2017年全美数学竞赛AMC12A试题
美国数学竞赛amc12
2002 AMC 12A ProblemsProblem 1Compute the sum of all the roots ofProblem 2Cindy 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?Problem 3According to the standard convention for exponentiation,If the order in which the exponentiations are performed is changed, how many other values are possible?Problem 4Find the degree measure of an angle whose complement is 25% of its supplement.Problem 5Each 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.Problem 6For how many positive integers does there exist at least one positive integer n suchthat ?infinitely manyProblem 7A arc of circle A is equal in length to a arc of circle B. What is the ratio of circle A's area and circle B's area?Problem 8Betsy designed a flag using blue triangles, small white squares, and a red center square, as shown.Let be the total area of the blue triangles, the total area of the white squares,and the area of the red square. Which of the following is correct?Jamal wants to save 30 files onto disks, each with 1.44 MB space. 3 of the files take up 0.8 MB, 12 of the files take up 0.7 MB, and the rest take up 0.4 MB. It is not possible to split a file onto 2 different disks. What is the smallest number of disks needed to store all 30 files?Problem 10Sarah places four ounces of coffee into an eight-ounce cup and four ounces of cream into a second cup of the same size. She then pours half the coffee from the first cup to the second and, after stirring thoroughly, pours half the liquid in the second cup back to the first. What fraction of the liquid in the first cup is now cream?Problem 11Mr. Earl E. Bird gets up every day at 8:00 AM to go to work. If he drives at an average speed of 40 miles per hour, he will be late by 3 minutes. If he drives at an average speed of 60 miles per hour, he will be early by 3 minutes. How many miles per hour does Mr. Bird need to drive to get to work exactly on time?Problem 12Both roots of the quadratic equation are prime numbers. The number ofpossible values of isProblem 13Two different positive numbers and each differ from their reciprocals by . Whatis ?For all positive integers , let .Let . Which of the following relations is true?Problem 15The 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 isProblem 16Tina randomly selects two distinct numbers from the set {1, 2, 3, 4, 5}, and Sergio randomly selects a number from the set {1, 2, ..., 10}. What is the probability that Sergio's number is larger than the sum of the two numbers chosen by Tina?Problem 17Several sets of prime numbers, such as use each of the nine nonzero digits exactly once. What is the smallest possible sum such a set of primes could have?Problem 18Let and be circles definedby and respectively. What is the length ofthe shortest line segment that is tangent to at and to at ?The graph of the function is shown below. How many solutions does theequation have?Problem 20Suppose that and are digits, not both nine and not both zero, and the repeatingdecimal is expressed as a fraction in lowest terms. How many different denominators are possible?Problem 21Consider the sequence of numbers: For , the -th term of thesequence is the units digit of the sum of the two previous terms. Let denote the sum of thefirst terms of this sequence. The smallest value of for which is:Problem 22Triangle is a right triangle with as its rightangle, , and . Let be randomly chosen inside ,and extend to meet at . What is the probability that ?Problem 23In triangle , side and the perpendicular bisector of meet in point ,and bisects . If and , what is the area oftriangle ?Problem 1In the year, the United States will host the International Mathematical Olympiad.Let and be distinct positive integers such that the product .What is the largest possible value of the sum ?Problem 2Problem 3Each day, Jenny ate of the jellybeans that were in her jar at the beginning of that day.At the end of the second day, remained. How many jellybeans were in the jar originally?Problem 4The Fibonacci sequence starts with two 1s, and each term afterwards is the sum of its two predecessors. Which one of the ten digits is the last to appear in the units position of a number in the Fibonacci sequence?Problem 5If where thenProblem 6Two different prime numbers between and are chosen. When their sum is subtracted from their product, which of the following numbers could be obtained?Problem 7How many positive integers have the property that is a positive integer?Problem 8Figures, , , and consist of , , , and non-overlapping squares. If thepattern continued, how many non-overlapping squares would there be in figure?Problem 9Mrs. Walter gave an exam in a mathematics class of five students. She entered the scores in random order into a spreadsheet, which recalculated the class average after each score was entered. Mrs. Walter noticed that after each score was entered, the average was always an integer. The scores (listed in ascending order) were 71,76,80,82, and 91. What was the last score Mrs. Walters entered?Problem 10The point is reflected in the -plane, then its image is rotatedby about the -axis to produce , and finally, is translated by 5 units in thepositive-direction to produce . What are the coordinates of ?Problem 11Two non-zero real numbers, and satisfy. Which of the following is apossible value of?Problem 12Let A, M, and C be nonnegative integers such that . What is the maximumvalue of + + + ?Problem 13One morning each member of Angela’s family drank an 8-ounce mixture of coffee with milk. The amounts of coffee and milk varied from cup to cup, but were never zero. Angela drank a quarter of the total amount of milk and a sixth of the total amount of coffee. How many people are in the family?Problem 14When the mean,median, and modeof the listare arranged in increasing order, they form a non-constant arithmetic progression. What is the sum of all possible real values of?Problem 15Let be a function for which . Find the sum of all values of forwhich.Problem 16A checkerboard of rows and columns has a number written in each square, beginning inthe upper left corner, so that the first row is numbered , the secondrow , and so on down the board. If the board is renumbered so that the left column,top to bottom, is , the second column and so on across the board, some squares have the same numbers in both numbering systems. Find the sum of the numbers in these squares (under either system).Problem 17A centered at has radius and contains the point . The segment is tangent tothe circle at and . If point lieson and bisects , thenProblem 18In year , the day of the year is a Tuesday. In year , the day isalso a Tuesday. On what day of the week did th day of year occur?Problem 19triangle , , , . Let denote the midpointof and let denote the intersection of with the bisector of angle .Which of the following is closest to the area of the triangle ?Problem 20If and are positive numbers satisfyingThen what is the value of latex ?Problem 21Through a point on the hypotenuse of right triangle, lines are drawn parallel to the legs of the triangle so that the triangle is divided into asquare and two smaller right triangles. The area of one of the two small right triangles times the area of the square. The ratio of the area of the other small right triangle to the area of the square isProblem 22The graph below shows a portion of the curve defined by the quarticpolynomial. Which of the following is the smallest?Problem 23Professor Gamble buys a lottery ticket, which requires that he pick six different integersfrom through , inclusive. He chooses his numbers so that the sum of the base-ten logarithms of his six numbers is an integer. It so happens that the integers on the winning ticket have the same property— the sum of the base-ten logarithms is an integer. What is the probability that Professor Gamble holds the winning ticket?Problem 24If circular arcs and centers at and , respectively, then there exists acircletangent to both and , and to . If the length of is , then the circumference of the circle isProblem 25Eight congruent Equilateral triangle each of a different color, are used to construct a regular octahedron. How many distinguishable ways are there to construct the octahedron? (Two colored octahedrons are distinguishable if neither can be rotated to look just like the other.)。
AMC／AIME美国数学竞赛试题真题
AMC／AIME美国数学竞赛试题真题AMC／AIME美国数学竞赛试题真题考试信息AMC最新考试时间：●2010年第26届AMC8于11月16日，星期二●2011第12届AMC10A，第62届AMC12A 于2月8日，星期二●2011第12届AMC10B，第62届AMC12B 于2月23日，星期三●2011第29届AIME－1于3月17日，星期四2011第29届AIME－2于3月30日，星期三●2009年AMC8考试情况●2008年考试情况AMC/AIME中国历程：1983第1届AIME上海有76名同学获得参赛资格1984年第2届AIME有110人获得参赛资格1985年第3届AIME北京有118名同学获得参赛资格1986年第4届AIME上海有154名同学获得参赛资格，我国首次参加IMO的上海向明中学吴思皓就是在第四届AIME中获得满分1992年第10届AIME上海有一千多名同学获得参赛资格，其中格致中学潘毅明，交大附中张觉，上海中学葛建庆均获满分1993年第11届AIME上海有一千多名同学获得参赛资格，其中华东师大二附中高一王海栋，格致中学高二（女）黄静，市西中学高二张亮，复旦附中高三韩志刚四人获得满分，前三名总分排名复旦附中41分，华东师大二附中41分，上海中学40分。
