AMC10数学竞赛是美国高中数学竞赛中的一项，是针对高中一年级及初中三年级学生的数学测试，该竞赛开始于2000年，分A赛和B赛，于每年的2月初和2月中举行，学生可任选参加一项即可。不管是对高校申请还是今后在数学领域的发展都极其有利！那么接下来跟随小编来看一下AMC10的官方真题以及官方解答吧：Problem 21Cozy the Cat and Dash the Dog are going up a staircase with a certain number of steps. However, instead of walking up the steps one at a time, both Cozy and Dash jump. Cozy goes two steps up with each jump (though if necessary, he will just jump the last step). Dash goes five steps up with each jump (though if necessary, he will just jump the last steps if there are fewer than steps left). Suppose the Dash takes fewer jumps than Cozy to reach the top of the staircase. Let denote the sum of all possible numbers of steps this staircase can have. What is the sum of the digits of ?SolutionsSolution 1We can translate this wordy problem into this simple equation:We will proceed to solve this equation via casework.Case 1: Our equation becomes , where Using the fact that is an integer, we quickly find that and yield and , respectively.Case 2: Our equation becomes , where Using the fact that is an integer, we quickly find that yields . Summing up we get . The sum of the digits is .Solution 2We know from the problem that Dash goes 3 steps further than Cozy per jump (assuming they aren't within 4 steps from the top). That means that if Dash takes 19 fewer jumps than Cozy to get to the top of the staircase, the staircase must be at least 57 steps high (3*19=57). We then start using guess-and-check: steps: jumps for Cozy, and jumps for Dash, giving a difference of jumps. steps: jumps for Cozy, and jumps for Dash, giving a difference of jumps. steps: jumps for Cozy, and jumps for Dash, giving a difference of jumps.By the time we test steps, we notice that when the number of steps exceeds a multiple of , the difference in jumps increases. So, we have to find the next number that will increase the difference. doesn't because both both Cozy's and Dash's number of jumps increases, but does, and . actually gives a difference of 20 jumps, but goes back down to 19 (because Dash had to take another jump when Cozy didn't). We don't need to go any further because the difference will stay above 19 onward.Therefore, the possible numbers of steps in the staircase are , , and , giving a sum of . The sum of those digits is , so the answer is Solution 3We're looking for natural numbers such that .Let's call . We now have , or.Obviously, since , this will not work for any value under 6. In addition, since obviously , this will not work for any value over six, so we have and This can be achieved when and , or when and .Case One:We have and , so .Case Two:We have and , so .We then have , which has a digit sum of .Solution 4Translate the problem into following equation:Since , we havei.e.,We then have when or (dog's last jump has 2 steps and cat's last jump has 1 step), which yields and respectively.Another solution is when , which yields .Therefore, with , the digit sum is .以上就是小编对AMC10数学竞赛试题以及解析的介绍，希望对你有所帮助，更多学习资料请持续关注AMC数学竞赛网！
AMC10数学竞赛是美国高中数学竞赛中的一项，是针对高中一年级及初中三年级学生的数学测试，该竞赛开始于2000年，分A赛和B赛，于每年的2月初和2月中举行，学生可任选参加一项即可。不管是对高校申请还是今后在数学领域的发展都极其有利！那么接下来跟随小编来看一下AMC10的官方真题以及官方解答吧：Problem 19In , and . Squares and are constructed outside of the triangle. The points , and lie on a circle. What is the perimeter of the triangle?SolutionThe center of the circle lies on the perpendicular bisectors of both chords and . Therefore we know the center of the circle must also be the midpoint of the hypotenuse. Let this point be . Draw perpendiculars to and from , and connect and . . Let and . Then . Simplifying this gives . But by Pythagorean Theorem on , we know , because . Thus . So our equation simplifies further to . However , so , which means , or . Aha! This means is just an isosceles right triangle, so , and thus the perimeter is .Problem 20Erin the ant starts at a given corner of a cube and crawls along exactly 7 edges in such a way that she visits every corner exactly once and then finds that she is unable to return along an edge to her starting point. How many paths are there meeting these conditions? SolutionWe label the vertices of the cube as different letters and numbers shown above. We label these so that Erin can only crawl from a