2002AMC10 -A

(A) B = W
(B) W = R
(C) B = R
(D) 3B = 2R
(E) 2R = W
9. Suppose A, B , and C are three numbers for which 1001C − 2002A = 4004 and 1001B + 3003A = 5005. The average of the three numbers A, B , and C is (A) 1 (B) 3 (C) 6 (D) 9 (E) not uniquely determined
3. According to the standard convention for exponentiation, 22
22
=2
2(
22
)
= 216 = 65, 536.
If the order in which the exponentiations are performed is changed, how many other values are possible? (A) 0 (B) 1 (C) 2 (D) 3 (E) 4
Students who score in the top 1% on this AMC 10 will be invited to take the 20th annual American Invitational Mathematics Examination (AIME) on Tuesday, March 26, 2002 or on Tuesday, April 9, 2002. More details about the AIME and other information are on the back page of this test booklet.
Tuesday, FEBRUARY 12, 2002
Contest A
The MATHEMATICAL ASSOCIATION OF AMERICA American Mathematics Competitions
Presented by the Akamai Foundation
3rd Annual American Mathematics Contest 10
The Committee on the American Mathematics Competitions (CAMC) reserves the right to re-examine students before deciding whether to grant official status to their scores. The CAMC also reserves the right to disqualify all scores from a school if it is determined that the required security procedures were not followed. The publication, reproduction, or communication of the problems or solutions of the AMC 10 during the period when students are eligible to participate seriously jeopardizes the integrity of the results. Duplication at any time via copier, telephone, eMail, World Wide Web or media of any type is a violation of the copyright law. Copyright © 2002, Committee on the American Mathematics Competitions, Mathematical Association of America
5. Each of the small circles in the ﬁgure has radius one. The innermost circle is tangent to the six circles that surround it, and each of those circles is tangent to the large circle and to its small-circle neighbors. Find the area of the shaded region.
(A) π
(B) 1.5π
(C) 2π
(D) 3π
(E) 3.5π
3rd AMC 10 AFra bibliotek2002
3
6. Cindy was asked by her teacher to subtract 3 from a certain number and then divide the result by 9. Instead, she subtracted 9 and then divided the result by 3, giving an answer of 43. What would her answer have been had she worked the problem correctly? (A) 15 (B) 34 (C) 43 (D) 51 (E) 138
10. Compute the sum of all the roots of (2x + 3)(x − 4) + (2x + 3)(x − 6) = 0. (A) 7/2 (B) 4 (C) 5 (D) 7 (E) 13
11. Jamal wants to store 30 computer ﬁles on ﬂoppy disks, each of which has a capacity of 1.44 megabytes (mb). Three of his ﬁles require 0.8 mb of memory each, 12 more require 0.7 mb each, and the remaining 15 require 0.4 mb each. No ﬁle can be split between ﬂoppy disks. What is the minimal number of ﬂoppy disks that will hold all the ﬁles? (A) 12 (B) 13 (C) 14 (D) 15 (E) 16
8. Betsy designed a ﬂag using blue triangles ( ), small white squares ( ), and a red center square( ), as shown. Let B be the total area of the blue triangles, W the total area of the white squares, and R the area of the red square. Which of the following is correct?
AMC 10
3rd AMC 10 A 1. The ratio (A) 0.1
102000 +102002 102001 +102001
2002
2
is closest to which of the following numbers? (C) 1 (D) 5 (E) 10
(B) 0.2
2. For the nonzero numbers a, b, and c, deﬁne (a, b, c) = Find (2, 12, 9). (A) 4 (B) 5 (C) 6 (D) 7 (E) 8 c a b + + . b c a
1. DO NOT OPEN THIS BOOKLET UNTIL TOLD TO DO SO BY YOUR PROCTOR. 2. This is a twenty-five question, multiple choice test. Each question is followed by answers marked A,B,C,D and E. Only one of these is correct. 3. The answers to the problems are to be marked on the AMC 10 Answer Form with a #2 pencil. Check the blackened circles for accuracy and erase errors and stray marks completely. Only answers properly marked on the answer form will be graded. 4. SCORING: You will receive 6 points for each correct answer, 2.5 points for each problem left unanswered, and 0 points for each incorrect answer. 5. No aids are permitted other than scratch paper, graph paper, ruler, compass, protractor, erasers and calculators that are accepted for use on the SAT. No problems on the test will require the use of a calculator. 6. Figures are not necessarily drawn to scale. 7. Before beginning the test, your proctor will ask you to record certain information on the answer form. When your proctor gives the signal, begin working the problems. You will have 75 MINUTES working time to complete the test. 8. When you finish the exam, sign your name in the space provided on the Answer Form.

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美国数学测评AMC10A-2019(中英双语)

1.What is the value of下方表达式的值是多少？(A) 0 (B) 1 (C) 2 (D) 3 (E) 42.What is the hundreds digit of (20! − 15!) ? (20! − 15!)的百位上的数字是多少？ (A) 0 (B) 1 (C) 2 (D) 4(E) 53.Ana and Bonita were born on the same date in different years, n years apart. Last year Ana was 5 times as old as Bonita. This year Ana’s age is the square of Bonita’s age. What is n ?Ana 和Bonita 出生在不同年份的同一天, 相隔n 年。

去年Ana 的年龄是Bonita 年龄的5倍。

今年Ana 的年龄是Bonita 年龄的平方。

问n 的值是多少？(A) 3 (B) 5 (C) 9 (D) 12 (E) 154.A box contains 28 red balls, 20 green balls, 19 yellow balls, 13 blue balls, 11 white balls, and 9 black balls. What is the minimum number of balls that must be drawn from the box without replacement to guarantee that at least 15 balls of a single color will be drawn?一个盒子中有28个红球，20个绿球，19个黄球，13个蓝球，11个白球和9个黑球。

为了保证至少取出15个单一颜色的球, 在不允许放回重取的情况下，须从盒子里至少取出多少个球？(A) 75(B) 76(C) 79(D) 84(E) 915.What is the greatest number of consecutive integers whose sum is 45 ? 最多可以有多少个连续整数的和是45?(A) 9 (B) 25 (C) 45 (D) 90 (E) 1206. For how many of the following types of quadrilaterals does there exist a point in the plane of thequadrilateral that is equidistant from all four vertices of the quadrilateral?2019 AMC 10A Problems • a square• a rectangle that is not a square• a rhombus that is not a square• a parallelogram that is not a rectangle or a rhombus•an isosceles trapezoid that is not a parallelogram 在以下类型的四边形中，有多少种在平面上存在一个与其所有四个顶点距离都相等的点？• 正方形• 不是正方形的长方形• 不是正方形的菱形• 不是长方形或菱形的平行四边形• 不是平行四边形的等腰梯形(A) 1(B) 2 (C) 3 (D) 4 (E) 57. Two lines with slopes and intersect at . What is the area of the triangle enclosed by these two lines and the line两条斜率分别为1/2和2的直线相交于(2, 2)。

