RM86Z
- 23
- 6
- Homework Statement
- number of positive integer divisors of 10!
- Relevant Equations
- 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!
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!