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product of the integers

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Re: product of the integers

by Vemuri » Tue May 05, 2009 9:16 am
From the info provided p = 30! (since it is a product of integers from 1 to 30 inclusive)

The question is asking what is the greatest integer k for which 3k is a factor of p.

if k = 18, then 3k=54 (2*3*3*3) is a factor of p, i.e. 30!

what am I missing here?
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by wishkaro » Tue May 05, 2009 6:32 pm
Hope this will help to understand

p = 30! , p = N 3 k
30! = N 3 K
So power of 3 in 30 ! = 14
there are 10 multiples of 3 , so 10 + 1(for 9) + 1 (for 18) 2(for 27) = 14
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by Vemuri » Tue May 05, 2009 7:21 pm
wishkaro wrote:Hope this will help to understand

p = 30! , p = N 3 k
30! = N 3 K
So power of 3 in 30 ! = 14
there are 10 multiples of 3 , so 10 + 1(for 9) + 1 (for 18) 2(for 27) = 14
aaahhh....now I understand what was missing. The question stem stated that 3k is a factor of p. It should actually be 3^k & not 3k.
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