Question

对某个正整数 $n$ ，数 $110n^3$ 有 $110$ 个正整数因数，包括 $1$ 和 $110n^3$ 本身。数 $81n^4$ 有多少个正整数因数？

For some positive integer $n,$ the number $110n^3$ has $110$ positive integer divisors, including $1$ and the number $110n^3.$ How many positive integer divisors does the number $81n^4$ have?

$110$

$191$

$261$

$325$

$425$

答案：D

知识点：因数个数质因数分解

难度评级：2110

解答：

质因数分解为 $110=2\cdot5\cdot11$ 。若 $110n^3$ 有 $110=2\cdot5\cdot11$ 个因数，则它恰有三个素因子，指数为 $1,4,10$ 的某种排列。

对于素数 $2,5,11$ ，它们在 $110n^3$ 中的指数都是 $1$ 加上 $3$ 的倍数。指数 $1,4,10$ 都满足这个形式，所以它们在 $n$ 中对应为 $0,1,3$ 的某种排列。

在 $81n^4=3^4n^4$ 中，来自 $n^4$ 的指数为 $0,4,12$ ，另有素数 $3$ 的指数 $4$ 。因数个数为 $\begin{aligned} &(0+1)\cdot(4+1)\cdot(12+1) \\ &\quad {}\cdot(4+1)=325. \end{aligned}$

所以正确答案是 D。

The prime factorization is $110=2\cdot5\cdot11$ . If $110n^3$ has $110=2\cdot5\cdot11$ divisors, then it has exactly three prime factors, with exponents $1,4,10$ in some order.

For each of the primes $2,5,11$ , its exponent in $110n^3$ is $1$ more than a multiple of $3$ . The exponents $1,4,10$ all have this form, so the corresponding exponents in $n$ are $0,1,3$ in some order.

In $81n^4=3^4n^4$ , the exponents from $n^4$ are therefore $0,4,12$ , in some order, along with the exponent $4$ on prime $3$ . Hence the divisor count is $\begin{aligned} &(0+1)\cdot(4+1)\cdot(12+1) \\ &\quad {}\cdot(4+1)=325. \end{aligned}$

Thus, the correct answer is D.

2016 AMC 10A 第 22 题

答案：D

难度评级：2110

解答：

其他年份的第 22 题