Question

记号 $n!$ （读作“ $n$ 的阶乘”）定义为前 $n$ 个正整数的乘积。（例如， $3! = 1 \cdot 2 \cdot 3 = 6$ 。）定义正整数的“超级阶乘” $n^!$ 为前 $n$ 个正整数的阶乘的乘积。（例如， $3^! = 1! \cdot 2! \cdot 3! = 12$ 。） $51^!$ （也就是 $51$ 的超级阶乘）的质因数分解中含有多少个因数 $7$ ？

The notation $n!$ (read “ $n$ factorial”) is defined as the product of the first $n$ positive integers. (For example, $3! = 1 \cdot 2 \cdot 3 = 6.$ ) Define the superfactorial of a positive integer, denoted by $n^!,$ to be the product of the factorials of the first $n$ integers. (For example, $3^! = 1! \cdot 2! \cdot 3! = 12.$ ) How many factors of $7$ appear in the prime factorization of $51^!,$ the superfactorial of $51?$

$147$

$150$

$156$

$168$

$171$

答案：E

知识点：勒让德公式阶乘

难度评级：1820

视频讲解：

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文字解答：

对每个 $n!$ ，因数 $7$ 的个数是 $\lfloor n/7\rfloor+\lfloor n/49\rfloor$ 。因此 $51^{!}$ 中 $7$ 的指数是 $\sum_{n=1}^{51}\lfloor n/7\rfloor+\sum_{n=1}^{51}\lfloor n/49\rfloor$ 。第一个和为 $7(1+2+3+4+5+6)+3\cdot7$ $=168$ ，第二个和为 $3$ 。总数是 $171$ 。

For each $n!$ , the number of factors of $7$ is $\lfloor n/7\rfloor+\lfloor n/49\rfloor$ . Therefore the exponent of $7$ in $51^{!}$ is $\sum_{n=1}^{51}\lfloor n/7\rfloor+\sum_{n=1}^{51}\lfloor n/49\rfloor$ . The first sum is $7(1+2+3+4+5+6)+3\cdot7$ $=168$ , and the second sum is $3$ . The total is $171$ .

2026 AMC 8 第 24 题

答案：E

难度评级：1820

视频讲解：

文字解答：

其他年份的第 24 题