北京地区参加2006年AMC的共有7所市重点学校的842名学生，有515名学生获得参加AIME资格，其中，清华附中有61名学生参加AMC，45名学生获得AIME资格，20名学生获得荣誉奖章据悉中国大陆以下地区可以报名参加考试：北京地区：中国数学会奥林匹克委员会负责组织实施长春地区、哈尔滨地区也有参加考试在华举办的美国人子弟学校也有参加考试广州地区：《数学奥林匹克报》负责组织实施。
在中国大陆报名者就在中国大陆考试。
考题采用英文版。
2009年AMC中国地区参赛学校一览表more info。
2017 AMC 12B Problem and Solution
2017 AMC 12B2017 AMC 12A problems and solutions. The test was held on February 7, 2017.Problem 1Kymbrea's comic book collection currently has comic books in it, and she is adding to her collection at the rate of comic books per month. LaShawn's collection currently has comic books in it, and he is adding to his collection at the rate of comic books per month. After how many months will LaShawn's collection have twice as many comic books as Kymbrea's?SolutionKymbrea has comic books initially and every month, she adds two. This can be represented as where x is the number of months elapsed. LaShawn's collection, similarly, is . To find when LaShawn will have twice the number of comic books asKymbrea, we solve for x with the equation and get .Problem2Real numbers , , and satisfy the inequalities , , and . Which of the following numbers is necessarily positive?SolutionNotice that must be positive because . Therefore the answer is . The other choices:As grows closer to , decreases and thus becomes less than .can be as small as possible (), so grows close to as approaches .For all , , and thus it is always negative.The same logic as above, but when this time.Problem3Supposed that and are nonzero real numbers such that . What is the value of ?SolutionsSolution 1Rearranging, we find , or . Substituting, we canconvert the second equation into .Solution 2Substituting each and with , we see that the given equation holds true,as . Thus,Problem4Samia set off on her bicycle to visit her friend, traveling at an average speed of kilometers per hour. When she had gone half the distance to her friend's house, a tire went flat, and she walked the rest of the way at kilometers per hour. In all it took her minutes to reach her friend's house. In kilometers rounded to the nearest tenth, how far did Samia walk?Solution 1Let's call the distance that Samia had to travel in total as , so that we can avoid fractions. We know that the length of the bike ride and how far she walked are equal, so they areboth , or .She bikes at a rate of kph, so she travels the distance she bikes in hours. She walks at a rate of kph, so she travels the distance she walks in hours.The total timeis . This is equal to of an hour. Solving for , we have:Since is the distance of how far Samia traveled by both walking and biking, and we want to know how far Samia walked to the nearest tenth, we have that Samia walked about .Problem 5The data set has median , firstquartile , and third quartile . An outlier in a data set is a value that is more than times the interquartile range below the first quartle () or more than times the interquartile range above the third quartile (), where the interquartile range is definedas . How many outliers does this data set have?SolutionThe interquartile range is defined as , which is . times this value is , so all values more than below = is an outlier. The only one that fits this is . All values more than above = are also outliers, of whichthere are none so there is onlyProblem 6The circle having and as the endpoints of a diameter intersects the -axis at a second point. What is the -coordinate of this point?SolutionBecause the two points are on a diameter, the center must be halfway between them at the point (4,3). The distance from (0,0) to (4,3) is 5 so the circle has radius 5. Thus, the equationof the circle is .To find the x-intercept, y must be 0, so ,so , , .Problem 7The functions and are periodic with least period . What is the least period of the function ?The function is not periodic. Solutionhas values at its peaks and x-intercepts. Increase them to . Then we plug theminto . and .So, isSolution by TheUltimate123 (Eric Shen)Solution IIStart by noting that . Then realize that under this function the negative sine values yield the same as their positive value, so take the absolute value of the sine function to get the new period. This has period , so the answer is surprisingly !!!!Problem 8The ratio of the short side of a certain rectangle to the long side is equal to the ratio of the long side to the diagonal. What is the square of the ratio of the short side to the long side of this rectangle?Solution 1: Cross-MultiplicationLet be the short side of the rectangle, and be the long side of the rectangle. The diagonal,therefore, is . We can get the equation . Cross-multiplying, weget . Squaring both sides of the equation, we get , which simplifies to . Solving for a quadratic in , using the quadratic formulawe get which gives us . We know that the square of the ratio must be positive (the square of any real number is positive), so the solutionis .Solution by: vedadehhcSolution 2: SubstitutionSolution by HydroQuantumLet the short side of the rectangle be and let the long side of the rectangle be . Then, the diagonal, according to the Pythagorean Theorem, is . Therefore, we can write the equation:.We are trying to find the square of the ratio of to . Let's let our answer, , be . Then, squaring the above equation,.Thus, .Multiplying each side of the equation by ,.Adding each side by ,.Solving for using the Quadratic Formula,.Since the ratio of lengths and diagonals of a rectangle cannot be negative, and ,the symbol can only take on the . Therefore,.Problem 9A circle has center and has radius . Another circle has center andradius . The line passing through the two points of intersection of the two circles has equation . What is ?Solution 1The equations of the two circlesare and . Rearrange themto and , respectively. Their intersection points are where these two equations gain equality. The two points lie on the line with the equation . We cansimplify this like thefollowing.. Thus, .Solution by TheUltimate123 (Eric Shen)Solution 2: Shortcut with right trianglesNote the specificity of the radii, and , and that specificity is often deliberately added tosimplify the solution to a problem.One may recognize as the hypotenuse of the right triangle and as thehypotenuse of the right triangle with legs and . We can suppose that the legs of these triangles connect the circles' centers to their intersection along the gridlines of the plane.If we suspect that one of the intersections lies units to the right of and units above the center of the first circle, we find the point , which is infact unit to the left of and units below the center of the second circle at .Plugging into gives us .A similar solution uses the other intersection point, .Problem10At Typico High School, of the students like dancing, and the rest dislike it. Of those who like dancing, say that they like it, and the rest say that they dislike it. Of those who dislike dancing, say that they dislike it, and the rest say that they like it. What fraction of students who say they dislike dancing actually like it?SolutionWLOG, let there be students. of them like dancing, and do not. Of those who like dancing, , or of them say they dislike dancing. Of those who dislike dancing, ,or of them say they dislike it. Thus,Problem11Call a positive integer if it is a one-digit number or its digits, when read from left to right, form either a strictly increasing or a strictly decreasing sequence. Forexample, , , and are monotonous, but , , and are not. How many monotonous positive integers are there?Solution 1Case 1: monotonous numbers with digits in ascending orderThere are ways to choose n digits from the digits 1 to 9. For each of these ways, we can generate exactly one monotonous number by ordering the chosen digits in ascending order. Note that 0 is not included since it will always be a leading digit and that is not allowed. Also, (the empty set) isn't included because it doesn't generate a number. The sum isequivalent toCase 2: monotonous numbers with digits in descending orderThere are ways to choose n digits from the digits 0 to 9. For each of these ways, we can generate exactly one monotonous number by ordering the chosen digits in descending order. Note that 0 is included since we are allowed to end numbers with zeros.However, (the empty set) still isn't included because it doesn't generate a number. The sumis equivalent to We discard the number 0 since it is not positive. Thus there are here.Since the 1-digit numbers 1 to 9 satisfy both case 1 and case 2, we have overcounted by 9. Thus there are monotonous numbers.Solution 2Like Solution 1, divide the problem into an increasing and decreasing case:Case 1: Monotonous numbers with digits in ascending order.Arrange the digits 1 through 9 in increasing order, and exclude 0 because a positive