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# ps

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ps [#permalink]  19 Oct 2010, 22:57
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Question Stats:

48% (02:16) correct 51% (01:37) wrong based on 4 sessions
If N is the product of all positive integers less than 31, than what is the greatest integer k for which N/18^k is an integer?

A. 3
B. 6
C. 7
D. 14
E. 26

[Reveal] Spoiler:
c
[Reveal] Spoiler: OA
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Joined: 08 Sep 2010
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WE 1: 6 Year, Telecom(GSM)
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Re: ps [#permalink]  20 Oct 2010, 02:59
TomB wrote:
if N is the product of all positive integers less than 31, than what is the greatest integer k for which is an integer?

3
6
7
14
26
[Reveal] Spoiler:
c

Something is missing in this question.
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Re: ps [#permalink]  20 Oct 2010, 03:47
If N is the product of all positive integers less than 31, than what is the greatest integer k for which N/18^k is an integer?

A. 3
B. 6
C. 7
D. 14
E. 26

Given: n=30!. Question: if \frac{30!}{18^k}=integer then k_{max}=?

We should determine the highest power of 18 in 30!.

18=2*3^2, so we should find the highest powers of 2 and 3 in 30!:

Highest power of 2 in 30!: \frac{30}{2}+\frac{30}{4}+\frac{30}{8}+\frac{30}{16}=15+7+3+1=26, --> 2^{26};

Highest power of 3 in 30!: \frac{30}{3}+\frac{30}{9}+\frac{30}{27}=10+3+1=14 --> 3^{14};

n=30!=2^{26}*3^{14}*p, where p is the product of other multiples of 30! (other than 2 and 3) --> n=30!=(2*3^{2})^7*2^{19}*p=18^7*2^{19}*p --> so the highest power of 18 in 30! is 7 --> \frac{30!}{18^k}=\frac{18^7*2^{19}*p}{18^k}=integer --> k=7.

Hope it's clear.
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Re: ps   [#permalink] 20 Oct 2010, 03:47
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