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If N is the product of all positive integers less than 31,

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If N is the product of all positive integers less than 31, [#permalink] New post 19 Oct 2010, 21:57
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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: OA
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Re: ps [#permalink] New post 20 Oct 2010, 02:47
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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

Check this: everything-about-factorials-on-the-gmat-85592.html

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 --> \(\frac{30!}{18^k}=\frac{18^7*2^{19}*p}{18^k}=integer\) --> \(k=7\).

Answer: C.

Hope it's clear.
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Re: If N is the product of all positive integers less than 31, [#permalink] New post 08 Oct 2013, 12:05
TomB wrote:
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


Hi all, this was my approach for solving this.

Basically the question is asking us what is the highest power of 18 that can give us a number that is a factor of 31!

So remember 18 after prime factorization is (3^2)(2). Now there are going to be less factors of 3 in 31!, than factors of 2.
Therefore, lets find how many factors of 3 in 31! We can use this quick method

31/3^1 = 10
31/3^2= 3
31/3^3=1
Sum = 14
Just ignore the remainders.

So we have that 3^14 must be the least. Now don't forget that 18 is 3^2k so k must be ONLY 7, because 2k will give us the 14.
Hence answer is (C)

Hope it helps

Bunuel could you please validate this one? Thank you
Cheers
J

Last edited by jlgdr on 12 Feb 2014, 06:40, edited 1 time in total.
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Re: If N is the product of all positive integers less than 31, [#permalink] New post 12 Oct 2013, 18:33
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According to Bunuel \(18=2*3^2\)

I listed 14 3's
14 3's- 3 3 3 3 3 3 3 3 3 3 3 3 3 3

26 2's- 2 2 2 2 2 2 2 2 2 2 2 2 2 2 and so on

notice that 7 such combinations of 18 are possible.
so answer= \(18^7\)
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Re: ps [#permalink] New post 13 Nov 2013, 12:43
Bunuel wrote:
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

Check this: everything-about-factorials-on-the-gmat-85592.html

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 --> \(\frac{30!}{18^k}=\frac{18^7*2^{19}*p}{18^k}=integer\) --> \(k=7\).

Answer: C.

Hope it's clear.

Hi Bunuel,
The logic here is the same as the logic in finding terminating "0" of a number right?
But instead of checking for the level of "5" we do it for "3^2", right?
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Re: If N is the product of all positive integers less than 31, [#permalink] New post 01 Jan 2015, 06:29
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Re: If N is the product of all positive integers less than 31,   [#permalink] 01 Jan 2015, 06:29
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