AMC 8 2026 Problem 24

AMC8 2026 年第 24 题

整除与因数★★★★☆
The notation n! (read "nn factorial") is defined as the product of the first nn positive integers. (For example, 3!=1⋅2⋅3=63! = 1 \cdot 2 \cdot 3 = 6.) Define the superfactorial of a positive number, denoted by n!n^!, to be the product of the first nn factorials. (For example, 3!=1!⋅2!⋅3!=123^! = 1! \cdot 2! \cdot 3! = 12.) How many factors of 7 appear in the prime factorization of 51!51^!, the superfactorial of 51?
记号 n!（读作"nn 阶乘"）定义为前 nn 个正整数的乘积（例如 3!=1⋅2⋅3=63! = 1 \cdot 2 \cdot 3 = 6）。定义超阶乘 n!n^! 为前 nn 个阶乘的乘积（例如 3!=1!⋅2!⋅3!=123^! = 1! \cdot 2! \cdot 3! = 12）。在 51!51^! 的质因数分解中，因子 7 出现了多少次？
  1. A.147
  2. B.150
  3. C.156
  4. D.168
  5. E.171

正确答案 · Correct Answer

E

解析 · Solution

解法1

常规解法·数每个数在超阶乘里被乘了几遍
51!^=1!⋅2!⋅3!⋯51!\widehat{51!}=1!\cdot2!\cdot3!\cdots51!。第一步换个角度看这个大乘积：整数 kk 出现在哪些阶乘里？只有编号 ≥k\ge k 的阶乘（k!,(k+1)!,…,51!k!,(k+1)!,\dots,51!）才含有 kk，共 52-k 个。所以在 51!^\widehat{51!} 里，每个整数 kk 被乘了 52-k 遍。
因子 7 只能来自 7 的倍数，逐个数：
7 被乘 52-7=45 遍，每遍给 1 个 7；14：52-14=38 遍；21：31 遍；28：24 遍；35：17 遍；42：10 遍；49=7249=7^2：被乘 52-49=3 遍，但每遍给 2 个 7，贡献 3×2=6。
合计 45+38+31+24+17+10+6=171，选 E。
考场速算：把 49 的贡献拆成「第一份 3＋第二份 3」，前一半 45、38、31、24、17、10、3 是公差 -7 的等差数列，和 =(45+3)×72=168=\dfrac{(45+3)\times7}{2}=168；再补上 49 作为 727^2 的第二份 3，得 171。
易错提醒：干扰项 D =168 正是「把 49 只当一个 7」的结果——等差求和收工前记得补那 3 个；这类题几乎必埋「恰好差一项」的干扰项。

Solution 1

*(Elementary — count how many times each integer gets multiplied in.)*
51!^=1!⋅2!⋅3!⋯51!\widehat{51!}=1!\cdot2!\cdot3!\cdots51!. First re-view the big product: which factorials contain the integer kk? Only those with index ≥k\ge k — namely k!,(k+1)!,…,51!k!,(k+1)!,\dots,51!, that is 52-k of them. So inside 51!^\widehat{51!}, each integer kk is multiplied in 52-k times.
Factors of 7 come only from multiples of 7. Count them:
7 appears 52-7=45 times, one factor of 7 each; 14: 52-14=38; 21: 31; 28: 24; 35: 17; 42: 10; 49=7249=7^2: appears 52-49=3 times but gives TWO factors each, contributing 3×2=6.
Total 45+38+31+24+17+10+6=171, choice E.
Exam shortcut: split 49's contribution into "first 3 + second 3." The first parts 45, 38, 31, 24, 17, 10, 3 form an arithmetic sequence with difference -7: sum =(45+3)×72=168=\dfrac{(45+3)\times7}{2}=168; then add 49's second 3 as a 727^2 to get 171.
Watch out: trap choice D =168 is exactly "treating 49 as a single 7" — after the arithmetic-series shortcut, remember to add those 3; problems like this almost always plant an "off by one term" trap.

继续练习

免费刷 2026 整卷 →

真题版权归属 MAA（AMC）。本站仅供学习用途。版权方如有异议，请联系本站予以下架。

Problems © MAA (AMC). This site is for study purposes only. Rightsholders may contact us for removal.