number to a letter or a letter to a number (this can be seen as a coloring argument). The starting point is labeled .If we define a "move" as each time Erin crawls along a single edge from 1 vertex to another, we see that after 7 moves, Erin must be on a numbered vertex. Since this numbered vertex cannot be 1 unit away from (since Erin cannot crawl back to ), this vertex must be .Therefore, we now just need to count the number of paths from to . To count this, we can work backwards. There are 3 choices for which vertex Erin was at before she moved to , and 2 choices for which vertex Erin was at 2 moves before . All of Erin's previous moves were forced, so the total number of legal paths from to is .以上就是小编对AMC10数学竞赛试题以及解析的介绍，希望对你有所帮助，更多学习资料请持续关注AMC数学竞赛网！
AMC10数学竞赛是美国高中数学竞赛中的一项，是针对高中一年级及初中三年级学生的数学测试，该竞赛开始于2000年，分A赛和B赛，于每年的2月初和2月中举行，学生可任选参加一项即可。不管是对高校申请还是今后在数学领域的发展都极其有利！那么接下来跟随小编来看一下AMC10的官方真题以及官方解答吧：Problem 17The centers of the faces of the right rectangular prism shown below are joined to create an octahedron. What is the volume of this octahedron?SolutionThe octahedron is just two congruent pyramids glued together by their base. The base of one pyramid is a rhombus with diagonals and , for an area . The height , of one pyramid, is , so the volume of one pyramid is . Thus, the octahedron has volume Problem 18Johann has fair coins. He flips all the coins. Any coin that lands on tails is tossed again. Coins that land on tails on the second toss are tossed a third time. What is the expected number of coins that are now heads?SolutionEvery time the coins are flipped, half of them are expected to turn up heads. The expected number of heads on the first flip is , on the second flip is is , and on the third flip it is . Adding these gives 以上就是小编对AMC10数学竞赛试题以及解析的介绍，希望对你有所帮助，更多学习资料请持续关注AMC数学竞赛网！
AMC10数学竞赛是美国高中数学竞赛中的一项，是针对高中一年级及初中三年级学生的数学测试，该竞赛开始于2000年，分A赛和B赛，于每年的2月初和2月中举行，学生可任选参加一项即可。不管是对高校申请还是今后在数学领域的发展都极其有利！那么接下来跟随小编来看一下AMC10的官方真题以及官方解答吧：ProblemThe town of Hamlet has people for each horse, sheep for each cow, and ducks for each person. Which of the following could not possibly be the total number of people, horses, sheep, cows, and ducks in Hamlet?Solution 15Solution 1Let the amount of people be , horses be , sheep be , cows be , and ducks be . We knowThen the total amount of people, horses, sheep, cows, and ducks may be written as . Looking through the options, we see is impossible to make for integer values of and . So the answer is .Solution 2As the solution above says, the total amount of people, horses, sheep, cows, and ducks may be written as . However, instead of going through each of the solutions and testing the the options, you can use the Chicken McNugget Theorem to find the greatest number of people, horses, sheep, cows, and ducks that cannot be written in the form .so our answer is .Problem 16Al, Bill, and Cal will each randomly be assigned a whole number from to , inclusive, with no two of them getting the same number. What is the probability that Al's number will be a whole number multiple of Bill's and Bill's number will be a whole number multiple of Cal's? SolutionWe can solve this problem with a brute force approach.If Cal's number is :If Bill's number is , Al's can be any of .If Bill's number is , Al's can be any of .If Bill's number is , Al's can be .If Bill's number is , Al's can be .Otherwise, Al's number could not be a whole number multiple of Bill's.If Cal's number is :If Bill's number is , Al's can be .Otherwise, Al's number could not be a whole number multiple of Bill's while Bill's number is still a whole number multiple of Cal's.Otherwise, Bill's number must be greater than , i.e. Al's number could not be a whole number multiple of Bill's.Clearly, there are exactly cases where Al's number will be a whole number multiple of Bill's and Bill's number will be a whole number multiple of Cal's. Since there are possible permutations of the numbers Al, Bill, and Cal were assigned, the probability that this is true is 以上就是小编对AMC10数学竞赛试题以及解析的介绍，希望对你有所帮助，更多学习资料请持续关注AMC数学竞赛网！
