How many positive integer divisors does 8 ! have?

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GMAT Panda · Accepted Answer

E We prime factorize 8! by expressing each integer factor from 1 to 8 in its prime form 8! = 8 7 6 5 4 3 2 1 We combine like prime factors by summing their exponents 8! = (2^3) 7 (2 3) 5 (2^2) 3 2 1 = 2^7 3^2 5^1 7^1 We count the number of positive divisors by adding one to each exponent and multiplying (7+1) (2+1) (1+1) (1+1) = 8 3 2 2 = 96

Explanation	Calculations	Help
$We prime factorize 8! by expressing each integer factor from 1 to 8 in its prime form$	$8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$
$We combine like prime factors by summing their exponents$	$8! = (2^{3}) \times 7 \times (2 \times 3) \times 5 \times (2^{2}) \times 3 \times 2 \times 1 = 2^{7} \times 3^{2} \times 5^{1} \times 7^{1}$
$We count the number of positive divisors by adding one to each exponent and multiplying$	$(7 + 1) \times (2 + 1) \times (1 + 1) \times (1 + 1) = 8 \times 3 \times 2 \times 2 = 96$

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