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How many positive integers n are there such that n^3 is a

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How many positive integers n are there such that n^3 is a [#permalink] New post 18 Aug 2006, 04:46
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How many positive integers n are there such that n^3 is a factor of 10! ?

Last edited by kevincan on 18 Aug 2006, 23:36, edited 1 time in total.
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Re: PS: Factors of 10! [#permalink] New post 18 Aug 2006, 13:13
you are correct. 1 is also a +ve integer. so altogather 5 +ve integers:

1, 2, 3, 4, 6.
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 [#permalink] New post 18 Aug 2006, 13:23
how do we get 1, 2, 3 , 4 and 6?
is there a shortcut?
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Re: PS: Factors of 10! [#permalink] New post 18 Aug 2006, 16:42
kevincan wrote:
How much positive integers n are there such that n^3 is a factor of 10! ?



10! = 10*9*8*7*6*5*4*3*2

n=1 Yes
n=2 Yes
n=3 Yes
n=4 Yes
n=5 No
n=6 Yes (3*2,6, 9=3*3, 4=2*2)
n=7 No
n=8 No

Hence 5??
Heman
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 [#permalink] New post 19 Aug 2006, 01:03
2^3 * 2^3 * 2^2 * 3^3 * 3 * 5^2 * 7

Take only those terms that have cubes.
We have 1,2,2,3 i.e 2^2 * 3
Factors possible = (2+1) (1+1) = 6
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SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

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 [#permalink] New post 19 Aug 2006, 08:33
OK, I totally follow this working...and i agree with it..so whats the sixth factor? whats the answer 5 or 6?

I believe that total factors is 6..but I can only find out 5...whts the 6th factor?

ps_dahiya wrote:
2^3 * 2^3 * 2^2 * 3^3 * 3 * 5^2 * 7

Take only those terms that have cubes.
We have 1,2,2,3 i.e 2^2 * 3
Factors possible = (2+1) (1+1) = 6
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Re: PS: Factors of 10! [#permalink] New post 19 Aug 2006, 11:44
heman wrote:
kevincan wrote:
How much positive integers n are there such that n^3 is a factor of 10! ?



10! = 10*9*8*7*6*5*4*3*2

n=1 Yes
n=2 Yes
n=3 Yes
n=4 Yes
n=5 No
n=6 Yes (3*2,6, 9=3*3, 4=2*2)
n=7 No
n=8 No

Hence 5??
Heman


I see if I keep going
n=9 No
n=10 No
n=11 No
n= 12 Yes(4*3,6*2,8=4*2,9=3*3

Hence 6 factors. Long approach though

Heman
Re: PS: Factors of 10!   [#permalink] 19 Aug 2006, 11:44
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