AMC10数学竞赛是美国高中数学竞赛中的一项，是针对高中一年级及初中三年级学生的数学测试，该竞赛开始于2000年，分A赛和B赛，于每年的2月初和2月中举行，学生可任选参加一项即可。不管是对高校申请还是今后在数学领域的发展都极其有利！那么接下来跟随小编来看一下AMC10的官方真题以及官方解答吧：Problem 13The line forms a triangle with the coordinate axes. What is the sum of the lengths of the altitudes of this triangle?SolutionWe find the x-intercepts and the y-intercepts to find the intersections of the axes and the line. If , then . If is, then . Our three vertices are , , and . Two of our altitudes are and , and since it is a 5-12-13 right triangle, the hypotenuse is . Since the area of the triangle is , so our final altitude is . The sum of our altitudes is . Note that there is no need to calculate the final answer after we know that the third altitude has length since is the only choice with a denominator of .Problem 14Let , , and be three distinct one-digit numbers. What is the maximum value of the sum of the roots of the equation ?Solution 1Expanding the equation and combining like terms results in . By Vieta's formula the sum of the roots is . To maximize this expression we want to be the largest, and from there we can assign the next highest values to and . So let , , and . Then the answer is .Solution 2Factoring out from the equation yields . Therefore the roots are and . Because must be the larger root to maximize the sum of the roots, letting and be and respectively yields the sum .Solution 3There are 2 cases. Case 1 is that and . Lets test that 1st. If , the maximum value for and is . Then and The next highest values are and so and . Therefore, .以上就是小编对AMC10数学竞赛试题以及解析的介绍，希望对你有所帮助，更多学习资料请持续关注AMC数学竞赛网！
AMC10数学竞赛是美国高中数学竞赛中的一项，是针对高中一年级及初中三年级学生的数学测试，该竞赛开始于2000年，分A赛和B赛，于每年的2月初和2月中举行，学生可任选参加一项即可。不管是对高校申请还是今后在数学领域的发展都极其有利！那么接下来跟随小编来看一下AMC10的官方真题以及官方解答吧：Problem 11Among the positive integers less than 100, each of whose digits is a prime number, one is selected at random. What is the probability that the selected number is prime?SolutionThe one digit prime numbers are , , , and . So there are a total of ways to choose a two digit number with both digits as primes and 4 ways to choose a one digit prime, for a total of ways. Out of these , , , , , , , and are prime. Thus the probability is .Problem 12For how many integers is the point inside or on the circle of radius 10 centered at ?SolutionThe equation of the circle is . Plugging in the given conditions we have . Expanding gives: , which simplifies to and therefore or . So ranges from to , for a total of values.以上就是小编对AMC10数学竞赛试题以及解析的介绍，希望对你有所帮助，更多学习资料请持续关注AMC数学竞赛网！
AMC10数学竞赛是美国高中数学竞赛中的一项，是针对高中一年级及初中三年级学生的数学测试，该竞赛开始于2000年，分A赛和B赛，于每年的2月初和2月中举行，学生可任选参加一项即可。不管是对高校申请还是今后在数学领域的发展都极其有利！那么接下来跟随小编来看一下AMC10的官方真题以及官方解答吧：Problem 9The shaded region below is called a shark's fin falcata, a figure studied by Leonardo da Vinci. It is bounded by the portion of the circle of radius and center that lies in the first quadrant, the portion of the circle with radius and center that lies in the first quadrant, and the line segment from to . What is the area of the shark's fin falcata?SolutionThe area of the shark's fin falcata is just the area of the quarter-circle minus the area of the semicircle. The quarter-circle has radius so it has area . The semicircle has radius so it has area . Thus, the shaded area is Problem 10What are the sign and units digit of the product of all the odd negative integers strictly greater than ?SolutionSince , the product must end with a .The multiplicands are the odd negative integers from to . There are of these numbers. Since , the product is negative.Therefore, the answer must be 以上就是小编对AMC10数学竞赛试题以及解析的介绍，希望对你有所帮助，更多学习资料请持续关注AMC数学竞赛网！
AMC10数学竞赛是美国高中数学竞赛中的一项，是针对高中一年级及初中三年级学生的数学测试，该竞赛开始于2000年，分A赛和B赛，于每年的2月初和2月中举行，学生可任选参加一项即可。不管是对高校申请还是今后在数学领域的发展都极其有利！那么接下来跟随小编来看一下AMC10的官方真题以及官方解答吧：Problem 7Consider the operation "minus the reciprocal of," defined by . What is ?Solution, so . Also, , so . Thus, Problem 8The letter F shown below is rotated clockwise around the origin, then reflected in the -axis, and then rotated a half turn around the origin. What is the final image?SolutionThe first rotation moves the base of the to the negative -axis, and the stem to the positive -axis. The reflection then moves the stem to the negative -axis, with the base unchanged. Then the half turn moves the stem to the positive axis and the base to the positive -axis, choice .以上就是小编对AMC10数学竞赛试题以及解析的介绍，希望对你有所帮助，更多学习资料请持续关注AMC数学竞赛网！
