AMC 10 · 2004 · #6

Grade 8 number-theory
Archetype Small-Domain Constrained Search └ Sub-archetype Answer-Choice Plug-In
factorialperfect-squaresprime-factorization easier-related-problemsystematic-enumeration ↑ Prerequisites: perfect-squares
📏 Medium solution 💡 2 insights
Problem
Each of the five answer choices is a product of two of the factorials 98!, 99!, 100!, and 101!, where n! means the product of all whole numbers from 1 through n. Decide which one of the five choices is a perfect square.

Pick an answer.

(A)
$98! \cdot 99!$
(B)
$98! \cdot 100!$
(C)
$99! \cdot 100!$
(D)
$99! \cdot 101!$
(E)
$100! \cdot 101!$

AMC 10 2004 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Solve an Easier Related Problem

The numbers here are too big to ever write down, the signature trigger for Tool #9 (Solve an Easier Related Problem): trade the giant factorials for a small, manageable structure and answer the question there. Tool #4 (Introduce a Variable) supplies that structure — call the smaller index n and use (n+1)!=(n+1) · n! to fold each product into a perfect square times one tiny leftover. Tool #3 (Eliminate Possibilities) then sweeps the five choices, keeping the one whose leftover is a perfect square and discarding the rest.

1STEP 1

Shrink the giants to a leftover

Never compute the giants — (n+1)!=(n+1)⋅n!(n+1)!=(n+1)\cdot n! folds n!⋅(n+1)!n!\cdot(n+1)! into (n!)2⋅(n+1)(n!)^2\cdot(n+1): a square times one tiny leftover.

(n+1)!=(n+1) · n! → n!·(n+1)!=(n!)²·(n+1)
2STEP 2

A square times what is still a square

A square factor keeps every prime's count even, so (n!)2⋅k(n!)^2\cdot k is a perfect square exactly when the leftover kk is one.

(n!)² · k is a perfect square ⇔ k is a perfect square
3STEP 3

Test the neighboring pairs

Consecutive pairs (A), (C), (E) leave 9999, 100100, 101101; only 100=102100=10^2 is a square, so (A) and (E) are out.

99!·100!=(99!)²·100=(99!)²·10²; 99, 101 are not squares
4STEP 4

Clear the gapped pairs and conclude

The gapped pairs leave 9900=302⋅119900=30^2\cdot 11 and 10100=102⋅10110100=10^2\cdot 101, each with a lone prime, so the square is (C).

98!·100!=(98!)²·9900, 99!·101!=(99!)²·10100 → only (C)
Answer
99! · 100!
The five leftovers are 99, 9900, 100, 10100, and 101, and exactly one of them, 100=10², is a perfect square — matching the promise of a single correct choice. The traps line up with real mistakes: the prime leftovers 99 (which is 9·11), 101, and the lone-prime products 9900=30²·11 and 10100=10²·101 each hide one prime to an odd power, so none can be a square. Landing on (C), whose leftover is exactly 10², is a strong confirmation.
💡Key takeaway

A product of two factorials is a perfect square exactly when the small leftover that bridges them is itself a perfect square, so pair them up and just check that leftover.

  • Shrink the giants to a leftover
  • A square times what is still a square
  • Test the neighboring pairs
  • Clear the gapped pairs and conclude

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