AMC8 2026

AMC8 2026 · Q24

AMC8 2026 · Q24. It mainly tests Primes & prime factorization, Counting divisors.

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 $51$?

符号 $n!$（读作“n 的阶乘”）定义为前 $n$ 个正整数的乘积。（例如，$3! = 1 \cdot 2 \cdot 3 = 6$）。定义正整数的超阶乘，记为 $n^!$，为前 $n$ 个整数的阶乘的乘积。（例如，$3^! = 1! \cdot 2! \cdot 3! = 12$）。$51^!$（51 的超阶乘）在素因数分解中包含多少个因子 7？

(A) 147 147

(B) 150 150

(D) 168 168

(E) 171 171

Answer

Correct choice: (E)

正确答案：（E）

Solution

The superfactorial $51!^1$ is defined as \[ 51!^1 = 1! \times 2! \times 3! \times \cdots \times 51!. \] The exponent of 7 in the prime factorization of $51!^1$ is \[ \sum_{k=1}^{51} v_7(k!), \] where $v_7(k!)$ is the exponent of 7 in $k!$. We know that \[ v_7(k!) = \left\lfloor \frac{k}{7} \right\rfloor + \left\lfloor \frac{k}{49} \right\rfloor + \left\lfloor \frac{k}{343} \right\rfloor + \cdots. \] Since $343 > 51$, only the first two terms matter. Thus, the total exponent is \[ \sum_{k=1}^{51} \left( \left\lfloor \frac{k}{7} \right\rfloor + \left\lfloor \frac{k}{49} \right\rfloor \right) = \sum_{k=1}^{51} \left\lfloor \frac{k}{7} \right\rfloor + \sum_{k=1}^{51} \left\lfloor \frac{k}{49} \right\rfloor. \] **First sum:** $\sum_{k=1}^{51} \left\lfloor k/7 \right\rfloor$ - $k=7$ to $13$: $\left\lfloor k/7 \right\rfloor = 1$ (7 terms) $\to 7 \times 1 = 7$ - $k=14$ to $20$: $=2$ (7 terms) $\to 14$ - $k=21$ to $27$: $=3$ (7 terms) $\to 21$ - $k=28$ to $34$: $=4$ (7 terms) $\to 28$ - $k=35$ to $41$: $=5$ (7 terms) $\to 35$ - $k=42$ to $48$: $=6$ (7 terms) $\to 42$ - $k=49$ to $51$: $=7$ (3 terms) $\to 21$ Total: $7 + 14 + 21 + 28 + 35 + 42 + 21 = 168$. **Second sum:** $\sum_{k=1}^{51} \left\lfloor k/49 \right\rfloor$ Only $k=49,50,51$ give 1, so the sum is $3$. **Overall exponent:** $168 + 3 = 171$. Therefore, the answer is $\boxed{\textbf{(E) 171}}$.

超阶乘 $51^!$ 定义为 \[ 51^! = 1! \times 2! \times 3! \times \cdots \times 51!。 \] $51^!$ 在素因数分解中 7 的指数为 \[ \sum_{k=1}^{51} v_7(k!)， \] 其中 $v_7(k!)$ 表示 $k!$ 中因子 7 的指数。已知 \[ v_7(k!) = \left\lfloor \frac{k}{7} \right\rfloor + \left\lfloor \frac{k}{49} \right\rfloor + \left\lfloor \frac{k}{343} \right\rfloor + \cdots。 \] 因为 $343 > 51$，所以只需考虑前两项。总指数为 \[ \sum_{k=1}^{51} \left( \left\lfloor \frac{k}{7} \right\rfloor + \left\lfloor \frac{k}{49} \right\rfloor \right) = \sum_{k=1}^{51} \left\lfloor \frac{k}{7} \right\rfloor + \sum_{k=1}^{51} \left\lfloor \frac{k}{49} \right\rfloor。 \] **第一部分:** $\sum_{k=1}^{51} \left\lfloor \frac{k}{7} \right\rfloor$ - $k=7$ 到 $13$：$\left\lfloor \frac{k}{7} \right\rfloor = 1$ （7 个数）$\to 7 \times 1 = 7$ - $k=14$ 到 $20$：$=2$ （7 个数）$\to 14$ - $k=21$ 到 $27$：$=3$ （7 个数）$\to 21$ - $k=28$ 到 $34$：$=4$ （7 个数）$\to 28$ - $k=35$ 到 $41$：$=5$ （7 个数）$\to 35$ - $k=42$ 到 $48$：$=6$ （7 个数）$\to 42$ - $k=49$ 到 $51$：$=7$ （3 个数）$\to 21$ 总和：$7 + 14 + 21 + 28 + 35 + 42 + 21 = 168$。 **第二部分:** $\sum_{k=1}^{51} \left\lfloor \frac{k}{49} \right\rfloor$ 只有 $k=49, 50, 51$ 时为 1，和为 $3$。 **总指数：** $168 + 3 = 171$。因此，答案为 $\boxed{\textbf{(E) 171}}$。

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