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

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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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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.

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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]
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