2002 AMC 10B Problems

2002 AMC 10B Problems Problem 1 The ratio 200220032001632• is: （A ）61 （B ）31 （C ）21 （D ）32 （E ）23 Problem 2For the nonzero numbers and define cb a abc c b a ++=),,(Find (2,4,6) . （A ）1 （B ）2 （C ）4 （D ）6 （E ）24Problem 3The arithmetic mean of the nine numbers in the set {9,99,999,9999,…,999999999} is a 9-digit number M, all of whose digits are distinct. The number M does not contain the digit（A ）0 （B ）2 （C ）4 （D ）6 （E ）8Problem 4What is the value of 14)23()14)(23(+--+-x x x x , when x=4 ?（A ）0 （B ）1 （C ）10 （D ）11 （E ）12Problem 5Circles of radius and are externally tangent and are circumscribed by a third circle, as shown in the figure. Find the area of the shaded region.（A ）π3 （B ）π4 （C ）π6 （D ）π9 （E ）π12Problem 6For how many positive integers is 232+-n n a prime number?（A ）none （B ）one （C ）two （D ）more than two, but finitely many （E ）infinitely manyProblem 7 Let be a positive integer such that n 1713121+++ is an integer. Which of the following statements is not true?（A ）2 divides n （B ）3 divides n （C ）6 divides n （D ）7 divides n （E ）n>84 Problem 8Suppose July of year N has five Mondays. Which of the following must occur five times in the August of year N ? (Note: Both months have 31 days.)（A ）Monday （B ）Tuesday （C ）Wednesday （D ）Thursday （E ）Friday Problem 9Using the letters A,M,O,S and U , we can form five-letter "words". If these "words" are arranged in alphabetical order, then the "word" U S A M O occupies position（A ）112 （B ）113 （C ）114 （D ）115 （E ）116 Problem 10Supposethat and are nonzero real numbers, and that the equation 02=++b ax x has solutions and . Then the pair (a,b) is（A ）(-2,1) （B ）(-1,2) （C ）(1,-2) （D ）(2,-1) （E ）(4,4) Problem 11The product of three consecutive positive integers is times their sum. What is the sum of the squares?（A ）50 （B ）77 （C ）110 （D ）149 （E ）194 Problem 12For which of the following values of does the equation 621--=--x k x x x have no solution for ?（A ）1 （B ）2 （C ）3 （D ）4 （E ）5 Problem 13Find the value(s) of such that 032128=-+-x y xy is true for all values of .（A ）32 （B ）or 2341- （C ）or 32-41- （D ）23 （E ）or 23-41- Problem 14The number is the square of a positive integer . In decimal representation,the sum of the digits of is （A ）7 （B ）14 （C ）21 （D ）28 （E ）35Problem 15The positive integers A, B, A-B , and A+B are all prime numbers. The sum of these four primes is（A ）even （B ）divisible by 3 （C ）divisible by 5 （D ）divisible by 7 （E ）prime Problem 16For how many integers is nn -20 the square of an integer? （A ）1 （B ）2 （C ）3 （D ）4 （E ）10 Problem 17A regular octagon ABCDEFGH has sides of length two. Find the area of △ADG .（A ）224+ （B ）26+ （C ）234+ （D ）243+ （E ）28+ Problem 18Four distinct circles are drawn in a plane. What is the maximum number of points where at least two of the circles intersect?（A ）8 （B ）9 （C ）10 （D ）12 （E ）16Problem 19Suppose that is an arithmetic sequence with 100...10021=+++a a a and 200...200102101=+++a a a . What is the value of（A ）0.0001 （B ）0.001 （C ）0.01 （D ）0.1 （E ）1Problem 20Let a, b, and c be real numbers such that a-7b+8c=4 and 8a+4b-c=7 Then 222c b a +- is（A ）0 （B ）1 （C ）4 （D ）7 （E ）8Problem 21Andy's lawn has twice as much area as Beth's lawn and three times as much as Carlos' lawn. Carlos' lawn mower cuts half as fast as Beth's mower and one third as fast as Andy's mower. If they all start to mow their lawns at the same time, who will finish first?（A ）Andy （B ）Beth （C ）Carlos （D ）Andy and Carlos tie for first （E ）All three tieProblem 22m =90º. Let M and N be the midpoints Let △XOY be a right-angled triangle with XOYof the legs OX and OY, respectively.Given XN=19 and YM=22, find XY .（A）24 （B）26 （C）28 （D）30 （E）32Problem 23Let be a sequence of integers such that and for all positive integers and Then is（A）45 （B）56 （C）67 （D）78 （E）89Problem 24Riders on a Ferris wheel travel in a circle in a vertical plane. A particular wheel has radius feet and revolves at the constant rate of one revolution per minute. How many seconds does it take a rider to travel from the bottom of the wheel to a point vertical feet above the bottom?（A）5 （B）6 （C）7.5 （D）10 （E）15Problem 25When is appended to a list of integers, the mean is increased by . When is appended to the enlarged list, the mean of the enlarged list is decreased by . How many integers were in the original list?（A）4 （B）5 （C）6 （D）7 （E）82002 AMC 10B Solution Problem 1 Problem 2 (C) Problem 3 The following problem is from both the 2002 AMC 12B #1 and 2002 AMC 10B #3, so both problems redirect to this page. We wish to find , or ,or. This does not have the digit 0, so the answer is (A) 0Problem 4 The following problem is from both the 2002 AMC 12B #2 and 2002 AMC 10B #4, so both problems redirect to this page. (D) 11Problem 5A line going through the centers of the two smaller circles also go through the diameter. The length of this line within the circle is 3+3+2+2=10 Because this is the length of the larger circle's diameter, the length of its radius is 5The area of the large circle is 25π, and the area of the two smaller circles is 9π+4π=13π. To find the area of the shaded region, subtract the area of the two smaller circles from the area of the large circle. 25π-13π=12π (E)Problem 6 The following problem is from both the 2002 AMC 12B #3 and 2002 AMC 10B #6, so both problems redirect to this page.Factoring, we get )1)(2(232--=+-n n n n . Either n-1 or n-2 is odd, and the other is even. Their product must yield an even number. The only prime that is even is 2, which is when is 3. The answer is one. (B) . Problem 7 The following problem is from both the 2002 AMC 12B #4 and 2002 AMC 10B #7, so both problems redirect to this page.Since 4241713121=++, 211424114241)14241(lim 0<+<+<+<∞→n n n From which it follows that 114241=+nand n=42. Thus the answer is Problem 8 The following problem is from both the 2002 AMC 12B #8 and 2002 AMC10B #8, so both problems redirect to this page.If there are five Mondays, there are only three possibilities for their dates: (1,8,15,22,29),(2,9,16,23,30) , and (3,10,17,24,31) .In the first case August starts on a Thursday, and there are five Thursdays, Fridays, and Saturdays in August.In the second case August starts on a Wednesday, and there are five Wednesdays, Thursdays, and Fridays in August.In the third case August starts on a Tuesday, and there are five Tuesdays, Wednesdays,and Thursdays in August. The only day of the week that is guaranteed to appear five times is therefore Thursday. D Problem 9 There are "words" beginning with each of the first four letters alphabetically. Fromthere, there are with as the first letter and each of the first three lettersalphabetically. After that, the next "word" is USAMO , hence our answeris . Problem 10 The following problem is from both the 2002 AMC 12B #6 and 2002 AMC 10B #10, so both problems redirect to this page.Solution 1Since 0)())((22=++=++-=--b ax x ab x b a x b x a x , it follows by comparing coefficients that –a-b=a and that ab=bSince b is nonzero, a=1, and -1-b=1, so b=-2 . Thus (a,b)=(1,-2) CSolution 2Another method is to use Vieta's formulas . The sum of the solutions to this polynomial is equal to the opposite of the coefficient, since the