美国数学竞赛AMC8专为8年级及以下的初中学生设计，但近年来的数据显示，越来越多小学4-6年级的考生加入到美国数学竞赛AMC8级别的考试行列中，而当这些学生能在成绩中取得“A”类标签，则是对孩子数学天赋的优势证明，不管是对美高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美国数学竞赛AMC8的官方真题以及官方解答吧：Problem 22Three members of the Euclid Middle School girls' softball team had the following conversation.Ashley: I just realized that our uniform numbers are all -digit primes.Brittany : And the sum of your two uniform numbers is the date of my birthday earlier this month.Caitlin: That's funny. The sum of your two uniform numbers is the date of my birthday later this month.Ashley: And the sum of your two uniform numbers is today's date.What number does Caitlin wear?SolutionThe maximum amount of days any given month can have is 31, and the smallest two digit primes are 11, 13, and 17. There are a few different sums that can be deduced from the following numbers, which are 24, 30, and 28, all of which represent the three days. Therefore, since Brittany says that the other two people's uniform numbers are earlier, so that means Caitlin and Ashley's numbers must add up to 24. Similarly, Caitlin says that the other two people's uniform numbers is later, so the sum must add up to 30. This leaves 28 as today's date. From this, Caitlin was referring to the uniform wearers 13 and 17, telling us that her number is 11, giving our solution as Problem 23One day the Beverage Barn sold cans of soda to customers, and every customer bought at least one can of soda. What is the maximum possible median number of cans of soda bought per customer on that day?SolutionIn order to maximize the median, we need to make the first half of the numbers as small as possible. Since there are people, the median will be the average of the and largest amount of cans per person. To minimize the first 49, they would each have one can. Subtracting these cans from the cans gives us cans left to divide among people. Taking gives us and a remainder of . Seeing this, the largest number of cans the person could have is , which leaves to the rest of the people. The average of and is . Thus our answer is .以上就是小编对AMC 8数学竞赛官方真题以及解析的介绍，希望对你有所帮助，更多AMC官方真题及学习资料请持续关注AMC数学竞赛网！
美国数学竞赛AMC8专为8年级及以下的初中学生设计，但近年来的数据显示，越来越多小学4-6年级的考生加入到美国数学竞赛AMC8级别的考试行列中，而当这些学生能在成绩中取得“A”类标签，则是对孩子数学天赋的优势证明，不管是对美高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美国数学竞赛AMC8的官方真题以及官方解答吧：Problem 21The -digit numbers and are each multiples of . Which of the following could be the value of ?SolutionThe sum of a number's digits is congruent to the number . must be congruent to 0, since it is divisible by 3. Therefore, is also congruent to 0. , so . As we know, , so , and therefore . We can substitute 2 for , so , and therefore . This means that C can be 1, 4, or 7, but the only one of those that is an answer choice is .Problem 22A -digit number is such that the product of the digits plus the sum of the digits is equal to the number. What is the units digit of the number?SolutionWe can think of the number as , where a and b are digits. Since the number is equal to the product of the digits () plus the sum of the digits (), we can say that . We can simplify this to , and factor to . Dividing by , we have that . Therefore, the units digit, , is 以上就是小编对AMC 8数学竞赛官方真题以及解析的介绍，希望对你有所帮助，更多AMC官方真题及学习资料请持续关注AMC数学竞赛网！
美国数学竞赛AMC8专为8年级及以下的初中学生设计，但近年来的数据显示，越来越多小学4-6年级的考生加入到美国数学竞赛AMC8级别的考试行列中，而当这些学生能在成绩中取得“A”类标签，则是对孩子数学天赋的优势证明，不管是对美高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美国数学竞赛AMC8的官方真题以及官方解答吧：Problem 19A cube with -inch edges is to be constructed from smaller cubes with -inch edges. Twenty-one of the cubes are colored red and are colored white. If the -inch cube is constructed to have the smallest possible white surface area showing, what fraction of the surface area is white?SolutionFor the least possible surface area that is white, we should have 1 cube in the center, and the other 5 with only 1 face exposed. This gives 5 square inches of white surface area. Since the cube has a surface area of 54 square inches, our answer is .Problem 20Rectangle has sides and . A circle of radius is centered at , a circle of radius is centered at , and a circle of radius is centered at . Which of the following is closest to the area of the region inside the rectangle but outside all three circles?SolutionThe area in the rectangle but outside the circles is the area of the rectangle minus the area of all three of the quarter circles in the rectangle.The area of the rectangle is . The area of all 3 quarter circles is . Therefore the area in the rectangle but outside the circles is . is approximately and substituting that in will give 以上就是小编对AMC 8数学竞赛官方真题以及解析的介绍，希望对你有所帮助，更多AMC官方真题及学习资料请持续关注AMC数学竞赛网！
