AMC10 2018 A

AMC10 2018 A · Q7

AMC10 2018 A · Q7. It mainly tests Exponents & radicals, 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?

有整数 n（不一定是正整数），使得 $4000 \cdot \left(\frac{2}{5}\right)^n$ 为整数的有多少个不同的 n？

(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 $4000 = 2^5 \cdot 5^3$， $4000 \cdot \left(\frac{2}{5}\right)^n = 2^{5+n} \cdot 5^{3-n}。$ 该乘积为整数当且仅当两个因子 $2^{5+n}$ 和 $5^{3-n}$ 都是整数；这当且仅当两个指数都为非负时成立。因此，所给表达式为整数当且仅当 $5+n \ge 0$ 且 $3-n \ge 0$。解恰好是满足 $-5 \le n \le 3$ 的所有整数。这样的取值共有 $3-(-5)+1=9$ 个。

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