AMC12 2005 B

AMC12 2005 B · Q21

AMC12 2005 B · Q21. It mainly tests Primes & prime factorization, Counting divisors.

A positive integer $n$ has $60$ divisors and $7n$ has $80$ divisors. What is the greatest integer $k$ such that $7^k$ divides $n$?

一个正整数 $n$ 有 $60$ 个因数，且 $7n$ 有 $80$ 个因数。求最大的整数 $k$ 使得 $7^k$ 整除 $n$？

(A) 0 0

(B) 1 1

(D) 3 3

(E) 4 4

Answer

Correct choice: (C)

正确答案：（C）

Solution

We may let $n = 7^k \cdot m$, where $m$ is not divisible by 7. Using the fact that the number of divisors function $d(n)$ is multiplicative, we have $d(n) = d(7^k)d(m) = (k+1)d(m) = 60$. Also, $d(7n) = d(7^{k+1})d(m) = (k+2)d(m) = 80$. These numbers are in the ratio 3:4, so $\frac{k+1}{k+2} = \frac{3}{4} \implies k = 2 \Rightarrow \boxed{\mathrm{C}}$.

令 $n = 7^k \cdot m$，其中 $m$ 不被 $7$ 整除。利用因数个数函数 $d(n)$ 的乘法性，有 $d(n)=d(7^k)d(m)=(k+1)d(m)=60$。同时，$d(7n)=d(7^{k+1})d(m)=(k+2)d(m)=80$。这两个数的比为 $3:4$，所以 $\frac{k+1}{k+2}=\frac{3}{4}\implies k=2 \Rightarrow \boxed{\mathrm{C}}$。

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