leading coefficient is 1; in other words, a+b=-a and the product of the solutions is equal to the constant term (i.e, a*b=b ). Since is nonzero, it follows that and therefore (from the first equation), b=-2a=-2 Problem 11 The following problem is from both the 2002 AMC 12B #7 and 2002 AMC 10B #11, so both problems redirect to this page.Let the three consecutive positive integers be a-1,a,a+1. so , a(a-1)(a+1)=24a , so (a-1)(a+1)=24, 24=4*6 , so a=5. Hence, the sum of the squares is 77654222=++ B. Problem 12The domain over which we solve the equation is .We can now cross-multiply to get rid of the fractions, we get (x-1)(x-6)=(x-k)(x-2)Simplifying that, we get 7x-6=(k+2)x-2k . Clearly for k=5 we get the equation -6=-10 which is never true. The answer is EFor other , one can solve for : k k x 26)5(-=-, hence kk x --=526 . We can easily verify that for none of the other four possible values of is this equal to or , hence there is a solution for in each of the other cases. Problem 13We have 8xy-12y+2x-3=4y(2x-3)+(2x-3)=(4y+1)(2x-3), as (4y+1)(2x-3)=0must be true for all , we must have , hence x=3/2 D .Problem 14Since, 756425364256425)2(585•=•=•=N .Combing the 's and 's gives us,This is 2048 with sixty-four, 's on the end. So, the sum of the digits of is 2+4+8=14 B Problem 15 The following problem is from both the 2002 AMC 12B #11 and 2002 AMC 10B #15, so both problems redirect to this page.Since A-B and A+B must have the same parity , and since there is only one even prime number, it follows that A-B and A+B are both odd. Thus one of A,B is odd and the other even. Since A+B>A>A-B>2, it follows that A (as a prime greater than 2) is odd. Thus B=2, and A-2, A, A+2 are consecutive odd primes. At least one of A-2, A, A+2 is divisible by 3, from which it follows that A-2=3 and A=5. The sum of these numbers is thus 17, a prime, so the answer is prime. E .Problem 16 The following problem is from both the 2002 AMC 12B #12 and 2002 AMC 10B #16, so both problems redirect to this page.Solution 1Let n n x -=202 , with (note that the solutions do not give any additionalsolutions for ). Then rewriting,12022+=x x n . Since , it follows thatdivides . Listing the factors of , we find that x=0,1,2,3 are the only 4 D solutions (respectively yielding n=0,10,16,18 ).Solution 2Forand the fraction is negative, for it is not defined, and forit is between 0 and 1. Thus we only need to examine and . For and we obviously get the squares and respectively.For prime the fraction will not be an integer, as the denominator will not contain the prime in the numerator.This leaves, and a quick substitution shows that out of these onlyn=16 and n=18 yield a square. Problem 17The area of the triangle ADG can be computed as 2AP DG • . We will now find DG and AP . Clearly, PFG is a right isosceles triangle with hypotenuse of length 2, hence PG=2. The same holds for triangle QED and its leg QD . The length of PQ is equal toFE=2 . Hence GD=2+22, and AP=PD=2+2. Then the area of ADG equals 2342)22)(222(2+=++=•AP DG C Problem 18 The following problem is from both the 2002 AMC 12B #14 and 2002 AMC 10B #18, so both problems redirect to this page.For any given pair of circles, they can intersect at most 2 times. Since there are pairs of circles, the maximum number of possible intersections is 6*2=12 . We can construct such a situation as below, so the answer is D .Problem 19Solution 1We should realize that the two equations are 100 terms apart, so by subtracting the two equations in a form likeWe can find the value of the common difference every hundred terms. But we forgot that it happens hundred times. So we have a divide the answer by hundred 100/100=1The answer yields us the common difference of every hundred terms. So you has to simply divide the answer by hundred again to find the common difference between one term 1/100=0.01 CSolution 2Adding the two given equations together givesNow, let the common difference be . Notice that, so we merely need to find to get the answer. The formula for an arithmetic sum is ))1(2(21-+n d a n where is the first term, is the number of terms, and is the common difference. Now we use this formula to find a closed form for the first given equation and the sum of the given equations. For the first equation, we have n=100 . Therefore, we have 50(2a 1+99d)=100, or *(1). For the sum of the equations (shown at the beginning of the solution) we have n=200, so 100(2a 1+199d)=300 or *(2) Now we have a system of equations in terms of and . Subtracting (1) from (2) eliminates, yielding 100d=1, and 100112=-=a a d C Solution 3Subtracting the 2 given equations yieldsNow express each a_n in terms of first term a_1 and common difference x between consecutive terms 100))99()199((...))2()102(())()101(())()100((11111111=+-++++-+++-++-+x a x a x a x a x a x a a x a Simplifying and canceling a_1 and x terms gives100x=1Problem 20a+8c=7b+4 and 8a-c=7-4b Squaring both,andare obtained. Adding the two equations and dividing by gives , soProblem 21 The following problem is from both the 2002 AMC 12B #17 and 2002 AMC 10B #21, so both problems redirect to this page.We say Andy's lawn has an area of . Beth's lawn thus has an area of2x , and Carlos's lawn has an area of 3x . We say Andy's lawn mower cuts at a speed of . Carlos's cuts at a speed of 3y , and Beth's cuts at a speed 32y . Each person's lawn is cut at a speed of rate area , so Andy's is cut in yx time, as is Carlos's. Beth's is cut in yx ⨯43, so the first one to finish is Beth. B . Problem 22 The following problem is from both the 2002 AMC 12B #20 and 2002 AMC 10B #22, so both problems redirect to this page.Let OM=x , ON=y . By the Pythagorean Theorem on △XON, MOYrespectively, ,19)2(222=+y x 22222)2(=+y xSumming these gives 8455522=+y x , so 16922=+y x . By the Pythagorean Theorem again, we have222)2()2(XY y x =+ so 26)169(4)(422==+=y x XY BProblem 23Solution 1First of all, write and in terms of a 2.can be represented by in different ways.Since both are equal toyou can set them equal to each other. 15212322+=+a a 32=a Substitute the value of back intoand substitute that intoSolution 2 Substituting into : . Since , . Therefore,and so on until. Adding the Left Hand Sides of all of these equations gives; adding the Right Hand Sides of these equations gives These two expressions must be equal; henceand and Substituting :Thus we have a general formula forand substituting m=12:Problem 24We can let this circle represent the ferris wheel with center and represent the desired point feet above the bottom. Draw a diagram like the one above. We find out is a triangle. That means and the ferris wheel has made of a revolution. Therefore, the time it takes to travel that much of a distance is of a minute, or seconds. The answer is . Alternatively, we could also say that is congruent to by SAS, so is 20, and is equilateral, andProblem 25 Let be the sum of the integers and be the number of elements in the list. Then2002 AMC10Bwe get the equations and . With a little work, the solution is found to be .2002 AMC 10B Answer Key1.E2. C3. A4. D5.E6. B7. E8. D9. D 10.C11.B 12. E 13. D 14. B 15. E 16.D 17.C 18.D 19.C 20.B21.B 22. B 23. D 24. D 25.A。

