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If the greatest integer k for which 3^k is a factor of n! is 8,
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23 May 2021, 23:45
23 May 2021, 23:45
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If the greatest integer k for which \(3^{k}\) is a factor of n! is 8, what is the largest possible value of p so that \(5^{p}\) is a factor of n!?
(A) 2
(B) 3
(C) 4
(D) 5
(E) 6
(A) 2
(B) 3
(C) 4
(D) 5
(E) 6
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If the greatest integer k for which 3^k is a factor of n! is 8,
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24 May 2021, 01:38
24 May 2021, 01:38
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GeminiHeat wrote:
If the greatest integer k for which \(3^{k}\) is a factor of n! is 8, what is the largest possible value of p so that \(5^{p}\) is a factor of n!?
(A) 2
(B) 3
(C) 4
(D) 5
(E) 6
(A) 2
(B) 3
(C) 4
(D) 5
(E) 6
arjunbir Thank you for pointing out my mistake, here's the edited version.
If n! is divisible by \(3^8\), then n! can have a maximum of eight 3s
n! could be between 18! and 20!
Why between 18 and 20?
Because, 18! will have eight 3s and so does 19! and 20!
[NOTE: It cannot be 21!, as we will have nine 3s then!]
i.e. (1)(2)(3)(4)(5)(6)(7)(8)(9) .... (18)(19)(20)
one 3 in 3, 6, 12, and 15
two 3s in 9, and 18
Now, if 20! is divisible by \(5^P\), p could only be 4 [5, 10, 15, 20]
Hence, option C
Re: If the greatest integer k for which 3^k is a factor of n! is 8,
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20 Aug 2021, 17:43
20 Aug 2021, 17:43
n!/(3^k) has to be an integer, and k is 8. This means, n! must have at least 8 3's. Since 8 3's are relatively easier to get through brute force: 21! has 8 3's.
21! also has 4 5's : 21/5 = 4 steps of 5, and there are no squares of 5 between 1 and 21, which leaves us with a final answer of 4 (C).
21! also has 4 5's : 21/5 = 4 steps of 5, and there are no squares of 5 between 1 and 21, which leaves us with a final answer of 4 (C).
