AMC12 2018 A

AMC12 2018 A · Q7

AMC12 2018 A · Q7. It mainly tests Fractions, Primes & prime factorization.

For how many (not necessarily positive) integer values of $n$ is the value of $4000 \cdot \left(\frac{2}{5}\right)^n$ an integer?

$4000 \cdot \left(\frac{2}{5}\right)^n$ 的值为整数的整数 $n$（不一定是正整数）有多少个？

(A) 3 3

(B) 4 4

(D) 8 8

(E) 9 9

Answer

Correct choice: (E)

正确答案：（E）

Solution

Answer (E): Because $4000=2^5\cdot 5^3$, $$4000\cdot \left(\frac{2}{5}\right)^n = 2^{5+n}\cdot 5^{3-n}.$$ This product will be an integer if and only if both of the factors $2^{5+n}$ and $5^{3-n}$ are integers, which happens if and only if both exponents are nonnegative. Therefore the given expression is an integer if and only if $5+n\ge 0$ and $3-n\ge 0$. The solutions are exactly the integers satisfying $-5\le n\le 3$. There are $3-(-5)+1=9$ such values.

答案（E）：因为 $4000=2^5\cdot 5^3$， $$4000\cdot \left(\frac{2}{5}\right)^n = 2^{5+n}\cdot 5^{3-n}.$$ 当且仅当因子 $2^{5+n}$ 和 $5^{3-n}$ 都是整数时，该乘积才是整数；而这当且仅当两个指数都非负时成立。因此，当且仅当 $5+n\ge 0$ 且 $3-n\ge 0$ 时，给定表达式为整数。解恰为满足 $-5\le n\le 3$ 的整数。这样的取值共有 $3-(-5)+1=9$ 个。

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