AMC 美国数学竞赛 2003 AMC 10A 试题及答案解析培训资料

2003 AMC 10A1、What is the difference between the sum of the first even counting numbers and the sum of the first odd counting numbers?SolutionThe first even counting numbers are .The first odd counting numbers are .Thus, the problem is asking for the value of..Alternatively, using the sum of an arithmetic progression formula, we can write .2、Members of the Rockham Soccer League buy socks and T-shirts. Socks cost per pair and each T-shirt costs more than a pair of socks. Each member needs one pair of socks and a shirt for home games and another pair of socks and a shirt for away games. If the total cost is , how many members are in the League?SolutionSince T-shirts cost dollars more than a pair of socks, T-shirts cost dollars.Since each member needs pairs of socks and T-shirts, the total cost for member is dollars.Since dollars was the cost for the club, and was the cost per member, the number of members in the League is .3、A solid box is cm by cm by cm. A new solid is formed by removing a cube cm on a side from each corner of this box. What percent of the original volume is removed?SolutionThe volume of the original box isThe volume of each cube that is removed isSince there are corners on the box, cubes are removed.So the total volume removed is .Therefore, the desired percentage is .4、It takes Mary minutes to walk uphill km from her home to school, but it takes her only minutes to walk from school to home along the same route. What is her average speed, in km/hr, for the round trip?SolutionSince she walked km to school and km back home, her total distance is km.Since she spent minutes walking to school and minutes walking back home, her total time is minutes = hours.Therefore her average speed in km/hr is5、Let and denote the solutions of . What is the value of ?SolutionUsing factoring:orSo and are and .Therefore the answer isOR we can use sum and product.6、Define to be for all real numbers and . Which of the following statements is not true?SolutionExamining statement C:when , but statement C says that it does for all . Therefore the statement that is not true is "for all " Alternatively, consider that the given "heart function" is actually the definition of the distance between two points. Examining all of the statements, only C is not necessarily true; if c is negative, the distance between and is the absolute value of , not itself, because distance is always nonnegative.7、How many non-congruent triangles with perimeter have integer side lengths?SolutionBy the triangle inequality, no one side may have a length greater than half the perimeter, which isSince all sides must be integers, the largest possible length of a side isTherefore, all such triangles must have all sides of length , , or . Since , at least one side must have a length of Thus, the remaining two sides have a combined length of . So, the remaining sides must be either and or and . Therefore, the number of triangles is .8、What is the probability that a randomly drawn positive factor ofis lessthan ?SolutionFor a positive number which is not a perfect square, exactly half of the positive factors will be less than .Since is not a perfect square, half of the positive factors of will be less than .Clearly, there are no positive factors of between and . Therefore half of the positive factors will be less than .So the answer is .9、SimplifySolution.Therefore:10、The polygon enclosed by the solid lines in the ﬁgure consists of congruent squares joined edge-to-edge. One more congruent square is attached to an edge at one of the nine positions indicated. How many of the nine resulting polygons can be folded to form a cube with one face missing?SolutionLet the squares be labeled , , , and .When the polygon is folded, the "right" edge of square becomes adjacent to the "bottom edge" of square , and the "bottom" edge of square becomes adjacent to the "bottom" edge of square .So, any "new" square that is attached to those edges will prevent the polygon from becoming a cube with one face missing.Therefore, squares , , and will prevent the polygon from becoming a cube with one face missing.Squares , , , , , and will allow the polygon to become a cube with one face missing when folded.Thus the answer is .Another way to think of it is that a cube missing one face has of it's faces. Since the shape has faces already, we need another face. The only way to add anopther face is if the added square does not overlap any of the others. ,, and overlap, while 9 do not. The answer is11、The sum of the two -digit numbers and is . What is ?SolutionSince , , and are digits, , , .Therefore, .12、A point is randomly picked from inside the rectangle withvertices , , , and . What is the probability that ?SolutionThe rectangle has a width of and a height of .The area of this rectangle is .The line intersects the rectangle at and .The area which is the right isosceles triangle with side length that has vertices at , , and .The area of this triangle isTherefore, the probability that is13、The sum of three numbers is . The ﬁrst is times the sum of the other two. The second is seven times the third. What is the product of all three?SolutionSolution 1Let the numbers be , , and in that order. The given tells us thatTherefore, the product of all three numbers is.Solution 2Alternatively, we can set up the system in matrix form:Or, in matrix formTo solve this matrix equation, we can rearrange it thus:Solving this matrix equation by using inverse matrices and matrix multiplication yieldsWhich means that , , and . Therefore,14、Let be the largest integer that is the product of exactly distinct prime numbers, , , and , where and are single digits. What is the sum of the digits of ?SolutionSince is a single digit prime number, the set of possible values of is .Since is a single digit prime number and is the units digit of the prime number , the set of possible values of is .Using these values for and , the set of possible values of isOut of this set, the prime values areTherefore the possible values of are:The largest possible value of is .So, the sum of the digits of is15、What is the probability that an integer in the set is divisible by and not divisible by ?SolutionThere are integers in the set.Since every 2nd integer is divisible by , there are integers divisible by in the set.To be divisible by both and , a number must be divisible by .Since every 6th integer is divisible by , there are integers divisible by both and in the set.So there are integers in this set that are divisible by and not divisible by .Therefore, the desired probability is16、What is the units digit of ?SolutionSince :Therefore, the units digit is17、The number of inches in the perimeter of an equilateral triangle equals the number of square inches in the area of its circumscribed circle. What is the radius, in inches, of the circle?SolutionLet be the length of a side of the equilateral triangle and let be the radius of the circle.In a circle with a radius the side of an inscribed equilateral triangle is .So .The perimeter of the triangle isThe area of the circle isSo:18、What is the sum of the reciprocals of the roots of the equationSolutionMultiplying both sides by :Let the roots be and .The problem is asking forBy Vieta's formulas:So the answer is .19、A semicircle of diameter sits at the top of a semicircle of diameter , as shown. The shaded area inside the smaller semicircle and outside the larger semicircle is called a lune. Determine the area of this lune.SolutionThe shaded area is equal to the area of the smaller semicircle minus the area of a sector of the larger circle plus the area of a triangle formed by two radii of the larger semicircle and the diameter of the smaller semicircle.The area of the smaller semicircle is .Since the radius of the larger semicircle is equal to the diameter of the smaller semicircle, the triangle is an equilateral triangle and the sector measures .The area of the sector of the larger semicircle is . The area of the triangle isSo the shaded area is20、A base-three-digit number is selected at random. Which of the following is closest to the probability that the base-representation and the base-representation of are both three-digit numerals?SolutionTo be a three digit number in base-10:Thus there are three-digit numbers in base-10To be a three-digit number in base-9:To be a three-digit number in base-11:So,Thus, there are base-10 three-digit numbers that are three digit numbers in base-9 and base-11.Therefore the desired probability is .21、Pat is to select six cookies from a tray containing only chocolate chip, oatmeal, and peanut butter cookies. There are at least six of each of these three kinds of cookies on the tray. How many different assortments of six cookies can be selected?SolutionSolution 1Let the ordered triplet represent the assortment of chocolate chip cookies, oatmeal cookies, and peanut butter cookies.Using casework:Pat selects chocolate chip cookies:Pat needs to select more cookies that are either oatmeal or peanut butter.The assortments are:assortments.Pat selects chocolate chip cookie:Pat needs to select more cookies that are either oatmeal or peanut butter.The assortments are:assortments.Pat selects chocolate chip cookies:Pat needs to select more cookies that are either oatmeal or peanut butter.The assortments are:assortments.Pat selects chocolate chip cookies:Pat needs to select more cookies that are either oatmeal or peanut butter.The assortments are:assortments.Pat selects chocolate chip cookies:Pat needs to select more cookies that are either oatmeal or peanut butter.The assortments are: assortments.Pat selects chocolate chip cookies:Pat needs to select more cookies that are either oatmeal or peanut butter.The assortments are: assortments.Pat selects chocolate chip cookies:Pat needs to select more cookies that are either oatmeal or peanut butter.The only assortment is: assortment.The total number of assortments of cookies that can be collected isSolution 2It is given that it is possible to select at least 6 of each. Therefore, we can make a bijection to the number of ways to divide the six choices into three categories, since it is assumed that their order is unimportant. Using the ball and urns formula, the number of ways to do this is22、In rectangle , we have , , is on with, is on with , line intersects line at , and is on line with . Find the length .SolutionS olution 1(Opposite angles are equal).(Both are 90 degrees).(Alt. Interior Angles are congruent).Therefore and are similar. and are also similar.is 9, therefore must equal 5. Similarly, must equal 3. Because and are similar, the ratio of and , must also hold true for and . , so is of . By Pythagorean theorem, ..So ..Therefore .Solution 2Since is a rectangle, .Since is a rectangle and ,.Since is a rectangle, .So, is a transversal, and .This is sufficient to prove that and .Using ratios:Since can't have 2 different lengths, both expressions for must be equal.Solution 3Since is a rectangle, , , and . From the Pythagorean Theorem, .LemmaStatement:Proof: , obviously.Since two angles of the triangles are equal, the third angles must equal each other. Therefore, the triangles are similar.Let .Also, , thereforeWe can multiply both sides by to get that is twice of 10, orSolution 4We extend BC such that it intersects GF at X. Since ABCD is a rectangle, it follows that CD=8, therefore, XF=8. Let GX=y. From the similarity of triangles GCH and GEA, we have the ratio 3:5 (as CH=9-6=3, and EA=9-4=5). GX and GF are the altitudes of GCH andGEA, respectively. Thus, y:y+8 = 3:5, from which we have y=12, thus GF=y+8=12+8=20. B.23、A large equilateral triangle is constructed by using toothpicks to create rows of small equilateral tr iangles. For example, in the ﬁgure we have rows of small congruent equilateral triangles, with small triangles in the base row. How many toothpicks would be needed to construct a large equilateral triangle if the base row of the triangle consists of small equilateral triangles?SolutionSolution 1There are small equilateral triangles.Each small equilateral triangle needs toothpicks to make it.But, each toothpick that isn't one of the toothpicks on the outside of the large equilateral triangle is a side for small equilateral triangles.So, the number of toothpicks on the inside of the large equilateral triangle isTherefore the total number of toothpicks isSolution 2We see that the bottom row of small triangles is formed from downward-facing triangles and upward-facing triangles. Since each downward-facing triangle uses three distinct toothpicks, andsince the total number of downward-facing triangles is, we have that the total number of toothpicks is24、Sally has ﬁve red cards numbered through and four blue cards numbered through . She stacks the cards so that the colors alternate and so that the number on each red card divides evenly into the number on each neighboring blue card. What is the sum of the numbers on the middle three cards?SolutionLet and designate the red card numbered and the blue card numbered , respectively.is the only blue card that evenly divides, so must be at one end of the stack and must be the card next to it.is the only other red card that evenly divides , so must be the other card next to .is the only blue card that evenly divides, so must be at the other end of the stack and must be the card next to it.is the only other red card that evenly divides , so must be the other card next to .doesn't evenly divide , so must be next to , must be next to , and must be in the middle.This yields the following arrangement from top to bottom:Therefore, the sum of the numbers on the middle three cards is.25、Let be a -digit number, and let and be the quotient and remainder, respectively, when is divided by . For how many values of is divisible by ?SolutionSolution 1When a -digit number is divided by , the first digits become the quotient, , and the last digits become the remainder, .Therefore, can be any integer from to inclusive, and can be any integer from to inclusive.For each of the possible values of , there are at least possible values of such that .Since there is "extra" possible value of that is congruent to , each of the values of that are congruent tohave more possible value of such that .Therefore, the number of possible values of such that is .Solution 2Let equal , where through are digits. Therefore,We now take :The divisor trick for 11 is as follows:"Let be an digit integer. Ifis divisible by , then is also divisible by ."Therefore, the five digit number is divisible by 11. The 5-digit multiples of 11 range from to . There aredivisors of 11 between those inclusive.学习资料Solution 3Since q is a quotient and r is a remainder when n is divided by 100. So we have . Since we are counting choices where q+r is divisible by 11, we have for some k. This means that n is the sum of two multiples of 11 and would thus itself be a divisor of 11. Then we can count all the four digit divisors of 11 as in Solution 2. (This solution is essentially the same as solution 2, but it does not necessarily involve mods and so could potentially be faster.)NotesThe part labeled "divisor trick" actually follows from the same observation we made in the previous step: , therefore and for all . For a digit numberwe get, as claimed.Also note that in the "divisor trick" we actually want to assign the signs backwards - if we make sure that the last sign is a , the result will have the same remainder modulo as the original number.精品文档。

