1973 AMC 12 第 19 题

Question

对正数

n

与

a

，定义

n_a!

为

\begin{aligned} n_a!={}&n(n-a)(n-2a)\\ &\cdot(n-3a)\cdots(n-ka) \end{aligned}\text{，}

其中

k

是满足

n\gt ka

的最大整数。则商

\frac{72_8!}{18_2!}

等于

Define

n_a!

for positive

n

and

a

to be

\begin{aligned} n_a!={}&n(n-a)(n-2a)\\ &\cdot(n-3a)\cdots(n-ka), \end{aligned}

where

k

is the greatest integer for which

n\gt ka.

Then the quotient

\frac{72_8!}{18_2!}

is equal to

4^5

4^6

4^8

4^9

4^{12}

将两个广义阶乘都写成九个明确的因子

Write both generalized factorials as nine explicit factors

从分子的每个因子中提出

8

，从分母的每个因子中提出

2

Factor

8

from every numerator term and

2

from every denominator term

两个乘积为

\begin{aligned} 72_8!&=72\cdot64\cdots8 =8^9(9!),\\ 18_2!&=18\cdot16\cdots2 =2^9(9!) \end{aligned}\text{。}

因此它们的商为

\frac{8^9(9!)}{2^9(9!)}=4^9\text{。}

所以正确答案是 D。

The two products are

\begin{aligned} 72_8!&=72\cdot64\cdots8 =8^9(9!),\\ 18_2!&=18\cdot16\cdots2 =2^9(9!). \end{aligned}

Their quotient is therefore

\frac{8^9(9!)}{2^9(9!)}=4^9.

Therefore, the correct answer is D.

其他年份的第 19 题