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Precalculus Mathematics Homework Help
Compute the number of positive integer divisors of 10
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[QUOTE="RM86Z, post: 6518039, member: 676298"] [B]Homework Statement:[/B] number of positive integer divisors of 10! [B]Relevant Equations:[/B] 10! Compute the number of positive integer divisors of 10!. By the fundamental theorem of arithmetic and the factorial expansion: 10! = 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 2 x 5 x 3^2 x 7 x 2 x 3 x 5 x 2^2 x 3 x 2 x 1 = 2^8 x 3^4 x 5^2 x 7 Then there are 9 possibilities for 2, 5 for 3, 3 for 5 and for 7 giving 9 x 5 x 3 = 135. The book gives 270 as the answer, where am I going wrong? Thank you! EDIT:Oops, I should have counted 7 as two giving 9 x 5 x 3 x 2 = 270! [/QUOTE]
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Precalculus Mathematics Homework Help
Compute the number of positive integer divisors of 10
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