深圳国际初中课程排名前十

各位期待已久的下半场2018AMC8真题解析来啦。本文将为大家分享2018AMC816-25题的中英文双语解答，AMC备赛不易，希望这份分析可以帮助大家查漏补缺数学知识储备情况。

（点击阅读上半场2018AMC8真题解析）

Problem 16

Answer: C

Solution:

Since the Arabic books and Spanish books have to be kept together, we can treat them both as just one book. That means we're trying to find the number of ways you can arrange one Arabic book, one Spanish book, and three German books, which is just 5 factorial. Now we multiply this product by 2! because there are 2 factorial ways to arrange just the Arabic books, and 4! ways to arrange just the Spanish books. Multiplying all these together, we have 5!*2!*4!=5760.

中文解析：

用捆绑法， 把2本Arabic 书看做一个整体， 4本Spanish书看做一个整体。这样Arabic， Spanish, 以及3本German书做排列，是5！。此外，2本Arabic内部有2！个排列方式， 4本Spanish书有4！个排列方式。因此本题的排列方法数是：5!*2!*4!=5760. 答案是C。

Problem 17

Answer: A

Solution:

Since Ella rides 5 times as fast as Bella, Ella rides at a rate of 25/2. Together, they move 15 feet towards each other every unit. Dividing 10560 by 15 to find the number of steps Bella takes results in the answer of 704.

中文解析：

Ella的速度是Bella的5倍，Bella每步走5/2 feet, Ella虽然骑自行车，我们可以理解为Ella每步走25/2 feet. 假设走x步两人相遇，因此有等式: 5/2 *x +25/2 * x =10560. 求得x=704. 答案是A。

Problem 18

Answer: E

Solution:

We can first find the prime factorization of 23232, which is 2^6 * 3^1 * 11^2 . Now, we just add one to our powers and multiply. Therefore, the answer is (6+1)*(1+1)*(2+1)=7*2*3=42.

中文解析：

写出23，232的质因数序列：23232=2*2*2*2*2*2*3*11*11. 因此她的因子的个数是: (6+1)*(1+1)*(2+1)=42. 答案是E。

Problem 19

Answer: C

Solution:

The sign of the next row on the pyramid depends on previous row. There are two options for the previous row, - or +. There are three rows to the pyramid that depend on what the top row is. Therefore, the ways you can make a + on the top is 2*2*2=8, so the answer is 8

中文解析：

第一行（最上面的一行）是+， 则第二行可以有两种方案：两个+ 或两个-。第二行的两个+和两个-分别对应第三行两种符号安排， 即第三行有2*2=4种安排。第三行的每种符号安排，都对应第四行的两种安排，因此第四行有4*2=8种安排。答案是C。

Problem 20

Answer: A

Solution:

By similar triangles, we have ADE=1/9ABC. Similarly, we see that BEF=4/9ABC. Using this information, we get ACFE=5/9 ABC. Then, since ADE=1/9 ABC, it follows that the CDEF=4/9 ABC. Thus, the answer would be 4/9.

中文解析:

由于DE//BC， 且AE：AB=1：3， 因此三角形ADE和三角形ABC相似， 其面积的比值是 1：9.有与EF//AC， 且BE：AB=2：3， 因此三角形BEF和三角形ABC相似，其面积的比值是4：9.平行四边形CDEF的面积=三角形ABC的面积-三角形BEF的面积-三角形AED的面积，可以认为是9-1-4=5.因此，题目所求四边形CDEF的面积和三角形ABC的面积的比值是：5：9. 答案是C。

Problem 21

Answer: E

Solution:

Looking at the values, we notice that 11-7=4, 9-5=4, 6-2=4. This means we are looking for a value that is four less than a multiple of 11, 9, 6. The least common multiple of (11,9,6) is 198. So the numbers that fulfill this can be written as 198k-4, where k is a positive integer. This value is only a three digit integer when k is 1,2,3,4,5, which gives 194, 392, 590, 788, 986 respectively. Thus we have 5 values, so our answer is 5.

