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What is the greatest value of d such that 6^d is a factor of 18! ?

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What is the greatest value of d such that 6^d is a factor of 18! ?  [#permalink]

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New post 14 Dec 2016, 05:55
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A
B
C
D
E

Difficulty:

  25% (medium)

Question Stats:

72% (01:13) correct 28% (01:03) wrong based on 102 sessions

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Re: What is the greatest value of d such that 6^d is a factor of 18! ?  [#permalink]

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New post 14 Dec 2016, 06:33
2
2
\(6 = 2*3\)

18! has definitely more 2s than 3s, hence exponent of 3 will play determinant role in finding highest power of 6 in 18!.

\([\frac{18}{3}] + [\frac{18}{3^2}] = 6 + 2 = 8\)

Answer E
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Re: What is the greatest value of d such that 6^d is a factor of 18! ?  [#permalink]

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New post 15 Dec 2016, 17:52
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Bunuel wrote:
What is the greatest value of d such that 6^d is a factor of 18! ?

A. 3
B. 4
C. 5
D. 6
E. 8


We need to determine the maximum value of d such that 18!/(6^d) is an integer. We must remember that an integer is divisible by 6 if it’s divisible by both 3 and 2. Thus, we must determine the number of factors of 2 and 3 in 18!. However, since we know there are fewer factors of 3 than factors of 2 in 18!, we can find the number of factors of 3 and thus be able to determine the maximum value of d.

To determine the number of factors of 3 within 18!, we can use the following shortcut in which we divide 18 by 3, and then divide the quotient of 18/3 by 3 and continue this process until we no longer get a nonzero quotient:

18/3 = 6

6/3 = 2

Since 2/3 does not produce a nonzero quotient, we can stop.

The final step is to add up our quotients; that sum represents the number of factors of 3 within 18!. Thus, there are 6 + 2 = 8 factors of 3 within 18!, and the maximum value of d is 8.

Answer: E
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Re: What is the greatest value of d such that 6^d is a factor of 18! ?  [#permalink]

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New post 14 Dec 2016, 08:11
Bunuel wrote:
What is the greatest value of d such that 6^d is a factor of 18! ?

A. 3
B. 4
C. 5
D. 6
E. 8


6 = 2 x 3

Highest part of 2 in 18! is 16

18/2 = 9
9/2 = 4
4/2 = 2
2/2 = 1

Highest part of 3 in 18! is 8

18/3 = 6
6/3 = 2

Since there are fewer 3's , correct answer will be there are 8 sixes...

Correct answer will be (E)

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Re: What is the greatest value of d such that 6^d is a factor of 18! ?  [#permalink]

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Re: What is the greatest value of d such that 6^d is a factor of 18! ?   [#permalink] 03 Feb 2019, 22:44
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