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# If 15!/3^m is an integer, what is the greatest possible value of m?

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Joined: 02 Sep 2009
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If 15!/3^m is an integer, what is the greatest possible value of m?  [#permalink]

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13 Mar 2019, 00:45
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15% (low)

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72% (00:45) correct 28% (01:29) wrong based on 81 sessions

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If 15!/3^m is an integer, what is the greatest possible value of m?

A 4
B 5
C 6
D 7
E 8

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Re: If 15!/3^m is an integer, what is the greatest possible value of m?  [#permalink]

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13 Mar 2019, 00:50
Bunuel wrote:
If 15!/3^m is an integer, what is the greatest possible value of m?

A 4
B 5
C 6
D 7
E 8

15/3 = 5

15/9 = 1

Total number of 3's in 15! = 6.

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If 15!/3^m is an integer, what is the greatest possible value of m?  [#permalink]

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10 Apr 2019, 08:11
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What is the question really asking here?

It's simply asking: For the quotient to be an integer, we need the same number of factors of 3 in the numerator as we need in the denominator. So our task here is to find how many factors of 3 are in the numerator, which means the denominator could have just as many to maintain that the quotient is an integer.

So $$\frac{15!}{3^m}$$ is really $$\frac{15*14*13*12*11*10*9.....}{3*3*3*3.....}$$

The question asks for the max number of 3's in the denominator. We already established, for the quotient to be an integer, we need to figure out how many 3's are in the numerator!

So simply count the factors of 3. That's 3, 6, 9, 12, 15... So that's 5 factors!! Nope, 9 has 2 factors of 3, so 6 factors.
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Re: If 15!/3^m is an integer, what is the greatest possible value of m?  [#permalink]

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13 Apr 2019, 18:35
Bunuel wrote:
If 15!/3^m is an integer, what is the greatest possible value of m?

A 4
B 5
C 6
D 7
E 8

Breaking down the primes of 3 within 15! we have:

3

6 = 3 x 2

9 = 3^2

12 = 2^2 x 3

15 = 3 x 5

We see that there are 6 factors of 3 in 15!, and thus the maximum value of m is 6.

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Re: If 15!/3^m is an integer, what is the greatest possible  [#permalink]

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02 May 2019, 01:27
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Re: If 15!/3^m is an integer, what is the greatest possible   [#permalink] 02 May 2019, 01:27
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