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 11A square-shaped floor is covered with congruent square tiles. If the total number of tiles that lie on the two diagonals is 37, how many tiles cover the floor?SolutionSince the number of tiles lying on both diagonals is , counting one tile twice, there are tiles on each side. Hence, our answer is .Problem 12The smallest positive integer greater than 1 that leaves a remainder of 1 when divided by 4, 5, and 6 lies between which of the following pairs of numbers?SolutionSince the remainder is the same for all numbers, then we will only need to find the lowest common multiple of the three given numbers, and add the given remainder (No fancy Chinese Remainder Theorem) . The is . Since , and that is in the range of 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 9All of Marcy's marbles are blue, red, green, or yellow. One third of her marbles are blue, one fourth of them are red, and six of them are green. What is the smallest number of yellow marbles that Marcy could have?SolutionThe green marbles and yellow marbles form of the total marbles. Now suppose the total number of marbles is . We know the number of yellow marbles is and a positive integer. Therefore, must divide . Trying the smallest multiples of for , we see that when , we get there are yellow marbles, which is impossible. However when , there are yellow marbles, which must be the smallest possible.Problem 10A box contains five cards, numbered 1, 2, 3, 4, and 5. Three cards are selected randomly without replacement from the box. What is the probability that 4 is the largest value selected?SolutionThere are possible groups of cards that can be selected. If is largest card selected, then the other two cards must be either , , or , for a total groups of cards. Then the probability is just 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 7Let be a 6-digit positive integer, such as 247247, whose first three digits are the same as its last three digits taken in the same order. Which of the following numbers must also be a factor of ?Solution 1Let Clearly, is divisible by .Solution 2We can see that numbers like can be written as . We can see that the alternating sum of digits is , which is . Because is a multiple of , any number is a multiple of , so the answer is -Baolan (hi MVMS)Solution 3The most important step is to realize that any number in the form . Thus every number in this form is divisible by , and the answer is , because it is the only choice that is a factor of .Problem 8Malcolm wants to visit Isabella after school today and knows the street where she lives but doesn't know her house number. She tells him, "My house number has two digits, and exactly three of the following four statements about it are true."(1) It is prime.(2) It is even.(3) It is divisible by 7.(4) One of its digits is 9.This information allows Malcolm to determine Isabella's house number. What is its units digit?SolutionNotice that (1) cannot be true. Otherwise, the number would have to prime and either be even or divisible by 7. This only happens if the number is 2 or 7, neither of which are two-digit numbers, so we run into a contradiction. Thus, we must have (2), (3), and (4) true. By (2), the -digit number is even, and thus the digit in the tens place must be . The only even -digit number starting with and divisible by is , which has a units digit of 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，更多学习资料请持续关注AMC美国数学竞赛网！
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 5What is the value of the expression ?Solution 1Directly calculating:We evaluate both the top and bottom: . This simplifies to .Solution 2It is well known that the sum of all numbers from to is . Therefore, the denominator is equal to . Now we can cancel the factors of , , and from both the numerator and denominator, only leaving . This evaluates to .Problem 6If the degree measures of the angles of a triangle are in the ratio , what is the degree measure of the largest angle of the triangle?SolutionThe sum of the ratios is . Since the sum of the angles of a triangle is , the ratio can be scaled up to . The numbers in the ratio represent the angles of the triangle. We want the largest, 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，更多学习资料请持续关注AMC美国数学竞赛网
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 3What is the value of the expression √16√8√4?SolutionProblem 4When 0.000315 is multiplied by 7928564 the product is closest to which of the following?SolutionWe can approximate 7928564 to 8000000 and 0.000315 to 0.0003 Multiplying the two yields 2400 This gives our answer to be （D）2400.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，更多学习资料请持续关注AMC美国数学竞赛网！
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 1Which of the following values is largest?Solution 1We compute each expression individually according to the order of operations. We get 2+0+1+7=10,2*0+1+7=8,2+0*1+7=9,2+0+1*7=9,2*0*1*7=0,. Since 10 is the greatest out of these numbers, (A)2+0+1+7 is the answer.Problem 2Alicia, Brenda, and Colby were the candidates in a recent election for student president. The pie chart below shows how the votes were distributed among the three candidates. If Brenda received 36 votes, then how many votes were cast all together?Solution 1Let x be the total amount of votes casted. From the chart, Brenda received 30% of the votes and had 36 votes. We can express this relationship as 30/100x=36. Solving for x, we get x=(E)120.Solution 2We're being asked for the total number of votes cast -- that represents 100% of the total number of votes (obviously). Brenda received 36 votes, which is 30/100=3/10 of the total number of votes. Multiplying 36 by 10/3 we get the total number of votes, which is (E)120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数学竞赛试题及答案吧：考题1：需要多少张地毯才能将12英尺长9英尺宽的举行地板覆盖？（地毯为边长3英尺的正方形）解决方案1：首先，我们可以算出地板的面积12*9。要做到这一点你需要108平方英尺的地毯。既然有3*3=9平方英尺地毯，你把108/9=12张地毯，所以我们的答案是(A)12。解决方案2：首先，地毯边长为3，地板宽为9，所以需要9/3=3；地板长为12，所以需要12/3=4，因此总共需要3*4=12张地毯，所以我们的答案是(A)12。考题2：点O是正八边形的中心ABCDEFGH，X是侧线AB的中点八边形区域阴影部分占了总表面积的多少？解决方案1：由于八边形ABCDEFGH是正八边形，因此它被分成8相等的部分，例如三角形△ABO,△BCO,△CDO等。这些部分由于它们都相等，所以八边形的每个部分都是1/8。阴影区域由3个这些相等的部分加上另一半的部分组成，因此阴影的八边形部分是1/8+1/8+1/8+1/16=（D）7/16。解决方案2：八边形被划分为16个相同的三角形（因此它们各自具有相等的面积）。由于阴影区域占据7了16总三角形，答案是（D）7/16。解决方案3：对于初学者我觉得有用的是将整个八边形分成三角形，如下所示：现在只需要计算较大的三角形就可以记住△BOX与△XOA不是完整的三角形，只是这些三角形的一半。我们将它计算在内，我们总共得到了3.5/8阴影形状。然后我们简化它以获得我们的答案(D)7/16。更多学习资料请持续关注AMC数学竞赛网！
前一段时间，复赛AIME已经考完了，2017~2018年AMC这一轮比赛已经结束。那么接下来要准备的就是2018~2019年的AMC考试了，想到有一些新加入的家长对整个考试流程不是很熟悉，今天AMC美国数学竞赛网来给大家送上一份AMC12的全年备考计划，值得一看：AMC12考试时间为：每年的2月份AMC12备考条件在AMC10要求的基础上，再深入学习中学竞赛内容，进行难度和解题技巧的训练AMC12备考计划建议10年级-12年级学完高中课内数学知识的学生参赛基础一般的学生建议从6月开始按部就班地复习，逐步解决英文数学词汇、长难句理解、指数函数与对数函数、数列、不等式、多项式、解析几何、平面几何、立体几何、排列组合、概率统计、数论综合等各个模块的知识盲区，反复多次的把至少近十年的真题吃透基础较好的学生建议从10月开始有计划的把最近十年的真题刷一遍，计时做完后对照答案进行订正，独立思考找出自己做题时犯的错误，自己培养批判性的竞赛思维以上就是AMC美国数学竞赛网为大家分享的关于2018~2019年的AMC12考试备考计划的详细介绍，更多AMC资讯、AMC真题、AMC培训请持续关注AMC美国数学竞赛网。
