2026 AMC 8 第 24 题

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

小提示：

先数每个阶乘中有多少个因数

7

。

Count how many times multiples of

7

appear in the factorials

大提示：

需要求和

\sum_{n=1}^{51}(\lfloor \frac{n}{7}\rfloor+\lfloor \frac{n}{49}\rfloor)

。

Use

\sum_{n=1}^{51}(\lfloor \frac{n}{7}\rfloor+\lfloor \frac{n}{49}\rfloor)

视频讲解：

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文字解答：

对每个

n!

，因数

7

的个数是

\lfloor \frac{n}{7}\rfloor+\lfloor \frac{n}{49}\rfloor

。因此

51^{!}

中

7

的指数是

\sum_{n=1}^{51}\lfloor \frac{n}{7}\rfloor+\sum_{n=1}^{51}\lfloor \frac{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 \frac{n}{7}\rfloor+\lfloor \frac{n}{49}\rfloor

. Therefore the exponent of

7

in

51^{!}

is

\sum_{n=1}^{51}\lfloor \frac{n}{7}\rfloor+\sum_{n=1}^{51}\lfloor \frac{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

.