What is the greatest value of x such that 3^x is a factor...

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What is the greatest value of x such that 3^x is a factor of 18!?

A. 9
B. 8
C. 6
D. 3
E. 2

The OA is B.

I need help to solve this PS question. Please, can any expert explain it for me? Thanks.

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by GMATGuruNY » Wed Nov 01, 2017 10:57 am
LUANDATO wrote:What is the greatest value of x such that 3^x is a factor of 18!?

A. 9
B. 8
C. 6
D. 3
E. 2
The prompt implies the following:
18!/(3^x) = integer.
To determine the greatest possible value of x, we need to count how many 3's can divide into 18!.
Put another way:
We need to count how many 3's are contained within 18!.

The following multiples of 3 are contained with 18!:
3, 5, 9, 12, 15, 18.
If we factor out all of the 3's contained within the values above, we get:
3
6=3*2
9=3*3
12=3*4
15=3*5
18=3*3*2.

The blue values indicate that there are eight 3's contained within 18!.
Implication:
Up to eight 3's can divide into 18!.
Thus, the greatest possible value of x = 8.

The correct answer is B.
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by Scott@TargetTestPrep » Fri Nov 01, 2019 7:03 pm
BTGmoderatorLU wrote:What is the greatest value of x such that 3^x is a factor of 18!?

A. 9
B. 8
C. 6
D. 3
E. 2

The OA is B.

I need help to solve this PS question. Please, can any expert explain it for me? Thanks.
Prime factorizing all the multiples of 3 up to 18, we have:

18 = 3^2 * 2^1

15 = 3^1 x 5

12 = 3^1 x 2^2

9 = 3^2

6 = 3^1 x 2

3 = 3^1

Thus, we see that there are 8 factors of 3 in 18! and thus the maximum value of n is 8.

Answer: B

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