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.)。

AMC10美国数学竞赛讲义全

AMC10美国数学竞赛讲义全AMC 中的数论问题1：Remember the prime between 1 to 100：2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 91 2：Perfect number:Let is the prime number.if21p - is also the prime number. then1(21)2p p --is the perfect number. For example:6,28,496.3: Let ,0n abc a =≠ is three digital integer .if 333n a b c =++Then the number n is called Daffodils number . There are only four numbers: 153 370 371 407 Let ,0n abcd a =≠ is four digital integer .if 4444d n a b c +=++Then the number n is called Roses number . There are only three numbers: 1634 8208 94744：The Fundamental Theorem of ArithmeticEvery natural number can be written as a product of primes uniquely up to order.5：Suppose that a and b are integers with b =0. Then there exists unique integers q a d such that 0 ≤ < |b| a d a = bq + .6：(1)G eatest Commo D v so : Let gcd (a, b) = max {d ∈ Z: d | a a d d | b}. For any integers a and b, we havegcd(a, b) = gcd(b, a) = gcd(±a, ±b) = gcd(a, b ? a) = gcd(a, b + a). For example: gcd(150, 60) = gcd(60, 30) = gcd(30, 0) = 30 (2)Least commo mult le:Let lcm(a,b)=m {d∈Z: a | d a d b | d }. (3)We have that: ab= gcd(a, b) lcm(a,b) 7：Congruence modulo nIf ,0a b mq m -=≠,then we call a congruence b modulo m and we rewritemod a b m ≡.(1)Assume a b c d m Z 0 m 0 .If a b mod m c d mod m then we havemod a c b d m ±≡± , mod ac bd m ≡ , mod k k a b m ≡(2) The equat o ax ≡ b (mod m) has a solut o f a d o ly f gcd(a, m) d v des b.8：How to find the unit digit of some special integers (1)How many zero at the end of !nFor example, when 100n =, Let N be the number zero at the end of 100!then10010010020424525125N=++=+=(2) ,,a n Z ∈Find the unit digit n a . For example, when 100,3n a ==9：Palindrome, such as 83438, is a number that remains the same when its digits are reversed.There are some number not only palindrome but 2 2 ，222 ， (1)Some special palindrome n that 2n is also palindrome. For example :222221111121111123211111123432111111111112345678987654321=====(2)How to create a palindrome? Almost integer plus the number of its reverseddigits and repeat it again and again. Then we get a palindrome. For example: 87781651655617267266271353135335314884+=+=+=+=But whether any integer has this Property has yet to prove(3) The palindrome equation means that equation from left to right and right to left it all set up. For example ：1242242112231132211121241388888831421211====Let ab and cde are two digital and three digital integers. If the digits satisfy the,,9a c b e d c e d ?=?=+≤, then ab cde edc ba ?=?.10: Features of an integer divisible by some prime number If n is even ，then 2|n⼀个整数n 的所有位数上的数字之和是3（或者9）的倍数，则n 被3（或者9）整除⼀个整数n 的尾数是零，则n 被5整除⼀个整数n 的后三位与截取后三位的数值的差被7、11、13整除，则n 被7、11、13整除⼀个整数n 的最后两位数被4整除，则n 被4整除⼀个整数n 的最后三位数被8整除，则n 被8整除⼀个整数n 的奇数位之和与偶数位之和的差被11整除，则n 被11整除 11. The number Theoretic functions If 312123t rr rrt n p p p p =(1) {}12()#0:|(1)(1)(1)t n a a n r r r χ=>=+++(2) 12222111222|()(1)(1)(1)t r r r t t t a nn a p pp p p p p p p δ==+++++++++∑(3) {}11221111122()#:,gcd(,)1()()()t t r r rr rr t t n a N a n a n p p p p p p φ---=∈≤==---For example: 2(12)(23)(21)(11)6χχ=?=++= 22(12)(23)(122)(13)28δδ=?=+++= 22(12)(23)(22)(31)4φφ=?=--= Exercise1. The sums of three whole numbers taken in pairs are 12, 17, and 19. What is the middle number?(A) 4 (B) 5 (C) 6 (D) 7(E) 83. For the positive integer n, let denote the sum of all the positive divisors of n with the exception of n itself. For example,<4>=1+2=3 and <12>=1+2+3+4+6=16. What is <<<6>>>?(A) 6 (B) 12 (C) 24 (D) 32 (E) 36 8. What is the sum of all integer solutions to 21<(x-2)<25? (A) 10 (B) 12(C) 15(D) 19(E) 5(A) 6 (B) 7 (C) 8 (D) 9 (E) 10(A) 1(B) 2(C) 3(D) 4(E) 515．The figures 123,,F F F and 4F shown are the first in a sequence of figures. For3n ≥, n F is constructed from -1n F by surrounding it with a square and placing one more diamond on each side of the new square than -1n F had on each side of its outside square. For example, figure 3F has 13 diamonds. How many diamonds are there in figure 20F ?18. Positive integers a, b, and c are randomly and independently selected with replacement from the set {1, 2, 3,…, 2010}. What is the probability that abc ab a ++ is divisible by 3? (A)13(B)2981(C)3181(D)1127(E)132724. Let ,a b and c be positive integers with >>a b c such that 222-b -c +=2011a ab and222+3b +3c -3-2-2=-1997a ab ac bc . What is a ?(A) 249 (B) 250 (C) 251 (D) 252 (E)2535. In multiplying two positive integers a and b, Ron reversed the digits of the two-digit number a. His erroneous product was 161. What is the correct value of the product of a and b?(A) 116 (B) 161 (C) 204 (D) 214 (E) 224 23. What is the hundreds digit of 20112011?(A) 1 (B) 4 (C) 5 (D) 6 (E) 99. A palindrome, such as 83438, is a number that remains the same when its digits are reversed. The numbers x and x+32 are three-digit and four-digit palindromes, respectively. What is the sum of the digits of x?(A) 20 (B) 21 (C) 22 (D) 23 (E) 2421. The polynomial 322010x ax bx -+- has three positive integer zeros. What is the smallest possible value of a?(A) 78 (B) 88 (C) 98 (D) 108 (E) 11824. The number obtained from the last two nonzero digits of 90! Is equal to n. What is n?(A) 12 (B) 32 (C) 48 (D) 52 (E) 6825. Jim starts with a positive integer n and creates a sequence of numbers. Each successive number is obtained by subtracting the largest possible integer square less than or equal to the current number until zero is reached. For example, if Jim starts with n=55, then his sequence contain 5 numbers:55 55-72= 6 6-22= 2 2-12= 1 1-12= 0Let N be the smallest umbe fo wh ch J m’s seque ce has umbe s. What is the units digit of N?(A) 1 (B) 3 (C) 5 (D) 7 (E) 9 21．What is the remainder when 01220093+3+3++3is divided by 8?(A) 0 (B) 1 (C) 2 (D) 4 (E) 65．