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GMAT Quant — Number Properties

How many positive integer divisors does 9! have?

Explanation	Calculations	Help
$We prime factorize 9! by expressing each integer from 1 to 9 in its prime form$	$9! = 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$	Theory & Tactics Method Card DIV4-C Click to view full details
$We combine like prime factors by summing their exponents$	$9! = (3^{2}) \times (2^{3}) \times 7 \times (2 \times 3) \times 5 \times (2^{2}) \times 3 \times 2 \times 1 = 2^{7} \times 3^{4} \times 5^{1} \times 7^{1}$
$We count the number of positive divisors by adding one to each exponent and multiplying$	$(7 + 1) \times (4 + 1) \times (1 + 1) \times (1 + 1) = 8 \times 5 \times 2 \times 2 = 160$	Theory & Tactics Method Card DIV9-A Click to view full details

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C

GMAT Quant — Number Properties

How many positive integer divisors does 9! have?

Explanation	Calculations	Help
$We prime factorize 9! by expressing each integer from 1 to 9 in its prime form$	$9! = 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$	Theory & Tactics Method Card DIV4-C Click to view full details
$We combine like prime factors by summing their exponents$	$9! = (3^{2}) \times (2^{3}) \times 7 \times (2 \times 3) \times 5 \times (2^{2}) \times 3 \times 2 \times 1 = 2^{7} \times 3^{4} \times 5^{1} \times 7^{1}$
$We count the number of positive divisors by adding one to each exponent and multiplying$	$(7 + 1) \times (4 + 1) \times (1 + 1) \times (1 + 1) = 8 \times 5 \times 2 \times 2 = 160$	Theory & Tactics Method Card DIV9-A Click to view full details

Scroll horizontally to view all columns

C