product of the integers

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wishkaro: Junior | Next Rank: 30 Posts; Posts: 27; Joined: Wed Feb 18, 2009 3:13 am; Thanked: 3 times

by wishkaro » Tue May 05, 2009 8:59 am

If p is the product of the integers from 1 to 30, inclusive, what is the greatest integer k for which 3k is a factor of p?
A. 10
B. 12
C. 14
D. 16
E. 18

OA C

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Source: Beat The GMAT — Problem Solving | Tue May 05, 2009 8:59 am

Vemuri: Legendary Member; Posts: 682; Joined: Fri Jan 16, 2009 2:40 amFollowed by: 1 member; Thanked: 32 times

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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wishkaro: Junior | Next Rank: 30 Posts; Posts: 27; Joined: Wed Feb 18, 2009 3:13 am; Thanked: 3 times

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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Vemuri: Legendary Member; Posts: 682; Joined: Fri Jan 16, 2009 2:40 amFollowed by: 1 member; Thanked: 32 times

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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