2026 AMC 8 Problem 24: Solution

Work through 2026 AMC 8 Problem 24 on prime factorization with a written solution, hints, and a visual explanation.

Number TheoryLevel 5 · Final stretch

The notation n! ("n factorial") is defined as the product of the first n positive integers. (For example, 3! = 1 × 2 × 3 = 6.) Define the superfactorial of a positive integer n to be the product of the factorials of the first n integers. (For example, the superfactorial of 3 is 1! × 2! × 3! = 12.) How many factors of 7 appear in the prime factorization of the superfactorial of 51?

  • A. 147
  • B. 150
  • C. 156
  • D. 168
  • E. 171
Hints
  1. Use the fact that the power of 7 in k! is ⌊k/7⌋ + ⌊k/49⌋, and think about how this count changes as k runs from 1 to 51.
  2. The count of 7s in k! stays constant over each run of 7 consecutive k-values, then jumps up — figure out the width and height of each block.
  3. Multiply each block's width by its height and add all the blocks — don't forget the extra jump once k reaches 49.
Read the step-by-step solution

Answer: E · 171

  1. Count the 7s in one factorial

    The superfactorial of 51 is 1! × 2! × ⋯ × 51!. By Legendre's formula, k! contains ⌊k/7⌋ + ⌊k/49⌋ factors of 7. The second term only kicks in once k reaches 49 = 7².

    v₇(k!) = ⌊k/7⌋ + ⌊k/49⌋

  2. Build the staircase for k = 1…51

    As k grows, the 7-count climbs by 1 at each multiple of 7 (7, 14, 21, …), giving equal-height blocks of width 7. At k = 49 it jumps by 2 — from 6 up to 8 — because both ⌊k/7⌋ and ⌊k/49⌋ increase there.

  3. Sum the blocks

    Add each block's width × height: 7×1 + 7×2 + 7×3 + 7×4 + 7×5 + 7×6 + 3×8 = 7 + 14 + 21 + 28 + 35 + 42 + 24 = 171.

    7 + 14 + 21 + 28 + 35 + 42 + 24 = 171

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