough prove the second statement - Bobby is an巨 that is also a Q.We CA

45、N 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 isII only22One morning each member of Angela's fa

46、mily 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?(A) 3(B) 4(C) 5(D) 6(E) 7SolutionThe exact val

47、ue "8 ounces" is not important. We will only use the fact that each member of the family drank the same amount.Let 崎 I 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 如，-卜仁ounces of fluids.Let Libe t

48、he number of family members. Then each family memberdrank 飛 ounces of fluids.We know that Angela drank4bounces of fluids.As Angela is a family member, we haveMultiply both sides byQto getIf :, we haveIf 不 J , we have=回Therefore the only remaining option is 23When the mean, median, and mode of the li

49、st 1。'J 工-'一二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) 20SolutionAs .occurs three times and each of the three other values just once, regardless of whatfwe choose the mode will al

50、ways beThe sum of all numbers is；為=落，therefore the mean isThe six known values,in sorted order, are 2、今與業(yè)匚IQ. From this2. If 2 <工 V4, the .sequence we conclude: If HESj, the median will be median will be . Finally, if ? "the median will beWe will now examine each of these three cases separat

51、ely.In the case :二 2：,both the median and the mode are 2, therefore we can not get any non-constant arithmetic progression._.x < -一In the case 2 < / v 4 we have，, because25 +工 25-to 25-6-4 I 一z父=-z- > > 0.Therefore our three values in25 + 42 曲一-order are. We want this to be an arithmetic progression.From the first two terms the difference must be一 Therefore thethird term must be 工 +-2) = 2X - 2.Solvingwe get the only solution for this case:The case ± J - remains.2