中文解析：

假设这个数是n. n被6除的余数是2， 注意2+4=6；n被9除的余数是5，注意5+4=9；n被11除的余数是7， 注意7+4=11.我们可以发现n+4 能够被6，9， 11整除。则6，9，11的公倍数满足整除的要求。其最小公倍数是198， 三位数中的倍数有：198， 198*2， 198*3， 198*4, 198*5. 共5个数。这些公倍数-4之后即满足题目条件。因此答案是E。

Problem 22

Answer: B

Solution:

Let the area of CEF be x. Thus, the area of triangle ACD is 45+x and the area of the square is 2*(45+x)=90+2x.CEF is similar with ABF with a 1:2 ratio, so the area of triangle ABF is 4x. Now consider trapezoid ABED. Its area is 45+4x. which is 3/4 the area of the square. We set up an equation in x: 45+4x=3/4 *(90+2x) . Solving, we get x=9. The area of square ABCD is 90+2x=90+2*9=108.

中文解析：

假设三角形CEF的面积是x. 则三角形ADC的面积是45+x, 正方形ABCD的面积是2*(45+x）,即90+2x。三角形CEF和三角形ABF是相似三角形，由于CE：AB=1：2，因此其面积的比值是：1：4. 则三角形ABF的面积是4x.三角形CEB是正方形ABCD的1/4. 因此面积是：1/4*(90+2x).三角形CEB的面积+三角形ABF的面积+四边形AFED的面积=正方形ABCD的面积， 因此有等式: 1/4 *(90+2x)+4x+45=90+2x. 求得x=9. 因此正方形ABCD的面积是：90+2x=108. 答案是B。

Problem 23

Answer: D

Solution:

We can decide 2 adjacent points with 8 choices. The remaining point will have 6 choices. However, we have counted the case with 3 adjacent points twice, so we need to subtract this case once (8*6-8=40). The case with the 3 adjacent points has 8 arrangements(C(8,3)=56), so our answer is 40/56=5/7.

中文解析：

我们按照下面的额规则组成三角形：使用八边形的一条边，再任选一个其他的点（有6个点可选)。可以组成三角形的数量是：8*6=48. 但是相邻的三个点组成的三角形会被计入了2次. 因此需要48-8=40.由八边形上任意三个点可以组成C（8，3）=56个三角形。因此题目所求的概率是：40/56=5/7. 答案是D。

Problem 24

Answer: C

Solution:

Note that EJCI is a rhombus by symmetry. Let the side length of the cube be s. By the Pythagorean theorem, EC=sqrt(3)*s and JI=sqrt(2)*s. Since the area of a rhombus is half the product of its diagonals, the area of the cross section is s*s*sqrt(6)/2. This gives R=sqrt(6)/2. Thus R*R=3/2.

中文解析：

假设正方体的边长是s。注意到EJCI的四条边相等，是个菱形。其对角线分别是ＪＩ和CE。JI的长度等于BD的长度。BD=sqrt(2) *s。在直角三角形ECA中， AC=sqrt(2)*s, AE =s, 根据勾股定理， AC=sqrt(3)*s.根据菱形的面积公式: 1/2 * JI *AC =1/2 *sqrt(2)*s*sqrt(3)*s=1/2 sqrt(6) *s.正方体的一个面的面积是s*s.题目所求R*R=（1/4*6*s*s)/(s*s)=3/2. 答案是C。

Problem 25

Answer: E

Solution:

We compute 2^8+1=256+1=257. We're all familiar with what 6*6*6 is, namely 216, which is too small. The smallest cube greater than it is 7*7*7. We notice that 2^18= （2^6）^3=64^3 , which therefore clearly will be the largest cube less than 2^18+1. So, the required number of cubes is 64-7+1=58.

中文解析：

2的8次方 +1 =256+1=257. 我们熟悉的6*6*6=216， 比257小， 7*7*7=343 满足条件比257大。2的18次方可以写成2的6次方再3次方，即64的3次方。题目所求的perfect cube 应该包含7的三次方，8的三次方，。。。。64的三次方。共有：64-7+1=58个。答案是E。

上文就是完整版本的2018AMC8真题解答啦。看过比赛的真实难度，大家觉得明年参赛是游刃有余，还是忐忑不安呢？

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