integer cannot begin with 0.To get a monotonous number, we can either include or exclude each of the remaining 9 digits, and there are ways to do this. However, we cannot exclude every digit at once, so we subtract 1 to get monotonous numbers for this case.Case 2: Monotonous numbers with digits in descending order.This time, we arrange all 10 digits in decreasing order and repeat the process tofind ways to include or exclude each digit. We cannot exclude every digit at once, and we cannot include only 0, so we subtract 2 to get monotonous numbers for this case.At this point, we have counted all of the single-digit monotonous numbers twice, so we must subtract 9 from our total.Thus our final answer is .Problem12What is the sum of the roots of that have a positive real part?Solution 1The root of any polynomial of the form will have all of it roots will havemagnitude and be the vertices of a regular -gon in the complex plane (This concept is known as the Roots of Unity). For the equation , it is easy tosee and as roots. Graphing these in the complex plane, we have four vertices of a regular dodecagon. Since the roots must be equally spaced, besides , there are fourmore roots with positive real parts lying in the first and fourth quadrants. We also know that the angle between these roots is . We only have to find the real parts of the roots lying in the first quadrant, because the imaginary parts would cancel out with those from the fourth quadrant. We have two triangles (the triangles formed by connecting the origin to the roots, and dropping a perpendicular line from each root to the real-axis), both withhypotenuse . This means that one has base and the other has base . Adding these and multiplying by two, we get the sum of the four roots as . However, wehave to add in the original solution of , so the answer is .Solution by vedadehhcSolution 2has a factor of , so we need to remember to multiply our solution below, usingthe Roots of Unity. We notice that the sum of the complex parts of all these roots is , because the points on the complex plane are symmetric. The rootswith are and by the Roots of Unity. Their real partsare and . Their sumis . But, remember to multiply by . Theanswer is .Solution by TheUltimate123 (Eric Shen)Problem 13In the figure below, of the disks are to be painted blue, are to be painted red, and is to be painted green. Two paintings that can be obtained from one another by a rotation or a reflection of the entire figure are considered the same. How many different paintings are possible?SolutionLooking at the answer choices, we see that the possibilities are indeed countable. Thus, we will utilize that approach in the form of two separate cases, as rotation and reflection take care of numerous possibilities. First, consider the case that the green disk is in a corner. This yields possible arrangements for the blue disks and red disks in the remaining available slots. Now, consider the case that the green disk is on an edge. This yields more possiblearrangements for the blue disks and red disks in the remaining available slots. Thus, our answer isProblem14An ice-cream novelty item consists of a cup in the shape of a 4-inch-tall frustum of a right circular cone, with a 2-inch-diameter base at the bottom and a 4-inch-diameter base at the top, packed solid with ice cream, together with a solid cone of ice cream of height 4 inches, whose base, at the bottom, is the top base of the frustum. What is the total volume of the ice cream, in cubic inches?SolutionSolution 1:The top cone has radius 2 and height 4 so it has volume .The frustum is made up by taking away a small cone of radius 1, height 4 from a large cone of radius 2, height 8, so it has volume .Adding, we get .Solution by: SilverLionSolution 2:Find the area of the cone with the method in Solution 1. The area of the frustrumisAdding, we getProblem 15Let be an equilateral triangle. Extend side beyond to a point sothat . Similarly, extend side beyond to a point so that , and extend side beyond to a point so that . What is the ratio of the area of to the area of ?Solution 1: Law of CosinesSolution by HydroQuantumLet .Recall The Law of Cosines. Letting ,Since both and are both equilateral triangles, they must be similar dueto similarity. This means that .Therefore, our answer is .Solution 2: InspectionNote that the height and base of are respectively 4 times and 3 times thatof . Therefore the area of is 12 times that of .By symmetry, . Adding the areas of these three triangles and for the total area of gives a ratio of ,or .Solution 3: CoordinatesFirst we note that due to symmetry. WLOG,let and Therefore, . Using the conditionthat , we get and . It is easy to checkthat . Since the area ratios of two similar figures is the square of the ratio of their lengths, the ratio isProblem 16The number has over positive integer divisors. One of them is chosen at random. What is the probability that it is odd?SolutionIf a factor of is odd, that means it contains no factors of . We can find the number of factors of two in by counting the number multiples of , , , and that are less than or equal to .After some quick counting we find that this number is . If the prime factorization of has factors of , there are choices for each divisor for how many factors of should be included (to inclusive). The probability that a randomlychosen factor is odd is the same as if the number of factors of is which is . Solution by: vedadehhcSolution 2We can write as its prime factorization:Each exponent of these prime numbers are one less than the number of factors at play here. This makes sense; is going to have factors: , and the other exponents will behave identically.In otherwords, has factors.We are looking for the probability that a randomly chosen factor of will be odd--numbers that do not contain multiples of as factors.From our earlier observation, the only factors of that are even are ones with at least one multiplier of , so our probability of finding an odd factor becomes the following:Problem 17A coin is biased in such a way that on each toss the probability of heads is and the probability of tails is . The outcomes of the tosses are independent. A player has the choiceof playing Game A or Game B. In Game A she tosses the coin three times and wins if all three outcomes are the same. In Game B she tosses the coin four times and wins if both the outcomes of the first and second tosses are the same and the outcomes of the third and fourth tosses are the same. How do the chances of winning Game A compare to the chances of winning Game B?The probability of winning Game A is less than the probability of winning Game B.The probability of winning Game A is less than the probability of winning Game B.The probabilities are the same.The probability of winning Game A is greater than the probability of winning Game B.The probability of winning Game A is greater than the probability of winning Game B. SolutionThe probability of winning Game A is the sum of the probabilities of getting three tails andgetting three heads which is . The probability of winning Game B is the sum of the probabilities of getting two heads and getting two tailssquared. This gives us . The probability of winningGame A is and the probability of winning Game B is , so the answer isProblem18The diameter of a circle of radius is extended to a point outside the circle sothat . Point is chosen so that and line is perpendicular to line . Segment intersects the circle at a point between and . What is the areaof ?Solution 1Let be the center of the circle. Notethat .However, by Power of apoint ,, so .Now. Since .Solution 2: Similar triangles with Pythagoreanis the diameter of the circle, so is a right angle, and therefore by AA similarity, .Because of this, , so . Likewise, , so .Thus the area of .Solution 3: Similar triangles without PythagoreanOr, use similar triangles all the way, dispense with Pythagorean, and go for minimal calculation:Draw with on . ... ( ratio applied twice).Problem19Let be the -digit number that is formed by writing the integers from to in order, one after the other. What is the remainder when is divided by ?SolutionWe will consider this number and . By looking at the last digit, it is obvious that the number is . To calculate the number , note thatso it is equivalent toLet be the remainder when this number is divided by . We knowthat and , so by the Chinese remainder theorem,since , ,or . So the answer isProblem 20Real numbers and are chosen independently and uniformly at random from theinterval . What is the probability that , where denotes the greatest integer less than or equal to the real number ?SolutionFirst let us take the case that . In this case, both and lie in the interval . The probability of this is . Similarly, in the casethat , and lie in the interval , and