2003 AMC 12A Problem 23 — Solution | LIVE by Po-Shen Loh

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2003 AMC 12A Problem 23

Attempt Problem 23 of the 2003 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2003 AMC 12A solutions, or check the answer key.

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23.

How many perfect squares are divisors of the product $1! \cdot 2! \cdot 3! \cdots 9!\,?$

$504$

$672$

$864$

$936$

$1008$

Answer: B

Concepts:prime factorization perfect square factor counting

Difficulty rating: 2110

Solution:

The product is $1! \cdot 2! \cdots 9! = 2^{30}\cdot3^{13}\cdot5^{5}\cdot7^{3}.$

A perfect-square divisor has the form $2^{2a}3^{2b}5^{2c}7^{2d}$ with $0\le a\le15,$ $0\le b\le6,$ $0\le c\le2,$ and $0\le d\le1.$

The number of choices is $16\cdot7\cdot3\cdot2=672.$

Thus, the correct answer is B.

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