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

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2003 AMC 12A Problem 23

Below is the professionally curated solution for Problem 23 of the 2003 AMC 12A, 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.

All of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

Concepts:prime factorization perfect square factor counting

Difficulty rating: 2110

23.

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

$504$

$672$

$864$

$936$

$1008$

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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