the probabilityis . It is easy to see that the probabilitiesfor for are the infinite geometric series that startsat and with common ratio . Using the formula for the sum of an infinite geometric series, we get that the probability is .Problem 21Last year, Isabella took 7 math tests and received 7 different scores, each an integer between 91 and 100, inclusive. After each test she noticed that the average of her test scores was an integer. Her score on the seventh test was 95. What was her score on the sixth test?Solution 1Let us simplify the problem. Since all of Isabella's test scores can be expressed as the sumof and an integer between and , we rewrite the problem into receiving scoresbetween and . Later, we can add to her score to obtain the real answer.From this point of view, the problem states that Isabella's score on the seventh test was . We note that Isabella received integer scores out of to . Since is already given as the seventh test score, the possible scores for Isabella on the other six testsare .The average score for the seven tests must be an integer. In other words, six distinct integers must be picked from set above, and their sum with must be a multiple of . The interval containing the possible sums of the six numbers in S arefrom to . We must now find multiples of from the interval to . There are fourpossibilities: , , , . However, we also note that the sum of the six numbers (besides ) must be a multiple of as well. Thus, is the only valid choice.(The six numbers sum to .)Thus the sum of the six numbers equals to . We apply the logic above in a similar way for the sum of the scores from the first test to the fifth test. The sum must be a multiple of . The possible interval is from to . Since the sum of the five scores must be less than , the only possibilities are and . However, we notice that does not work because the seventh score turns out to be from the calculation. Therefore, the sum of Isabella's scores from test to is . Therefore, her score on the sixthtest is . Our final answer is .Solution 2Let be Isabella's average after tests. , so . The only integer between and that satisfies this condition is . Let be Isabella'saverage after tests, and let be her sixth test score. ,so is a multiple of . Since is the only choice that is a multiple of , the answeris .Solution 3Let be the total sum of Isabella's first five test scores, and let be her score on the sixthtest. It follows that , , and , since at each step, her average score was an integer. Using the lastequivalence, , so we have a system of equivalences for . Solving this using the Chinese Remainder Theorem, weget .Now let's put a bound on . Using the given information that each test score was a distinct integer from to inclusive and that the seventh score was 95, weget . Since , we get . Therefore,The last preparation step will involve calculating all the possible test scores . Heretheyare:. This means that . Note that is not in the previous list because it corresponds to a score of , which we cannot have.We must have , and using the possible values we found for and , the only two that sum to are and . This corresponds to an value of , so the answer is .Problem 22Abby, Bernardo, Carl, and Debra play a game in which each of them starts with four coins. The game consists of four rounds. In each round, four balls are placed in an urn---one green, one red, and two white. The players each draw a ball at random without replacement. Whoever gets the green ball gives one coin to whoever gets the red ball. What is the probability that, at the end of the fourth round, each of the players has four coins?SolutionIt amounts to filling in a matrix. Columns are the random draws each round; rows are the coin changes of each player. Also, let be the number of nonzero elements in .WLOG, let . Parity demands that and must equal or .Case 1: and . There are ways to place 's in , so there are ways.Case 2: and . There are ways to place the in , ways to place the remaining in (just don't put it under the on top of it!), and ways for one of the other two players to draw the green ball. (We know it's green because Bernardo drew the red one.) We can just double to cover the case of , for a total of ways.Case 3: . There are three ways to place the in . Now, there are two cases as to what happens next.Sub-case 3.1: The in goes directly under the in . There's obviously way for that to happen. Then, there are ways to permute the two pairs of in and . (Either the comes first in or the comes first in .)Sub-case 3.2: The in doesn't go directly under the in . There are ways to place the , and ways to do the same permutation as in Sub-case 3.1. Hence, thereare ways for this case.There's a grand total of ways for this to happen, along with total cases. The probabilitywe're asking for is thusSolution 2 (Less Casework)We will proceed by taking cases based on how many people are taking part in this "transaction." We can have 2, 3, or 4 people all giving/receiving coins during the 4 turns. Basically, (like the previous solution), we are thinking this as filling out a 4x2 matrix of letters, where a letter on the left column represents this person gave, and a letter on the right column means this person recieved. We need to make sure that for each person that gave a certain amount, they received in total from other people that same amount, or in other words there are an equal number of A's, B's, C's, and D's on both columns of the matrix.Case 1: people. In this case, we can 4C2 ways to choose the two people, and 6 ways to get order them to get a count of 6 * 6 = 36 ways.Case 2: people. In this case, we have 4*(3C2)*4! = 288 ways to order 3 people.Case 3: people. In this case, we have 3*3*4! = 216 ways to order 4 people.So we have a total of 36+288+126=540 ways to order the four pairs of people. Now we divide this by the total number of ways - (4*3)^4 ( 4 times, 4 ways to choose giver, 3 to choose receiver). So the answer is 5/192.~ccx09 (NOTE: Due to the poor quality of this solution, please PM me and I will explain the numbers, I have some diagrams but I can't show it here)Problem 23The graph of , where is a polynomial of degree , containspoints , , and . Lines , , and intersect the graph again at points , , and , respectively, and the sum of the -coordinates of , , and is 24. What is ?SolutionFirst, we can define , which contains points , ,and . Now we find that lines , , and are defined by theequations , , and respectively. Since we want to find the -coordinates of the intersections of these lines and , we set each of them to ,and synthetically divide by the solutions we already know exist (eg. if we were looking atline , we would synthetically divide by the solutions and , because wealready know intersects the graph at and , which have -coordinates of and ). After completing this process on all three lines, we get that the -coordinates of , ,and are , , and respectively. Adding these together, weget which gives us . Substituting this back into the original equation,we get ,andProblem 24Quadrilateral has right angles at and , , and . There is a point in the interior of such that and the areaof is times the area of . What is ?Solution 1Let , , and . Note that . By the PythagoreanTheorem, . Since , the ratios of sidelengths must be equal. Since , and . Let F be a point on such that is an altitude of triangle . Notethat . Therefore, and . Since and form altitudes of triangles and , respectively, the areas of these triangles can be calculated. Additionally, the area of triangle can be calculated, as it is a right triangle. Solving for each of these yields:Therefore, the answerisSolution 2Draw line through , with on and on , . WLOGlet , , . By weighted average .Meanwhile, .. We obtain , namely .The rest is the same as Solution 1.Solution 3Let . Then from the similar triangles condition, we compute and . Hence, the -coordinate of is just . Since lies on the unit circle, we can compute the coordinate as . By Shoelace, we wantFactoring out denominators and expanding by minors, this is equivalent toThis factorsas , so and so the answer is .ProblemA set of people participate in an online video basketball tournament. Each person may be a member of any number of -player teams, but no two teams may have exactly thesame members. The site statistics show a curious fact: The average, over all subsets of size of the set of participants, of the number of complete teams whose members are among those people is equal to the reciprocal of the average, over all subsets of size of the set of participants, of the number of complete teams whose members are among those people. How many values , , can be the number of participants?SolutionSolution by Pieater314159Let there be teams. For each team, there are different subsets of players including that full team, so the total number of team-(group of 9) pairs isThus, the expected value of the number of full teams in a random set of players isSimilarly, the expected value of the number of full teams in a random set of players isThe condition is thus equivalent to the existence of a positive integer such that。