What is the sum of the digits of the square of 111,111,111? (A) 18 (B) 27(C) 45(D) 63(E) 81ABC 100643a S 2(A) 6 (B) 7(C) 8(D) 9(E) 1024. Let 2200820082k =+. What is the units digit of 222k +? (A) 0 (B) 1(C) 4(D) 6(E) 8AMC about algebraic problems⼀、Linear relations(1) Slope y-intercept form: y kx b =+ (k is the slope, b is the y-intercept) (2)Standard form: 0Ax By C ++= (3)Slope and one point 0000(,),()()P x y k slope y y k x x -=- (4) Two points 1122(,),(,)P x y P x y12121212y y y y y y x x x x x x ---==--- (5)x,y-intercept form: (,0),(0,),(0,0)1x yP a Q b a b a b≠≠+= ⼆、the relations of the two lines 111222:0,:0l A x B y C l A x B y C ++=++=(1) 1l ∥2l 122112210,0A B A B C B C B ?-=-≠ (1) 1l ⊥2l 12120A A B B ?-=三、Special multiplication rules:222223322332212211222112222()()()2()()()()()()(2)()((1)(1))(1)()n n n n n n n n n n n n n n a b a b a b a b a ab b a b a b a ab b a b a b a ab b a b a b a a b ab b n a b a b a a b ab b n is odd n a b c ab bc ac a b -----------=-+±=±+-=-+++=+-+-=-++++≥+=+-++-+->++=++?-22()()0b c c a a b c+-+-=?==四、quadratic equations and PolynomialThe quadratic equations 2(0)y ax bx c a =++≠ has two roots 12,x x then we has1212b c x x x x a a+=-=More generally, if the polynomial 121210nn n n n x a x a x a x a ---+++++= has nroots 123,,,,n x x x x ,then we have:1231122312123(1)n n n n n nx x x x a x x x x x x a x x x x a -++++=-++==-开⽅的开⽅、估计开⽅数的⼤⼩绝对值⽅程Arithmetic Sequence123(1)(2)(3)()n m a a n d a n d a n d a n m d =+-=+-=+-==+-121321()()()()2222n n n m n m n n a a n a a n a a n a a s ---+++++=====1(1)2n n n ds na -=+If n=2k, then we have 1()n k k s k a a +=+ If n=2k+1, then we have 1n k s na += Geometric sequence123123n n n n m n m a a q a q a q a q ----=====1(1)1,1n n a q q s q-≠=-Some special sequence , , 2, 3, 5, ,… 9,99,999,9999，… 1，11，111，1111，… Exercise4 .When Ringo places his marbles into bags with 6 marbles per bag, he has 4 marbles left over. When Paul does the same with his marbles, he has 3 marbles left over. Ringo and Paul pool their marbles and place them into as many bags as possible, with 6 marbles per bag. How many marbles will be left over?7 For a science project, Sammy observed a chipmunk and a squirrel stashing acorns in holes. The chipmunk hid 3 acorns in each of the holes it dug. The squirrel hid 4 acorns in each of the holes it dug. They each hid the same number of acorns, although the squirrel needed 4 fewer holes. How many acorns did the chipmunk hide?21. Four distinct points are arranged on a plane so that the segments connecting them have lengths ,,,,, and . What is the ratio of to ?13. An iterative average of the numbers 1, 2, 3, 4, and 5 is computed the following way. Arrange the five numbers in some order. Find the mean of the first two numbers, and then find the mean of that with the third number, then the mean of that with the fourth number, and finally the mean of that with the fifth number. What is the difference between the largest and smallest possible values that can be obtained using this procedure?16. Three runners start running simultaneously from the same point on a 500-meter circular track. They each run clockwise around the course maintaining constant speeds of 4.4, 4.8, and 5.0 meters per second. The runners stop once they are all together again somewhere on the circular course. How many seconds do the runners run?24. Let ,a b and c be positive integers with >>a b c such that 222-b -c +=2011a ab and222+3b +3c -3-2-2=-1997a ab ac bc . What is a ?(A) 249 (B) 250(C) 251(D) 252(E) 2531. What is246135135246++++-++++? (A) -1 (B) 536(C) 712(D)14760(E)43310. Consider the set of numbers {1, 10, 102, 103(010)}. The ratio of the largest element of the set to the sum of the other ten elements of the set is closest to which integer?(A) 1 (B) 9 (C) 10 (D) 11 (E) 101=(A) -64 (B) -24 (C) -9 (D) 24 (E) 5764. Let X and Y be the following sums of arithmetic sequences: X= 10 + 12 + + …+ 00. Y= 2 + + + …+ 02. What is the value of Y X -?(A) 92 (B) 98 (C) 100 (D) 102 (E) 1127. Which of the following equations does NOT have a solution?(A) 2(7)0x +=(B) -350x += 20=80= (E) -340x -=(A)(B)(C)2(D) (E) 613. What is the sum of all the solutions of 2602x x x =--?(A) 32 (B) 60 (C) 92 (D) 120 (E) 12414. The average of the numbers 1, 2, 3… 9 , 99, and x is 100x. What is x?(A)49101(B)50101(C)12 (D)51101(E)509911. The length of the interval of solutions of the inequality 23a x b ≤+≤ is 10.What is b-a?(A) 6 (B) 10 (C) 15 (D) 20 (E) 3013. Angelina drove at an average rate of 80 kph and then stopped 20 minutes for gas. After the stop, she drove at an average rate of 100 kph. Altogether she drove 250 km in a total trip time of 3 hours including the stop. Which equation could be used to solve for the time t in hours that she drove before her stop? (A) 880100()2503t t +-=(B) 80250t = (C) 100250t =(D) 90250t =(E) 880()1002503t t -+=21. The polynomial 32-2010x ax bx +- has three positive integer zeros. What is the smallest possible value of a?(A) 78 (B) 88 (C) 98 (D) 108 (E) 118 15．When a bucket is two-thirds full of water, the bucket and water weigh kilograms. When the bucket is one-half full of water the total weight is kilograms. In terms of and , what is the total weight in kilograms when the bucket is full of water?13．Suppose thatand. Which of the following is equal tofor every pair of integers16．Let ,,, and be real numbers with,, and. What is the sum of all possible values of5. Which of the following is equal to the product?81216442008............481242004n n +? (A) 251(B) 502(C) 1004(D) 2008 (E) 40167. The fraction 20082200622007220052(3)(3)(3)(3)-- simplifies to which of the following? (A) 1 (B) 9/4 (C) 3 (D) 9/2 (E) 913. Doug can paint a room in 5 hours. Dave can paint the same room in 7 hours. Doug and Dave paint the room together and take a one-hour break for lunch. Let t be the total time, in hours, required for them to complete the job working together, including lunch. Which of the following equations is satisfied by t ?(A) 11()(1)157t ++=(B) 11()1157t ++= (C) 11()157t +=(D) 11()(1)157t +-=(E) (57)1t +=15. Yesterday Han drove 1 hour longer than Ian at an average speed 5 miles per hour faster than Ian. Jan drove 2 hours longer than Ian at an average speed 10 miles per hour faster than Ian. Han drove 70 miles more than Ian. How many more miles did Jan drive than Ian?(A) 120 (B) 130 (C) 140 (D) 150 (E) 160AMC 中的⼏何问题⼀、三⾓形有关知识点1.三⾓形的简单性质与⼏个⾯积公式①三⾓形任何两边之和⼤于第三边；②三⾓形任何两边之差⼩于第三边；③三⾓形三个内⾓的和等于 0°；④三⾓形三个外⾓的和等于3 0°；⑤三⾓形⼀个外⾓等于和它不相邻的两个内⾓的和；⑥三⾓形⼀个外⾓⼤于任何⼀个和它不相邻的内⾓。

