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19 Oct 2010, 22:57
Kudos
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
58% (01:46) correct
42%
(02:02)
wrong
based on 1886
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Date
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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
A. 3
B. 6
C. 7
D. 14
E. 26
20 Oct 2010, 03:47
Kudos
Bookmarks
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.
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.
jlgdr
Joined: 06 Sep 2013
Last visit: 24 Jul 2015
Posts: 1,311
Given Kudos: 355
Concentration: Finance
Schools: Wharton '17 (A) HBS '17 (WA$)
TomB
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