2017-2018年美国“数学大联盟杯赛”（中国赛区）初赛高中年级试卷
2017-2018年美国“数学⼤联盟杯赛”（中国赛区）初赛⾼中年级试卷2017-2018年度美国“数学⼤联盟杯赛”(中国赛区)初赛(⼗、⼗⼀、⼗⼆年级)(初赛时间：2017年11⽉26⽇，考试时间90分钟，总分300分)学⽣诚信协议：考试期间，我确定没有就所涉及的问题或结论，与任何⼈、⽤任何⽅式交流或讨论，我确定我所填写的答案均为我个⼈独⽴完成的成果，否则愿接受本次成绩⽆效的处罚。
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考前注意事项：1. 本试卷是⼗、⼗⼀、⼗⼆年级试卷，请确保和你的参赛年级⼀致；2. 本试卷共4页(正反⾯都有试题)，请检查是否有空⽩页，页数是否齐全；3. 请确保你已经拿到以下材料：本试卷(共4页，正反⾯都有试题)、答题纸、英⽂词汇⼿册、草稿纸。
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填空题(每⼩题10分，答对加10分，答错不扣分，共300分。
)1.Each pirate wants his own treasure chest, but there is 1 more pirate than thereare treasure chests. If the pirates would agree to pair up so each pirateshares a treasure chest with another pirate, then 1 treasure chest wouldnot be assigned to any pirate. How many treasure chests are there?Answer: ________________.2.If m and nare positive integers that satisfy 10=, what is the greatest possiblevalue of m + n?Answer: ________________.3.There are an infinite number of points with positive coordinates(x,y) the sum of whose coordinates is the square of an integer.Among all such points (x,y), which one satisfies y = 2x and hasx as small as possible?Answer: ________________.4.As shown, a small square is inscribed in one of the triangles formed whenboth diagonals of a larger square are drawn. If the area of the larger squareis 144, what is the area of the smaller square?Answer: ________________.5.Trisection points on opposite sides of a rectangle are joined, as shown. Ifthe area of the shaded region is 2018, what is the area of the rectangle?Answer: ________________.6. A unit fraction is a fraction whose numerator is 1 and whosedenominator is a positive integer. What is the largest rationalnumber that can be written as the sum of 3 different unitfractions?Answer: ________________.7.What is the greatest possible perimeter of a rectangle whose length and width are different prime numbers, each less than 120?Answer: ________________.8.Mom, Dad, and I each write a positive integer. My number is leastand Dad's is greatest. The average of all 3 numbers is 20. Theaverage of the 2 smallest numbers is 8. If Dad's number is d andif my number is m, what is the greatest possible value of d–m?Answer: ________________.9.If 8 different integers are chosen at random from the first 15 positive integers, what is the probability that an additional number chosen at random from the remaining 7 positive integers is smaller than every one of the 8 originally chosen positive integers?Answer: ________________.10.What sequence of 5 positive integers has these three properties:1) All but one of the numbers is a multiple of 5.2) Every number after the first is 1 more than the sum of all the preceding numbers.3) The first number is as small as possible.Answer: ________________.11.Three beavers (one not shown) take turns biting a tree until it falls. Thesecond beaver is twice as likely as the first to make the tree fall. Thethird is twice as likely as the second to make the tree fall. What isthe probability that a bite taken by the third beaver causes thetree to fall?Answer: ________________.12.What is the ratio, larger to smaller, of a rectangle's dimensions if halfof the rectangle is similar to the original rectangle?Answer: ________________.第1页，共4页第2页，共4页A rectangle is partitioned into 9 different squares, as shown at the right. The area of the smallest square, shown fully darkened, is 1. Two other squares have areas of 196 and 324, as shown. What is the area of the shaded square? Answer: ________________.When the square of an eight-digit integer is subtracted from the square of a differenteight-digit integer, the difference will sometimes have eight identical even digits. What are both possible values of the repeated digit in such a situation? Answer: ________________.If the perimeter of an isosceles triangle with integral sides is 2017, how many different lengthsare possible for the legs? Answer: ________________.What are all ordered triples of positive primes (p ,q ,r ) which satisfy p q + 1 = r ? Answer: ________________.The reflection of (6,3) across the line x = 4 is (2,3). If m ≠ 4, what is the reflection of (m ,n )across the line x = 4? Answer: ________________.The vertices of a triangle are (8,7), (0,1), and (8,1). What are thecoordinates of all points inside this triangle that have integralcoordinates and lie on the bisector of the smallest angle of the triangle? Answer: ________________.In a regular 10-sided polygon, two pairs of different vertices (four different verticesaltogether) are chosen at random, so that all points chosen are distinct from each other. What is the probability that the line segments determined by each pair of points do not intersect? Answer: ________________.A line segment is drawn from the upper right vertex of aparallelogram, as shown, dividing the opposite side into segments with lengths in a 2:1 ratio. If the area of the parallelogram is 90, what is the area of the shaded region?Answer: ________________.21. If 0 < a ≤ b ≤ 1, what is the maximum value of ab 2 – a 2b ? Answer: ________________.22. What are all ordered pairs of integers (x ,y ) that satisfy 5x 3 + 2xy – 23 = 0? Answer: ________________.23. If two altitudes of a triangle have lengths 10 and 15, what is the smallest integer that couldbe the length of the third altitude?Answer: ________________.24. If h is the number of heads obtained when 4 fair coins are each tossed once, what is theexpected (average) value of h 2? Answer: ________________.25. What is the largest integer N for which 7x + 11y = N has no solution in non-negativeintegers (x ,y )? Answer: ________________.26. There are only two six-digit integers n greater than 100 000 for which n 2 has n as its finalsix digits (or, equivalently, for which n 2 – n is divisible by 106). One of the integers is 890 625. What is the other? Answer: ________________.27. A hexagon is inscribed in a circle as shown. If lengths of three sidesof the hexagon are each 1 and the lengths of the other three sides are each 2, what is the area of this hexagon? Write your answer in its exact format or round to the nearest tenth. Answer: ________________.28. If x is a number chosen uniformly at random between 0 and 1, what is the probability thatthe greatest integer ≤ 21log x ??is odd?Answer: ________________.29. In the interval -1 < x < 1, sin θ is one root of x 4 – 4x 3 + 2x 2 – 4x + 1 = 0. In that sameinterval, for what ordered pair of integers (a ,b ) is cos 2θ one root of x 2 + ax + b = 0? Answer: ________________.30. Let P (x ) = 2x 10 + 3x 9 + 4x + 9. If z is a non-real solution of z 3 = 1, what is the numericalvalue of 23111P P P z z z++ ? ? ???????Answer: ________________.第3页，共4页第4页，共4页。
美国数学邀请赛2017AMC10试题及中文翻译.doc
8th AMC 10 A 20071. One ticket to a show costs $20 at full price. Susan buys 4 tickets using a coupon that gives her a 25% discount. Pam buys 5 tickets using a coupon that gives her a 30% discount. How many more dollars does Pam pay than Susan?(A) 2 (B) 5 (C) 10 (D) 15 (E) 20中文：一张展览票全价为20美元。
Susan 用优惠券买4张票打七五折。
Pam 用优惠券买5张票打七折。
Pam 比Susan 多花了多少美元？2. Define a@b=ab - b 2 and a#b=a + b - ab 2. What is 6@26#2? (A)－12 (B)－14 (C) 18 (D) 14 (E) 12中文：定义a @ b=ab-b 2 ，a # b=a + b - ab 2。
求6@26#2的值。
3. An aquarium has a rectangular base that measures 100cm by 40cm and has a height of 50cm. It is filled with water to a height of 40cm. A brick with a rectangular base that measures 40cm by 20cm and a height of 10cm is placed in the aquarium. By how many centimeters does that water rise?(A) 0.5 (B) 1 (C) 1.5 (D) 2 (E)2.5中文：一个养鱼缸有100cm ×40cm 的底，高为50cm 。
amc12真题及答案
