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OG中的新GRE数学题解析（13）
小编在此与大家分享新GRE OG中与新GRE数学的相关内容，希望各位在2015年能够搞定GRE数学真题。
17.Let S be the set of all positive integers n such that n2 is a multiple of both 24 and 108. Which of the following integers are divisors of every integer n in S ?
Indicate all such integers.
A 12
B 24
C 36
D 72
这个题有点小复杂，我先把OG上的解答贴上来。再写我自己的
To determine which of the integers in the answer choices is a divisor of every positive integer n in S, you must first understand the integers that are in S. Note that in this question you are given information about n2, not about n itself. Therefore, you must use the information about n2 to derive information about n. The fact that n2 is a multiple of both 24 and 108 implies that n2 is a multiple of the least common multiple of 24 and 108. To determine the least common multiple of 24 and 108, factor 24 and 108 into prime factors as (23)(3) and (22)(33), respectively. Because these are prime factorizations, you can conclude that the least common
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multiple of 24 and 108 is (23)(33). Knowing that n2 must be a multiple of (23)(33) does not mean that every multiple of (23)(33) is a possible value of n2, because n2 must be the square of an integer. The prime factorization of a square number must contain only even exponents. Thus, the least multiple of (23)(33) that is a square is (24)(34). This is the least possible value of n2, and so the least possible value of n is (22)(32), or 36. Furthermore, since every value of n2 is a multiple of (24)(34), the values of n are the positive multiples of 36; that is, S{36, 72, 108, 144, 180, . . .} .The question asks for integers that are divisors of every integer n in S, that is, divisors of every positive multiple of 36. Since Choice A, 12, is a divisor of 36, it is also a divisor of every multiple of 36. The same is true for Choice C, 36. Choices B and D, 24 and 72, are not divisors of 36, so they are not divisors of every integer in S. The correct answer consists of Choices A and C.
这个意思就是先求出24和108的最小公倍数，然后通过加倍使其成为一个整数的平方，这样就可以找出一系列的n了，这些n的公公因数应该有哪些?我找了前面的两个，36和72，所以AC可以选出来了，之后的所有数肯定包含了这两个选项，而BD因为不满足前面这两数，所以就排除了。
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