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Bunuel
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15! would have the following multiples of 3: 3,6,9,12 and 15.

3 would have 1 power of 3
6 would have 1 power of 3
9 would have 2 powers of 3
12 would have 1 power of 3
15 would have 1 power of 3

So, total powers = 1+1+2+1+1 = 6
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Bunuel
If k is the greatest positive integer such that 3^k is a divisor of 15! then k =

A. 3
B. 4
C. 5
D. 6
E. 7


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Answer=D
To calculate k, we need to find out highest power of 3 in 15!
15!/3+15!/3^2=5+1=6
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Bunuel
If k is the greatest positive integer such that 3^k is a divisor of 15! then k =

A. 3
B. 4
C. 5
D. 6
E. 7


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MAGOOSH OFFICIAL SOLUTION:
Attachment:
3kdivisor15factorial_text.PNG
3kdivisor15factorial_text.PNG [ 16.9 KiB | Viewed 24804 times ]
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Bunuel
If k is the greatest positive integer such that 3^k is a divisor of 15! then k =

A. 3
B. 4
C. 5
D. 6
E. 7


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15! contains six 3's

15/3 = 5
5/3 = 1

So 3^6 can divide 15!

Hence the correct answer must be (D) 6
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Bunuel
If k is the greatest positive integer such that 3^k is a divisor of 15! then k =

A. 3
B. 4
C. 5
D. 6
E. 7


Kudos for a correct solution.

Evaluating the multiples of 3 in 15!, we have:

3 = 3^1

6 = 3^1 x 2

9 = 3^2

12 = 3^1 x 2^2

15 = 5 x 3^1

Thus, there are, at most, 6 three’s in 15!, and thus the largest possible value of k is 6.

Answer: D
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15! / 3^k = 15*....12*...9*...6...*3*2*1
So,
15= 3*5
12= 3*4
9 = 3*3
6 = 3*2
3= 3*1
Thus no of 3 is 6
Answer is 6
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