Problem 1What is the value of ?SolutionProblem 2For what value of does ?SolutionProblem 3The remainder can be defined for all real numbers and with bywhere denotes the greatest integer less than or equal to . What is the value of ?SolutionProblem 4The mean, median, and mode of the data values are all equal to . What is the value of ?SolutionProblem 5Goldbach's conjecture states that every even integer greater than 2 can be written as the sum of two prime numbers (for example, ). So far, no one has been able to prove that the conjecture is true, and no one has found a counterexample to show that the conjecture is false. What would a counterexample consist of?SolutionProblem 6A triangular array of coins has coin in the first row, coins in the second row, coins in the third row, and so on up to coins in the th row. What is the sum of the digits of ?SolutionProblem 7Which of these describes the graph of ?SolutionProblem 8What is the area of the shaded region of the given rectangle?SolutionProblem 9The five small shaded squares inside this unit square are congruent and have disjoint interiors. The midpoint of each side of the middle square coincides with one of the vertices of the other four small squares as shown. The common side length is , where and are positive integers. What is ?SolutionProblem 10Five friends sat in a movie theater in a row containing seats,numbered to from left to right. (The directions "left" and "right" are from the point of view of the people as they sit in the seats.) During the movie Ada went to the lobby to get some popcorn. When she returned, she found that Bea had moved two seatsto the right, Ceci had moved one seat to the left, and Dee and Edie had switched seats, leaving an end seat for Ada. In which seat had Ada been sitting before she got up?SolutionProblem 11Each of the students in a certain summer camp can either sing, dance, or act. Some students have more than one talent, but no student has all three talents. Thereare students who cannot sing, students who cannot dance, and students who cannot act. How many students have two of these talents?SolutionProblem 12In , , , and . Point lies on , and bisects . Point lies on ,and bisects . The bisectors intersect at . What is the ratio : ?SolutionProblem 13Let be a positive multiple of . One red ball and green balls are arranged in a line in random order. Let be the probability that at least of the green balls are on the same side of the red ball. Observe that andthat approaches as grows large. What is the sum of the digits of the least value of such that ?SolutionProblem 14Each vertex of a cube is to be labeled with an integer from through , with each integer being used once, in such a way that the sum of the four numbers on the verticesof a face is the same for each face. Arrangements that can be obtained from each other through rotations of the cube are considered to be the same. How many different arrangements are possible?SolutionProblem 15Circles with centers and , having radii and , respectively, lie on the same side of line and are tangent to at and , respectively,with between and . The circle with center is externally tangent to each of the other two circles. What is the area of triangle ?SolutionProblem 16The graphs of and are plotted on the same set of axes. How many points in the plane with positive -coordinates lie on two or more of the graphs?SolutionProblem 17Let be a square. Let and be the centers, respectively, of equilateral triangles with bases and each exterior to the square. What is the ratio of the area of square to the area of square ?SolutionProblem 18For some positive integer the number has positive integer divisors, including and the number How many positive integer divisors does thenumber have?SolutionProblem 19Jerry starts at on the real number line. He tosses a fair coin times. When he gets heads, he moves unit in the positive direction; when he gets tails, he moves unit in the negative direction. The probability that he reaches at some time during this process is where and are relatively prime positive integers. What is (For example, he succeeds if his sequence of tosses is )SolutionProblem 20A binary operation has the properties that and that for all nonzero real numbers and (Here the dot represents the usual multiplication operation.) The solution to the equation can be written as where and are relatively prime positive integers. What isSolutionProblem 21A quadrilateral is inscribed in a circle of radius Three of the sides of this quadrilateral have length What is the length of its fourth side?SolutionProblem 22How many ordered triples of positive integers satisfy and ?SolutionProblem 23Three numbers in the interval are chosen independently and at random. What is the probability that the chosen numbers are the side lengths of a triangle with positive area?SolutionProblem 24There is a smallest positive real number such that there exists a positive real number such that all the roots of the polynomial are real. In fact, for this value of the value of is unique. What is the value ofSolutionProblem 25Let be a positive integer. Bernardo and Silvia take turns writing and erasing numbers on a blackboard as follows: Bernardo starts by writing the smallest perfect square with digits. Every time Bernardo writes a number, Silvia erases the last digits of it. Bernardo then writes the next perfect square, Silvia erases the last digits of it, and this process continues until the last two numbers that remain on the board differ by at least 2. Let be the smallest positive integer not written on the board. For example, if , then the numbers that Bernardo writes are , and the numbers showing on the board after Silvia erases are and , and thus . What is the sum of the digits of ?2016 AMC 12A Answer Key1 B2 C3 B4 D5 E6 D7 D8 D9 E10 B11 E12 C13 A14 C15 D16 D17 B18 D19 B20 A21 E22 A23 C24 B25 E。
AMC12历年试卷_2000-2008
AMC 12 2000
1 In the year 2001, the United States will host the International Mathematical Olympiad. Let I, M, and O be distinct positive integers such that the product I · M · O = 2001. What’s the largest possible value of the sum I + M + O? (A)23 (B)55 (C)99 (D)111 (E)671
4 The Fibonacci Sequence 1, 1, 2, 3, 5, 8, 13, 21, . . . 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? (A)0 (B)4 (C)6 (D)7 (E)9
16 A checkerboard of 13 rows and 17 columns has a number written in each square, beginning in the upper left corner, so that the rst row is numbered 1, 2, . . . , 17, the second row 18, 19, . . . , 34, and so on down the board. If the board is renumbered so that the left column, top to bottom, is 1, 2, . . . , 13, the second column 14, 15, . . . , 26 and so on across the board, some squares have the same numbers in both numbering systems. Find the sum of the numbers in these squares (under either system). (A) 222 (B) 333 (C) 444 (D) 555 (E) 666
美国数学测评AMC12A-2019(中英双语)
2019AMC 12A Problems1. The area of a pizza with radius 4 is N percent larger than the area of a pizza with radius 3 inches. What is the integer closest to N ?半径为4英寸的比萨饼面积比半径为3英寸的比萨饼面积大百分之N 。
最接近N 的整数是什么？A B C D E ()25()33()44()66()782.Suppose a is 150% of b . What percent of a is b 3?设a 是b 的150%。
那么a 的百分之多少是3b ?A B C D E +3()50()66()150()200()45023. 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 drawnfrom 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，最少须从盒子中取出多少个球？A B C D E ()75()76()79()84()914. What is the greatest number of consecutive integers whose sum is 45? 最多可以有多少个连续整数，它们的总和是45?A B C D E ()9()25()45()90()1205. Two lines with slopes21and 2 intersect at (2,2). What is the area of the triangle enclosed by these two lines and the line +=x y 10?两条斜率分别为21和2的直线相交于(2, 2)。
赏析几道美国AMC12数学竞赛题
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2017 AMC 10A
2017 AMC 10A Problems and SolutionsProblem 1What is the value of?（A）70 （B）97 （C）127 （D）159 （E）729Problem 2Pablo buys popsicles for his friends. The store sells single popsicles for $1each, 3-popsicle boxes for $2each, and5-popsicle boxes for$3. What is the greatest number of popsicles that Pablo can buy with$8?（A）8 （B）11 （C）12 （D）13 （E）15Problem3Tamara has three rows of two-feet by-feet flower beds in her garden. The beds are separated and also surrounded by-foot-wide walkways, as shown on the diagram. What is the total area of the walkways, in square feet?（A）72 （B）78 （C）90 （D）120 （E）150Problem 4Mia is “helping” her mom pick up 30toys that are strewn on the floor. Mia’s mom manages to put 3toys into the toy box every 30 seconds, but each time immediately after those 30 seconds have elapsed, Mia takes 2toys out of the box. How much time, in minutes, will it take Mia and her mom to put all 30 toys into the box for the first time?（A）13.5 （B）14 （C）14.5 （D）15 （E）15.5Problem 5The sum of two nonzero real numbers is 4 times their product. What is the sumof the reciprocals of the two numbers?（A）1 （B）2 （C）4 （D）8 （E）12Problem 6Ms. Carroll promised that anyone who got all the multiple choice questions right on the upcoming exam would receive an A on the exam. Which of these statements necessarily follows logically?Problem 7Jerry and Silvia wanted to go from the southwest corner of a square field to the northeast corner. Jerry walked due east and then due north to reach the goal, but Silvia headed northeast and reached the goal walking in a straight line. Which of the following is closest to how much shorter Silvia's trip was, compared to Jerry's trip?（A）30% （B）40% （C）50% （D）60% （E）70%Problem 8At a gathering of 30people, there are 20 people who all know each other and 10 people who know no one. People who know each other hug, and people who do not know each other shake hands. How many handshakes occur?