AMC 10 · 2003 · #23

Grade 8 number-theory
prime-factorizationdivisor-countperfect-squares identify-subproblemssystematic-enumeration ↑ Prerequisites: prime-factorization
📏 Long solution 💡 3 insights
Problem
Let N=1! · 2! · 3! · 4! · 5! · 6! · 7! · 8! · 9!. Count the positive integers d that divide N and are perfect squares, that is, d=m² for some integer m.

Pick an answer.

(A)
504
(B)
672
(C)
864
(D)
936
(E)
1008
How to solve
Strategy Identify Subproblems

Nothing can be counted until N is written in primes, and multiplying nine factorials out is hopeless. Tool #15 (Organize Information in More Ways) fixes that: instead of reading the product factorial by factorial, read it number by number and ask how many of the nine factorials each number 1 through 9 sits inside. That single regrouping turns N into nine simple powers. Tool #7 (Identify Subproblems) then handles the primes one at a time — the exponent of 2, of 3, of 5, of 7 — so a large bookkeeping job becomes four small additions, and later it separates the one hard question ('how many square divisors') into four independent easy ones. Tool #4 (Introduce a Variable) names a general divisor 2^a3^b5^c7^e so that 'divides N' and 'is a perfect square' both become plain conditions on four numbers. Tool #2 (Make a Systematic List) finishes by listing the allowed even exponents for each prime and multiplying the list lengths.

1STEP 1

Regroup by base, not by factorial

Regrouping by base, each k appears 10 minus k times across the nine factorials.

N=Π_n=1⁹n!=Π_k=1⁹k¹0-k=1⁹ · 2⁸ · 3⁷ · 4⁶ · 5⁵ · 6⁴ · 7³ · 8² · 9¹
2STEP 2

Break the nine bases into primes

Every base is at most nine, so only the primes 2, 3, 5 and 7 can occur.

N=2⁸ · 3⁷ · (2²)⁶ · 5⁵ · (2 · 3)⁴ · 7³ · (2³)² · (3²)¹
3STEP 3

Add up each prime's exponent

Collecting exponents prime by prime gives 2³⁰ · 3¹³ · 5⁵ · 7³.

N=2³⁰ · 3¹³ · 5⁵ · 7³
4STEP 4

Describe every divisor

A divisor is any choice of exponents within those bounds, by unique factorization.

d ∣ N⇔ d=2^a3^b5^c7^e with 0 ≤ a ≤ 30, 0 ≤ b ≤ 13, 0 ≤ c ≤ 5, 0 ≤ e ≤ 3
5STEP 5

Say exactly when a divisor is a square

It is a square exactly when every exponent is even, proved in both directions.

2^a3^b5^c7^e is a perfect square ⇔ a,b,c,e are all even
6STEP 6

Count the even choices and multiply

The even counts 16, 7, 3 and 2 multiply to 672, choice (B).

16 · 7 · 3 · 2=672→(B)
Answer
672
The prime exponents carry the whole answer, so recount them by different bookkeeping: go factorial by factorial instead of base by base. The number of factors of 2 inside n! for n=1,2,…,9 is 0,1,1,3,3,4,4,7,7, and those add to 30. For 3 the list is 0,0,1,1,1,2,2,2,4, adding to 13. For 5 it is 0,0,0,0,1,1,1,1,1, adding to 5, and for 7 it is 0,0,0,0,0,0,1,1,1, adding to 3. All four totals match Step 3 exactly, reached by a completely different route. A size check also fits: N has 31 · 14 · 6 · 4=10416 divisors in all, so square divisors are about 6.5% of them, which is the kind of small fraction squares should occupy. Finally, the near-miss choice is instructive: treating the exponent of 7 as though it allowed three even values instead of two gives 16 · 7 · 3 · 3=1008, which is choice (E). The bound 7³ permits only 7⁰ and 7², so the correct count stays at 672, choice (B).
💡Key takeaway

Rewrite the product so each number shows how many factorials it lives in, and then a square divisor is just an even helping of each prime.

  • Regroup by base, not by factorial
  • Break the nine bases into primes
  • Add up each prime's exponent
  • Describe every divisor
  • Say exactly when a divisor is a square
  • Count the even choices and multiply