2002AMC10-B

15. The positive integers A, B , A − B , and A + B are all prime numbers. The sum of these four primes is (A) even (E) prime (B) divisible by 3 (C) divisible by 5 (D) divisible by 7
16. For how many integers n is (A) 1 (B) 2 (C) 3
n 20−n
the square of an integer? (E) 10
3
2
(A) 3π
(B) 4π
(C) 6π
(D) 9π
(E) 12π
3rd AMC 10 B
2002
3
6. For how many positive integers n is n2 − 3n + 2 a prime number? (A) none (B) one (C) two (D) more than two, but ﬁnitely many (E) inﬁnitely many 7. Let n be a positive integer such that following statements is not true: (A) 2 divides n (E) n > 84 8. Suppose July of year N has ﬁve Mondays. Which of the following must occur ﬁve times in August of year N ? (Note: Both months have 31 days.) (A) Monday (B) Tuesday (C) Wednesday (D) Thursday (E) Friday (B) 3 dividesion

2000年-2002 美国AMC10

PABCD2000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) 671 。

2. 2000(20002000)为。

3. Jenny 每天早上都会吃掉她所剩下的聪明豆的20%，今知在第二天结束时，有32颗剩下，试问一开始聪明豆有颗。

(A) 40 (B) 50 (C) 55 (D) 60 (E) 75 。

4. 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) 8.77 。

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) 9 。

7. 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。

(A) 3+33 (B) 2+334 (C) 2+2(D)2533+ (E) 2+335 。

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）。

amc 10 2023 题目

【主题】解析AMC 10 2023年数学竞赛题目【内容】一、导言2023年AMC 10数学竞赛作为一项备受关注的数学竞赛，吸引了众多中学生的参与。

在本文中，我们将对2023年AMC 10数学竞赛的题目进行一一解析，帮助读者更好地理解和掌握其中的数学知识和解题技巧。

二、数学竞赛概况AMC（American Mathematics Competitions）是由美国数学协会（MAA）主办的一项全国性数学竞赛。

AMC 10是专门面向中学生的数学竞赛，旨在激发学生对数学的兴趣，培养其数学能力和解决问题的能力。

每年举办的AMC 10数学竞赛题目涵盖了数学中的各个领域，要求考生具有扎实的数学基础和灵活的思维能力。

三、2023年AMC 10数学竞赛题目解析1. 第一道题目题目：求下列无理数的整数部分：$\sqrt{129}+\sqrt{189}-\sqrt{144}-\sqrt{121}$解析：我们可以将每个无理数的平方根表示成对应的整数和余数的形式，然后进行合并化简，经过一系列运算，最终得到答案。

2. 第二道题目题目：设函数$f(x)=\frac{ax+b}{cx+d}$，其中$a,b,c,d$是常数，$ad-bc\neq0$，且$c\neq0$。

若对任意实数$x$恒有$f(f(x))=x$，则$f(x)$的表达式是什么？解析：通过设定函数$f(x)$，并根据题目给出的条件，我们可以利用函数的复合运算和方程的求解，来确定$f(x)$的表达式。

3. 第三道题目题目：在平面直角坐标系中，曲线$C$的方程为$x^2+y^2-4x-6y+9=0$，则曲线$C$的图像是什么？解析：通过将方程进行配方完全平方，消去一次项，得到标准方程，然后结合圆的性质，来描述曲线$C$的图像特征。

4. 第四道题目题目：若正实数$a,b,c$满足$a+b+c=1$，则$\frac{1+a}{1-a}+\frac{1+b}{1-b}+\frac{1+c}{1-c}$的值是多少？解析：题目中给出的条件可以让我们有一个开始，然后我们可以通过利用分式的化简和求和规律来得出最终的结果。