（A）240 （B）245 （C）250 （D）480 （E）490Problem 9Minnie rides on a flat road at20kilometers per hour (kph), downhill at30kph, and uphill at5kph. Penny rides on a flat road at kph, downhill at40kph, and uphill at kph. Minnie goes from town to town, a distance of km all uphill, then from town to town, a distance of km all downhill, and then back to town, a distance of km on the flat. Penny goes the other way around using the same route. How many more minutes does it take Minnie to complete the45-km ride than it takes Penny?（A）45 （B）60 （C）65 （D）90 （E）95Problem 10Joy has30thin rods, one each of every integer length from1cm through30cm. She places the rods with lengths cm,cm, and cm on a table. She then wants to choose a fourth rod that she can put with these three to form a quadrilateral with positive area. How many of the remaining rods can shechoose as the fourth rod?（A）16 （B）17 （C）18 （D）19 （E）20Problem 11The region consisting of all points in three-dimensional space within units of line segment has volume. What is the length?（A）6 （B）12 （C）18 （D）20 （E）24Problem 12Let be a set of points in the coordinate plane such that two of the three quantities and are equal and the third of the three quantities is no greater than this common value. Which of the following is a correct description forProblem 13Define a sequence recursively by and the remainder when is divided by3,for all Thus the sequence starts What is（A）6 （B）7 （C）8 （D）9 （E）10Problem 14Every week Roger pays for a movie ticket and a soda out of his allowance. Last week, Roger's allowance was dollars. The cost of his movie ticket was 20% of the difference between and the cost of his soda, while the cost of his soda was 5% of the difference between and the cost of his movie ticket. To the nearest whole percent, what fraction of did Roger pay for his movie ticket and soda?（A）9% （B）19% （C）22% （D）23% （E）25%Problem 15Chloéchooses a real number uniformly at random from the interval. Independently, Laurent chooses a real number uniformly at random from the interval . What is the probability that Laurent's number is greater than Chloé's number?Problem 16There are 10 horses, named Horse 1, Horse 2,, Horse 10. They get their names from how many minutes it takes them to run one lap around a circular race track: Horse runs one lap in exactly minutes. At time 0 all the horses are together at the starting point on the track. The horses start running in the same direction, and they keep running around the circular track at their constant speeds. The least time, in minutes, at which all 10 horses will again simultaneously be at the starting point is. Let be the least time, in minutes, such that at least 5 of the horses are again at the starting point. What is the sum of the digits of?（A）2 （B）3 （C）4 （D）5 （E）6Problem 17Distinct points,,,lie on the circle and have integer coordinates. The distances and are irrational numbers. What is the greatest possible value of the ratio?Problem 18Amelia has a coin that lands heads with probability 1/3, and Blaine has a coin that lands on heads with probability 2/5. Amelia and Blaine alternately toss their coins until someone gets a head; the first one to get a head wins. All coin tosses are independent. Amelia goes first. The probability that Amelia wins is, where and are relatively prime positive integers. What is?（A）1 （B）2 （C）3 （D）4 （E）5Problem 19Alice refuses to sit next to either Bob or Carla. Derek refuses to sit next to Eric. How many ways are there for the five of them to sit in a row of 5 chairs under these conditions?（A）12 （B）16 （C）28 （D）32 （E）40Problem 20Let S(n)equal the sum of the digits of positive integer. For example,S(1507)=13. For a particular positive integer,S(n)=1274. Which ofthe following could be the value of?（A）1 （B）3 （C）12 （D）1239 （E）1265Problem 21A square with side length is inscribed in a right triangle with sides of length 3,4, and5so that one vertex of the square coincides with the right-angle vertex of the triangle. A square with side length is inscribed in another right triangle with sides of length 3,4, and so that one side of the square lies on the hypotenuse of the triangle. What is?Problem 22Sides and of equilateral triangle are tangent to a circle at points and respectively. What fraction of the area of lies outside the circle?Problem 23How many triangles with positive area have all their vertices at points inthe coordinate plane, where and are integers between1and5, inclusive? （A）2128 （B）2148 （C）2160 （D）2200 （E）2300Problem 24For certain real numbers,, and, the polynomialhas three distinct roots, and each root of is also a root of the polynomialWhat is?（A）-9009 （B）-8008 （C）-7007 （D）-6006 （E）-5005Problem 25How many integers between 100and999, inclusive, have the property that some permutation of its digits is a multiple of11 between 100and 999For example, both121 and 211have this property.（A）226 （B）243 （C）270 （D）469 （E）4862017 AMC 10A SolutionsProblem 1Solution 1Notice this is the term in a recursive sequence, defined recursivelyas Thus:Solution 2Starting to compute the inner expressions, we see the results are 1,3 ,7,15,…. This is always1less than a power of2. The only admissible answer choice by this rule is thus C127.Solution 3Working our way from the innermost parenthesis outwards and directly computing, we have C.Problem 2$3boxes give us the most popsicles/dollar, so we want to buy as many of those as possible. After buying2, we have $2left. We cannot buy a third $3box, so we opt for the $2box instead (since it has a higher popsicles/dollar ratio than the $1pack). We're now out of money. We bought 5+5+3=13popsicles, so the answer is D13.Problem3Finding the area of the shaded walkway can be achieved by computing the total area of Tamara's garden and then subtracting the combined area of her six flower beds.Since the width of Tamara's garden contains three margins, the total width is feet.Similarly, the height of Tamara's garden is feet. Therefore, the total area of the garden is square feet.Finally, since the six flower beds each have an area of square feet, the area we seek is, and our answer isSolution1Every 30 seconds 3-2=1 toys are put in the box, so after seconds there will be 27 toys in the box. Mia's mom will then put 3 toys into to the box and we have our total amount of time to seconds, which equals 14 minutes. B14Solution 2Though Mia's mom places 3 toys every 30 seconds, Mia takes out 2 toys right after. Therefore, after 30 seconds, the two have collectively placed 1 toy into the box. Therefore by 13.5 minutes, the two would have placed 27 toys into the box. Therefore, at 14 minutes, the two would have placed 30 toys into the box. Though Mia may take 2 toys out right after, the number of toys in the box first reaches 30 by 14 minutes. B14Problem 5Solution1Let the two real numbers be . We are given that anddividing both sides by , ，Note: we can easily verify that this is the correct answer; for example, 1/2 and 1/2 work, and the sum of their reciprocals is 4.Solution 2Instead of using algebra, another approach at this problem would be to notice the fact that one of the nonzero numbers has to be a fraction. See for yourself. And by looking into fractions, we immediately see that and 1would fit the rule. C14Solution 3Rewriting the given statement: "if someone got all the multiple choice questions right on the upcoming exam then he or she would receive an A on the exam." If that someone is Lewis the statement becomes: "if Lewis got all the multiple choice questions right, then he got an A on the exam." The contrapositive: "If Lewis did not receive an A, then he got at least one of the multiple choice questions wrong (did not get all of them right)" must also be true leaving B as the correct answer. B is also equivalent to the contrapositive of the original statement, which implies that it must be true, so the answer is B Problem 7Let represent how far Jerry walked, and represent how far Sylvia walked. Since the field is a square, and Jerry walked two sides of it, while Silvia walked the diagonal, we can simply define the side of the square field to be one, and find the distances they walked. Since Jerry walked two sides,Since Silvia walked the diagonal, she walked the hypotenuse of a 45, 45, 90 triangle with leg length 1. Thus,We can thentakeProblem 8Solution 1Each one of the ten people has to shake hands with all the other people they don’t know. So. From there, we calculate how many handshakes occurred between the people who don’t know each other. This is simply counting how many ways to choose two people to shake hands, or. Thus the answer is.Solution 2We can also use complementary counting. First of all,handshakes or hugs occur. Then, if we can find the number of hugs, then we can subtract it from 435 to find the handshakes. Hugs only happen between the 20 peoplewho know each other, so there are hugs..Solution 3We can focus on how many handshakes the 10 people get.The 1st person gets 29 handshakes.2nd gets 28......And the 10th receives 20 handshakes.We can