Question

令 $D(n)$ 表示将正整数 $n$ 写成乘积的方法数，其中 $k \ge 1$ ，每个 $f_i$ 都是严格大于 $1$ 的整数，并且因子列出的顺序有区别（也就是说，只差因子顺序的两个表示也计为不同）。例如，数 $6$ 可以写成 $6$ 、 $2 \cdot 3$ 和 $3 \cdot 2$ ，所以 $D(6) = 3$ 。求 $D(96)$ 。 $n = f_1 \cdot f_2 \cdots f_k,$

Let $D(n)$ denote the number of ways of writing the positive integer $n$ as a product $n = f_1 \cdot f_2 \cdots f_k,$ where $k \ge 1,$ the $f_i$ are integers strictly greater than $1,$ and the order in which the factors are listed matters (that is, two representations that differ only in the order of the factors are counted as distinct). For example, the number $6$ can be written as $6,$ $2 \cdot 3,$ and $3 \cdot 2,$ so $D(6) = 3.$ What is $D(96)?$

$112$

$128$

$144$

$172$

$184$

答案：A

知识点：递推计数因数

难度评级：2220

解答：

第一个因子 $f_1$ 可以是任意因子 $d \gt 1$ 之后剩余部分是 $n/d$ 的有序分解。因此 $D(n) = \sum_{d \mid n,\, d \gt 1} D(n/d)$ 并设 $D(1) = 1$ 。

在 $96 = 2^5\cdot 3$ 的因子上逐步计算： $D(2) = D(3) = 1$ $D(4) = 2$ $D(6) = 3$ $D(8) = 4$ $D(12) = 8$ $D(16) = 8$ $D(24) = 20$ $D(32) = 16$ $D(48) = 48$ 。

最后 $\begin{gathered} D(96) = D(48) + D(32) \\ {}+ D(24) + D(16) + D(12) \\ {}+ D(8) + D(6) + D(4) \\ {}+ D(3) + D(2) + D(1) \\ = 48 + 16 + 20 + 8 \\ {}+ 8 + 4 + 3 + 2 \\ {}+ 1 + 1 + 1 = 112. \end{gathered}$

所以正确答案是 A。

The first factor $f_1$ can be any divisor $d \gt 1,$ after which the rest is an ordered factorization of $n/d.$ So $D(n) = \sum_{d \mid n,\, d \gt 1} D(n/d),$ with $D(1) = 1.$

Computing over the divisors of $96 = 2^5\cdot 3:$ $D(2) = D(3) = 1,$ $D(4) = 2,$ $D(6) = 3,$ $D(8) = 4,$ $D(12) = 8,$ $D(16) = 8,$ $D(24) = 20,$ $D(32) = 16,$ $D(48) = 48.$

Finally $\begin{gathered} D(96) = D(48) + D(32) \\ {}+ D(24) + D(16) + D(12) \\ {}+ D(8) + D(6) + D(4) \\ {}+ D(3) + D(2) + D(1) \\ = 48 + 16 + 20 + 8 \\ {}+ 8 + 4 + 3 + 2 \\ {}+ 1 + 1 + 1 = 112. \end{gathered}$

Thus, the correct answer is A.

2020 AMC 12B 第 24 题

答案：A

难度评级：2220

解答：

其他年份的第 24 题