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GRE Prep Club Team Member
Joined: 20 Feb 2017
Posts: 2509
Given Kudos: 1058
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If k is an integer and 33!/22! is divisible by 6^k, what is the maximu
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22 Mar 2021, 05:30
22 Mar 2021, 05:30
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If k is an integer and \(\frac{33!}{22!}\) is divisible by \(6^k\), what is the maximum possible value of k?
(A) 3
(B) 4
(C) 5
(D) 6
(E) 7
(A) 3
(B) 4
(C) 5
(D) 6
(E) 7
Re: If k is an integer and 33!/22! is divisible by 6^k, what is the maximu
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22 Mar 2021, 05:39
22 Mar 2021, 05:39
D
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Re: If k is an integer and 33!/22! is divisible by 6^k, what is the maximu
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22 Mar 2021, 06:05
22 Mar 2021, 06:05
1
GeminiHeat wrote:
If k is an integer and \(\frac{33!}{22!}\) is divisible by \(6^k\), what is the maximum possible value of k?
(A) 3
(B) 4
(C) 5
(D) 6
(E) 7
(A) 3
(B) 4
(C) 5
(D) 6
(E) 7
\(\frac{33!}{22!}\) is divisible by \(6^k\)
\(\frac{33!}{22!} = (3)(11)*(2^5)*(31)*(3)(2)(5)*(29)*(2^2)(7)*(3^3)*(2)(13)*(5^2)*(2^3)(3)*(23)\)
\(6^k = (2^k)(3^k)\)
We can notice that we more \(2\)s than \(3\)s - so we have have a constraint of \(3\)s
Therefore, maximum value of \(k\) would be the number of \(3\)s i.e. \(6\)
Hence, option D