write this as the sum of an arithmetic sequence.Therefore, the answer isProblem 9The distance from town to town is km uphill, and since Minnie rides uphill at a speed of kph, it will take her hours. Next, she will ride from town to town, a distance of km all downhill. Since Minnie rides downhill at a speed of kph, it will take her half an hour. Finally, she rides from town back to town, a flat distance of km. Minnie rides on a flat road at kph, so this will take her hour. Her entire trip takes her hours. Secondly, Penny will go from town to town, a flat distance of km. Since Penny rides on a flat road at kph, it will take her2/3of an hour. Next Penny will go from town to town, which is uphill for Penny. Since Penny rides at a speed of kph uphill, and town and are km apart, it will take her hours. Finally, Penny goes from Town back to town, a distance of km downhill. Since Penny rides downhill at kph, it will only take her 1/4of an hour. In total, it takes her 29/12hours, which simplifies to hours and minutes. Finally, Penny's Hour Minute trip was minutesless than Minnie's Hour Minute TripProblem 10The triangle inequality generalizes to all polygons, so and to get. Now, we know that there are numbers between and exclusive, but we must subtract to account for the 2 lengths already used that are between those numbers, which gives 19-2=17 BProblem 11In order to solve this problem, we must first visualize what the region contained looks like. We know that, in a three dimensional plane, the region consisting of all points within units of a point would be a sphere with radius. However, we need to find the region containing all points within 3 units of a segment. It can be seen that our region is a cylinder with two hemispheres on either end.We know the volume of our region, so we set up the following equation (the volume of our cylinder + the volume of our two hemispheres will equal):, where is equal to the length of our line segment. Solving, we find that.Problem 12Problem 13A pattern starts to emerge as the function is continued. The repeating pattern is The problem asks for the sum of eight consecutive terms in the sequence. Because there are eight numbers in the repeating sequence, we just need to find the sum of the numbers in the sequence, which isProblem 14Let = cost of movie ticket, Let = cost of sodaWe can create two equations:Problem 15 Solution 1Solution 2Solution3Scale down by 2017to get that Chloe picks from and Laurent picks from. There are an infinite number of cases for the number that Chloe picks, but they are all centered around the average of. Therefore, Laurent has a range of to to pick from, on average, which is a length of out of a total length of. Therefore, the probability is 1.5/2=3/4 CProblem 16Solution 1If we have horses,, then any number that is a multiple of the all those numbers is a time when all horses will meet at the starting point. The least of these numbers is the LCM. To minimize the LCM, we need the smallest primes, and we need to repeat them a lot. By inspection, we find that. Finally,.Solution 2We are trying to find the smallest number that has one-digit divisors. Therefore we try to find the LCM for smaller digits, such as,,, or. We quickly consider since it is the smallest number that is the LCM of,,and. Since has single-digit divisors, namely,,,, and, our answer isProblem 17Because,,, and are integers there are only a few coordinates that actually satisfy the equation. The coordinates are and We want to maximize and minimize They also have to be the square root of something, because they are both irrational. The greatest value of happens when it and are almost directly across from each other and are in different quadrants. For example, the endpoints of the segment could be (-4,3)and (3,-4)because the two points are almost across from each other. The least value of is when the two endpoints are in the same quadrant and are very close to each other. This can occur when, for example,is (3,4)and is (4,3)They are in the same quadrant and no other point on the circle with integer coordinates is closer to the point (3,4)than (4,3)Using the distanceformula, we get that is and that isProblem 18Solution 1Solution 2Problem 19Solution 1For notation purposes, let Alice be A, Bob be B, Carla be C, Derek be D, and Eric be E. We can split this problem up into two cases:A sits on an edge seat.Then, since B and C can't sit next to A, that must mean either D or E sits next to A. After we pick either D or E, then either B or C must sit next to D/E. Then, we can arrange the two remaining people in two ways. Since there are two different edge seats that A can sit in, there are a total of.A does not sit in an edge seat.In this case, then only two people that can sit next to A are D and E, and there are two ways to permute them, and this also handles the restriction that D can't sit next to E. Then, there are two ways to arrange B and C, the remaining people. However, there are three initial seats that A can sit in, so there are seatings in this case.Adding up all the cases, we have.Solution 2Label the seats through. The number of ways to seat Derek and Eric in the five seats with no restrictions is. The number of ways to seat Derek and Eric such that they sit next to each other is(which can be figure out quickly), so the number of ways such that Derek and Eric don't sit next to each other is. Note that once Derek and Eric are seated, there are three cases.The first case is that they sit at each end. There are two ways to seat Derek and Eric. But this is impossible because then Alice, Bob, and Carla would have to sit in some order in the middle three seats which would lead to Alice sitting next to Bob or Carla, a contradiction. So this case gives us ways.Another possible case is if Derek and Eric seat in seats and in some order. There are 2 possible ways to seat Derek and Eric like this. This leaves Alice, Bob, and Carla to sit in any order in the remaining three seats. Since no two of these three seats are consecutive, there are ways to do this. So the second case gives us total ways for the second case.The last case is if once Derek and Eric are seated, exactly one pair of consecutive seats is available. There are ways to seat Derek and Eric like this. Once they are seated like this, Alice cannot not sit in one of the two consecutive available seats without sitting next to Bob and Carla. So Alice has to sit in the other remaining chair. Then, there are two ways to seat Bob and Carla in the remaining two seats (which are consecutive). So this case gives us ways.So in total there are 12+16=28. So our answer is.Problem 20Solution 1Solution 2Solution 3Problem 21Problem 22Problem 23Problem 24 Solution 1Solution 2Solution 3Problem 25Solution 1Let the three-digit number be :If a number is divisible by , then the difference between the sums of alternating digits is a multiple of .There are two cases: andWe now proceed to break down the cases. Note: let so that we avoid counting the same permutations and having to subtract them later.Solution 2We note that we only have to consider multiples of 11 and see how many valid permutations each has. We can do casework on the number of repeating digits that the multiple of 11 has:Solution 3We can overcount and then subtract. We know there are multiples of. We can multiply by for each permutation of these multiples. (Yet some multiples don't have 6)Now divide by, because if a number with digits,, and is a multiple of 11, then is also a multiple of 11 so we have counted the same permutations twice.Basically, each multiple of 11 has its own 3 permutations (say has and whereas has and). We know that each multiple of 11 has at least 3 permutations because it cannot have 3 repeating digits.Hence we have permutations without subtracting for overcounting. Now note that we overcounted cases in which we have 0's at the start of each number. So, in theory, we could just answer and move on.If we want to solve it, then we continue.We overcounted cases where the middle digit of the number is 0 and the last digit is 0.Note that we assigned each multiple of 11 3 permutations.The last digit is gives possibilities where we overcounted by permutation for each of.The middle digit is 0 gives 8 possibilities where we overcount by 1.andSubtracting gives.Now, we may ask if there is further overlap (I.e if two of and and were multiples of) Thankfully, using divisibility rules, this can never happen as taking the divisibility rule mod 11 and adding we get that,, or is congruent to. Since are digits, this can never happen as none of them can equal 11 and they can't equal 0 as they are the leading digit of a 3 digit number in each of the cases.2017 AMC 10A Answer Key1. C2. D3. B4. B5. C6. B7. A8. B9. C10. B11. D12. E13. D14. D15. C16. B17. D18. D19. C20. D21. D22. E23. B24. C25. A。