AMC 美国数学竞赛 2002 AMC 10A 试题及答案解析

2002 AMC 10A1、The ratio is closest to which of the following numbers?SolutionWe factor as . As , ouranswer is .2、For the nonzero numbers , , , define.Find .Solution. Ouranswer is then .Alternate solution for the lazy: Without computing the answer exactly,we see that , , and . The sumis , and as all the options are integers, the correct one is obviously .3、According to the standard convention for exponentiation,.If the order in which the exponentiations are performed is changed, how many other values are possible?SolutionThe best way to solve this problem is by simple brute force.It is convenient to drop the usual way how exponentiation is denoted,and to write the formula as , where denotes exponentiation. We are now examining all ways to add parentheses to this expression. There are 5 ways to do so:1.2.3.4.5.We can note that . Therefore options 1 and 2 are equal, and options 3 and 4 are equal. Option 1 is the one given in the problem statement. Thus we only need to evaluate options 3 and 5.Thus the only other result is , and our answer is .4、For how many positive integers does there exist at least one positive integer such that ?infinitely manySolutionSolution 1For any we can pick , we get , therefore theanswer is .Solution 2Another solution, slightly similar to this first one would be using Simon's Favorite Factoring Trick.Let , thenThis means that there are infinitely many numbers that can satisfythe inequality. So the answer is .5、Each of the small circles in the figure has radius one. The innermost circle is tangent to the six circles that surround it, and each of those circles is tangent to the large circle and to its small-circle neighbors. Find the area of the shaded region.SolutionThe outer circle has radius , and thus area . The littlecircles have area each; since there are 7, their total area is . Thus,our answer is .6、Cindy was asked by her teacher to subtract from a certain numberand then divide the result by . Instead, she subtracted and thendivided the result by , giving an answer of . What would heranswer have been had she worked the problem correctly?SolutionWe work backwards; the number that Cindy started with is. Now, the correct result is . Ouranswer is .7、If an arc of on circle has the same length as an arc of oncircle , then the ratio of the area of circle to the area of circle isSolutionLet and be the radii of circles A and B, respectively.It is well known that in a circle with radius r, a subtended arc oppositean angle of degrees has length .Using that here, the arc of circle A has length . The arcof circle B has length . We know that they are equal,so , so we multiply through and simplify to get . As all circles are similar to one another, the ratio of the areas is just thesquare of the ratios of the radii, so our answer is .8、Betsy designed a flag using blue triangles, small white squares, anda red center square, as shown. Let be the total area of the bluetriangles, the total area of the white squares, and the area of thered square. Which of the following is correct?SolutionThe blue that's touching the center red square makes up 8 triangles, or 4 squares. Each of the corners is 2 squares and each of the edges is 1, totaling 12 squares. There are 12 white squares, thus we have.9、There are 3 numbers A, B, and C, such that ,and . What is the average of A, B, and C?More than 1SolutionNotice that we don't need to find what A, B, and C actually are, just their average. In other words, if we can find A+B+C, we will be done.Adding up the equations gives soand the average is . Our answer is .10、Compute the sum of all the roots of.SolutionSolution 1We expand to get which isafter combining like terms. Using the quadratic partof Vieta's Formulas, we find the sum of the roots is . Solution 2Combine terms to get, hence the rootsare and , thus our answer is .11、Jamal wants to store computer files on floppy disks, each ofwhich has a capacity of megabytes (MB). Three of his files requireMB of memory each, more require MB each, and theremaining require MB each. No file can be split between floppydisks. What is the minimal number of floppy disks that will hold all the files?SolutionA 0.8 MB file can either be on its own disk, or share it with a 0.4 MB. Clearly it is not worse to pick the second possibility. Thus we will have 3 disks, each with one 0.8 MB file and one 0.4 MB file.We are left with 12 files of 0.7 MB each, and 12 files of 0.4 MB each.Their total size is MB. The total capacity of 9 disks is MB, hence we need at least 10 more disks. And wecan easily verify that 10 disks are indeed enough: six of them will carry two 0.7 MB files each, and four will carry three 0.4 MB files each.Thus our answer is .12、Mr. Earl E. Bird leaves his house for work at exactly 8:00 A.M. every morning. When he averages miles per hour, he arrives at hisworkplace three minutes late. When he averages miles per hour, hearrives three minutes early. At what average speed, in miles per hour, should Mr. Bird drive to arrive at his workplace precisely on time?SolutionSolution 1Let the time he needs to get there in be t and the distance he travelsbe d. From the given equations, we know that and. Setting the two equal, we have andwe find of an hour. Substituting t back in, we find . From, we find that r, and our answer, is .Solution 2Since either time he arrives at is 3 minutes from the desired time, the answer is merely the harmonic mean of 40 and 60. The harmonicmean of a and b is . In this case, a and b are 40 and 60,so our answer is , so .Solution 3A more general form of the argument in Solution 2, with proof:Let be the distance to work, and let be the correct average speed.Then the time needed to get to work is .We know that and . Summing these twoequations, we get: .Substituting and dividing both sides by , we get ,hence .(Note that this approach would work even if the time by which he is late was different from the time by which he is early in the other case - we would simply take a weighed sum in step two, and hence obtaina weighed harmonic mean in step three.)13、Give a triangle with side lengths 15, 20, and 25, find the triangle's smallest height.SolutionSolution 1This is a Pythagorean triple (a 3-4-5 actually) with legs 15 and 20. Thearea is then . Now, consider an altitude drawn to anyside. Since the area remains constant, the altitude and side to which it is drawn are inversely proportional. To get the smallest altitude, it must be drawn to the hypotenuse. Let the length be x; we have, so and x is 12. Our answer is then.Solution 2By Heron's formula, the area is , hence the shortest altitude'slength is .14、Both roots of the quadratic equation are prime numbers. The number of possible values of isSolutionConsider a general quadratic with the coefficient of being and theroots being and . It can be factored as which is just. Thus, the sum of the roots is the negative of the coefficient of and the product is the constant term. (In general, this leads to Vieta's Formulas).We now have that the sum of the two roots is while the product is. Since both roots are primes, one must be , otherwise the sumwould be even. That means the other root is and the product mustbe . Hence, our answer is .15、Using the digits 1, 2, 3, 4, 5, 6, 7, and 9, form 4 two-digit prime numbers, using each digit only once. What is the sum of the 4 prime numbers?SolutionOnly odd numbers can finish a two-digit prime number, and a two-digit number ending in 5 is divisible by 5 and thus composite,hence our answer is .(Note that we did not need to actually construct the primes. If we had to, one way to match the tens and ones digits to form four primes is , , , and .)16、Let . What is?SolutionLet . Since one ofthe sums involves a, b, c, and d, it makes sense to consider 4x. We have. Rearranging, we have , so .Thus, our answer is .17、Sarah pours four ounces of coffee into an eight-ounce cup and fourounces of cream into a second cup of the same size. She then transfers half the coffee from the first cup to the second and, after stirring thoroughly, transfers half the liquid in the second cup back to the first. What fraction of the liquid in the first cup is now cream?SolutionWe will simulate the process in steps.In the beginning, we have:▪ounces of coffee in cup▪ounces of cream in cupIn the first step we pour ounces of coffee from cup to cup ,getting:▪ounces of coffee in cup▪ounces of coffee and ounces of cream in cupIn the second step we pour ounce of coffee and ounces of cream from cup to cup , getting:▪ounces of coffee and ounces of cream in cup▪the rest in cupHence at the end we have ounces of liquid in cup , and outof these ounces is cream. Thus the answer is .18、A cube is formed by gluing together 27 standard cubicaldice. (On a standard die, the sum of the numbers on any pair of opposite faces is 7.) The smallest possible sum of all the numbers showing on the surface of the cube isSolutionIn a 3x3x3 cube, there are 8 cubes with three faces showing, 12 with two faces showing and 6 with one face showing. The smallest sum with three faces showing is 1+2+3=6, with two faces showing is 1+2=3, and with one face showing is 1. Hence, the smallest possiblesum is . Our answer is thus.19、Spot's doghouse has a regular hexagonal base that measures oneyard on each side. He is tethered to a vertex with a two-yard rope.What is the area, in square yards, of the region outside of the doghouse that Spot can reach?SolutionPart of what Spot can reach is of a circle with radius 2, whichgives him . He can also reach two parts of a unit circle, whichcombines to give . The total area is then , which gives .20、Points and lie, in that order, on , dividing it intofive segments, each of length 1. Point is not on line . Point lieson , and point lies on . The line segments andare parallel. Find .SolutionAs is parallel to , angles FHD and FGA are congruent. Also,angle F is clearly congruent to itself. From SSS similarity,; hence . Similarly, . Thus,.21、The mean, median, unique mode, and range of a collection of eight integers are all equal to 8. The largest integer that can be an element of this collection isSolutionAs the unique mode is , there are at least two s.As the range is and one of the numbers is , the largest one can beat most .If the largest one is , then the smallest one is , and thus the meanis strictly larger than , which is a contradiction.If the largest one is , then the smallest one is . This means that wealready know four of the values: , , , . Since the mean of all thenumbers is , their sum must be . Thus the sum of the missing fournumbers is . But if is the smallest number,then the sum of the missing numbers must be at least ,which is again a contradiction.If the largest number is , we can easily find the solution. Hence, our answer is .NoteThe solution for is, in fact, unique. As the median must be , thismeans that both the and the number, when ordered by size,must be s. This gives the partial solution . For themean to be each missing variable must be replaced by the smallestallowed value.22、A sit of tiles numbered 1 through 100 is modified repeatedly by the following operation: remove all tiles numbered with a perfect square, and renumber the remaining tiles consecutively starting with 1. How many times must the operation be performed to reduce the number of tiles in the set to one?SolutionSolution 1The pattern is quite simple to see after listing a couple of terms.Solution 2Given tiles, a step removes tiles, leaving tiles behind. Now,, so in the next step tilesare removed. This gives , another perfect square.Thus each two steps we cycle down a perfect square, and insteps, we are left with tile, hence our answer is.23、Points and lie on a line, in that order, with and. Point is not on the line, and . The perimeterof is twice the perimeter of . Find .SolutionFirst, we draw an altitude to BC from E.Let it intersect at M. As triangle BEC is isosceles, we immediately get MB=MC=6, so the altitude is 8. Now, let . Using the Pythagorean Theorem on triangleEMA, we find . From symmetry,as well. Now, we use the fact that the perimeter of is twice the perimeter of .We have so. Squaring both sides, we havewhich nicely rearranges into. Hence, AB is 9 so our answer is .24、Tina randomly selects two distinct numbers from the setand Sergio randomly selects a number from the set. The probability that Sergio's number is larger than the sum of the two numbers chosen by Tina isSolutionThis is not too bad using casework.Tina gets a sum of 3: This happens in only one way (1,2) and Sergio can choose a number from 4 to 10, inclusive. There are 7 ways that Sergio gets a desirable number here.Tina gets a sum of 4: This once again happens in only one way (1,3). Sergio can choose a number from 5 to 10, so 6 ways here.Tina gets a sum of 5: This can happen in two ways (1,4) and (2,3). Sergio can choose a number from 6 to 10, so 2*5=10 ways here.Tina gets a sum of 6: Two ways here (1,5) and (2,4). Sergio can choose a number from 7 to 10, so 2*4=8 here.Tina gets a sum of 7: Two ways here (2,5) and (3,4). Sergio can choose from 8 to 10, so 2*3=6 ways here.Tina gets a sum of 8: Only one way possible (3,5). Sergio chooses 9 or 10, so 2 ways here.Tina gets a sum of 9: Only one way (4,5). Sergio must choose 10, so 1 way.In all, there are ways. Tina chooses twodistinct numbers in ways while Sergio chooses a number inways, so there are ways in all. Since , ouranswer is .25、In trapezoid with bases and , we have ,, , and . The area of isSolutionSolution 1It shouldn't be hard to use trigonometry to bash this and find the height, but there is a much easier way. Extend and to meet atpoint :Since we have , with the ratio ofproportionality being . Thus So the sides of are , which we recognize to be aright triangle. Therefore (we could simplify some of the calculation using that the ratio of areas is equal to the ratio of the sides squared),Solution 2Draw altitudes from points and :Translate the triangle so that coincides with . We getthe following triangle:The length of in this triangle is equal to the length of the original, minus the length of . Thus .Therefore is a well-known right triangle. Its area is, and therefore its altitude is.Now the area of the original trapezoid is.。

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