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# If a!/6^5 is an integer and a is a positive integer, what is the minim

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Joined: 02 Sep 2009
Posts: 58427
If a!/6^5 is an integer and a is a positive integer, what is the minim  [#permalink]

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11 Jan 2018, 23:13
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45% (medium)

Question Stats:

69% (01:39) correct 31% (01:30) wrong based on 44 sessions

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If a!/6^5 is an integer and a is a positive integer, what is the minimum possible value of a?

A. 30
B. 18
C. 16
D. 15
E. 12

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If a!/6^5 is an integer and a is a positive integer, what is the minim  [#permalink]

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Updated on: 12 Jan 2018, 10:06
Bunuel wrote:
If a!/6^5 is an integer and a is a positive integer, what is the minimum possible value of a?

A. 30
B. 18
C. 16
D. 15
E. 12

Since we're told that a!/6^5 is an integer we'll look for a solution dealing with integer properties.
This is a Logical approach.

Saying that a!/6^5 is an integer is the same as saying that a! is divisible by 6^5
This means it has 5 6's in its factorization meaning 5 3's and 5 2's.
Since the first 5 factors of 2 are in are 2,(2*2)=4,6,(2*2*2)=8 and the first 5 factors of 3 are 3,6,(3*3)=9,12
then any number larger or equal to 12! willl have 5 3's and 5 2's in its factorization, as required.

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Originally posted by DavidTutorexamPAL on 12 Jan 2018, 08:34.
Last edited by DavidTutorexamPAL on 12 Jan 2018, 10:06, edited 2 times in total.
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Re: If a!/6^5 is an integer and a is a positive integer, what is the minim  [#permalink]

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12 Jan 2018, 09:28
1
david: is it not E? as 9 can be written as 3^2.
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Re: If a!/6^5 is an integer and a is a positive integer, what is the minim  [#permalink]

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12 Jan 2018, 10:04
dhiyanesh wrote:
david: is it not E? as 9 can be written as 3^2.

You're right!
Was a bit too fast with that one.

Fixed.
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If a!/6^5 is an integer and a is a positive integer, what is the minim  [#permalink]

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12 Jan 2018, 10:36
We must have enough 2's and 3's in the numerator to cancel the denominator which is 6^5 = (2^5 x 3^5)

starting with the smallest value of 12!

12- 3x4
11
10
9 - 3x3
8
7
6- 3x2
5
4
3 - 3x1
2
1

thus, 12! have a sufficient number of 3s to eliminate 3^5 in the denominator leaving a positive integer in the numerator.

please correct me if I've explained improperly
If a!/6^5 is an integer and a is a positive integer, what is the minim   [#permalink] 12 Jan 2018, 10:36
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