美国数学竞赛AMC8专为8年级及以下的初中学生设计，但近年来的数据显示，越来越多小学4-6年级的考生加入到美国数学竞赛AMC8级别的考试行列中，而当这些学生能在成绩中取得“A”类标签，则是对孩子数学天赋的优势证明，不管是对美高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美国数学竞赛AMC8的官方真题以及官方解答吧：Problem 17George walks mile to school. He leaves home at the same time each day, walks at a steady speed of miles per hour, and arrives just as school begins. Today he was distracted by the pleasant weather and walked the first mile at a speed of only miles per hour. At how many miles per hour must George run the last mile in order to arrive just as school begins today?SolutionNote that on a normal day, it takes him hour to get to school. However, today it took hour to walk the first mile. That means that he has hours left to get to school, and mile left to go. Therefore, his speed must be , so is the answer.Problem 18Four children were born at City Hospital yesterday. Assume each child is equally likely to be a boy or a girl. Which of the following outcomes is most likelySolution 1We'll just start by breaking cases down. The probability of A occurring is . The probability of B occurring is .The probability of C occurring is , because we need to choose 2 of the 4 children to be girls.For D, there are two possible cases, 3 girls and 1 boy or 3 boys and 1 girl. The probability of the first case is because we need to choose 1 of the 4 children to be a boy. However, the second case has the same probability because we are choosing 1 of the 4 children to be a girl, so the total probability is .So out of the four fractions, D is the largest. So our answer is Solution 2The possibilities are listed out in the fourth row of Pascal's triangle, with the leftmost being the possibility of all boys and the rightmost being the possibility of all girls. Since the fourth row of Pascal's Triangle goesand are all the possibilities of two children from each gender, there are a total of possibilities of three children from one gender and one from the other. Since there are a total of total possibilities for the gender of the children, has the highest probability.以上就是小编对AMC 8数学竞赛官方真题以及解析的介绍，希望对你有所帮助，更多AMC官方真题及学习资料请持续关注AMC数学竞赛网！
美国数学竞赛AMC8专为8年级及以下的初中学生设计，但近年来的数据显示，越来越多小学4-6年级的考生加入到美国数学竞赛AMC8级别的考试行列中，而当这些学生能在成绩中取得“A”类标签，则是对孩子数学天赋的优势证明，不管是对美高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美国数学竞赛AMC8的官方真题以及官方解答吧：Problem 15The circumference of the circle with center is divided into equal arcs, marked the letters through as seen below. What is the number of degrees in the sum of the angles and ?SolutionFor this problem, it is useful to know that the measure of an inscribed angle is half the measure of its corresponding central angle. Since each unit arc is of the circle's circumference, each unit central angle measures . Then, we know that the inscribed arc of so ; and the inscribed arc of so . Problem 16The "Middle School Eight" basketball conference has teams. Every season, each team plays every other conference team twice (home and away), and each team also plays games against non-conference opponents. What is the total number of games in a season involving the "Middle School Eight" teams?SolutionWithin the conference, there are 8 teams, so there are pairings of teams, and each pair must play two games, for a total of games within the conference.Each team also plays 4 games outside the conference, and there are 8 teams, so there are a total of games outside the conference.Therefore, the total number of games is , so is our answer.以上就是小编对AMC 8数学竞赛官方真题以及解析的介绍，希望对你有所帮助，更多AMC官方真题及学习资料请持续关注AMC数学竞赛网！
美国数学竞赛AMC8专为8年级及以下的初中学生设计，但近年来的数据显示，越来越多小学4-6年级的考生加入到美国数学竞赛AMC8级别的考试行列中，而当这些学生能在成绩中取得“A”类标签，则是对孩子数学天赋的优势证明，不管是对美高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美国数学竞赛AMC8的官方真题以及官方解答吧：Problem 13If and are integers and is even, which of the following is impossible? and are even and are odd is even is odd none of these are impossibleSolutionSince is even, either both and are even, or they are both odd. Therefore, and are either both even or both odd, since the square of an even number is even and the square of an odd number is odd. As a result, must be even. The answer, then, is .Problem 14Rectangle and right triangle have the same area. They are joined to form a trapezoid, as shown. What is ?SolutionThe area of is . The area of is , which also must be equal to the area of , which, since , must in turn equal . Through transitivity, then, , and . Then, using the Pythagorean Theorem, you should be able to figure out that is a triangle, so , or .Solution 2The area of the rectangle is Since the parallel line pairs are identical, . Let be . is the area of the right triangle. Solving for , we get According to the Pythagorean Theorem, we have a 5-12-13 triangle. So, the hypotenuse has to be .以上就是小编对AMC 8数学竞赛官方真题以及解析的介绍，希望对你有所帮助，更多AMC官方真题及学习资料请持续关注AMC数学竞赛网！
美国数学竞赛AMC8专为8年级及以下的初中学生设计，但近年来的数据显示，越来越多小学4-6年级的考生加入到美国数学竞赛AMC8级别的考试行列中，而当这些学生能在成绩中取得“A”类标签，则是对孩子数学天赋的优势证明，不管是对美高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美国数学竞赛AMC8的官方真题以及官方解答吧：Problem 11Jack wants to bike from his house to Jill's house, which is located three blocks east and two blocks north of Jack's house. After biking each block, Jack can continue either east or north, but he needs to avoid a dangerous intersection one block east and one block north of his house. In how many ways can he reach Jill's house by biking a total of five blocks?SolutionWe can apply complementary counting and count the paths that DO go through the blocked intersection, which is . There are a total of paths, so there are paths possible. is the correct answer.Solution 2We can make a diagram of the roads available to Jack.Then, we can simply list the possible routes.There are 4 possible routes, so our answer is .Problem 12A magazine printed photos of three celebrities along with three photos of the celebrities as babies. The baby pictures did not identify the celebrities. Readers were asked to match each celebrity with the correct baby pictures. What is the probability that a reader guessing at random will match all three correctly?SolutionLet's call the celebrities A, B, and C. There is a chance that celebrity A's picture will be selected, and a chance that his baby picture will be selected. That means there are two celebrities left. There is now a chance that celebrity B's picture will be selected, and another chance that his baby picture will be selected. This leaves a chance for the last celebrity, so the total probability is . However, the order of the celebrities doesn't matter, so the final probability will be (B).Solution 2There is a chance that the reader will choose the correct baby picture for the first person. Next, the second person gives a chance, and the last person leaves only 1 choice. Thus, the probability is Solution 3There are ways assign the pictures to each of the celebrities. There is one favorable outcome where all of them are matched correctly, so the answer is .以上就是小编对AMC 8数学竞赛官方真题以及解析的介绍，希望对你有所帮助，更多AMC官方真题及学习资料请持续关注AMC数学竞赛网！
AMC 8数学竞赛专为8年级及以下的初中学生设计，但近年来的数据显示，越来越多小学4-6年级的考生加入到AMC 8级别的考试行列中，而当这些学生能在成绩中取得“A”类标签，则是对孩子数学天赋的优势证明，不管是对美高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AMC 8的官方真题以及官方解答吧：Problem 8Eleven members of the Middle School Math Club each paid the same amount for a guest speaker to talk about problem solving at their math club meeting. They paid their guest speaker . What is the missing digit of this -digit number?SolutionA number is divisible by if the difference between the sum of the digits in the odd-numbered slots (e.g. the ones slot, the hundreds slot, etc.) and the sum in the even-numbered slots (e.g. the tens slot, the thousands slot) is a multiple of . So is equivalent to . Clearly cannot be equal to or any multiple of greater than that. Also, if the expression is to be equal to a negative multiple of , must be 14 or greater, which violates the condition that A is a digit. So .Problem 9In , is a point on side such that and measures . What is the degree measure of ?SolutionBD = DC, so angle DBC = angle DCB = 70. Then CDB = 40. Since angle ADB and BDC are supplementary, .Problem 10The first AMC was given in and it has been given annually since that time. Samantha turned years old the year that she took the seventh AMC . In what year was Samantha born?SolutionThe seventh AMC 8 would have been given in 1991. If Samantha was 12 then, that means she was born 12 years ago, so she was born in 以上就是小编对AMC 8数学竞赛官方真题以及解析的介绍，希望对你有所帮助，更多AMC官方真题及学习资料请持续关注AMC数学竞赛网！