What is the largest power of 6! that can divide 60! ?

AKY13



60/32 should be 1

Thanks for pointing out. However the answer is same, I think finding out the no. of 5s in 60 fac is not the correct way of doing. 2^4 is larger than 5 hence would be less in nos.

Hi
Similar question for practice:
https://gmatclub.com/forum/what-is-the- ... fl=similar
For theory you can refer to the following blog:
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/0 ... actorials/
Hope it helps.

We note that 6! = 6 x 5 x 4 x 3 x 2 x 1 = 5^1 x 3^2 x 2^4 = 16 x 9 x 5. We need to determine the greatest power of these factors present in 60!.

To determine the number of 16s within 60!, we need to determine the number of 2s. We can use the following shortcut in which we divide 60 by 2, then divide the quotient of 60/2 by 2 and continue this process until we no longer get a nonzero quotient.

60/2 = 30

30/2 = 15

15/2 = 7 (we can ignore the remainder)

7/2 = 3 (we can ignore the remainder)

3/2 = 1 (we can ignore the remainder)

Since 1/2 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 2 within 60!.

Thus, there are 30 + 15 + 7 + 3 + 1 = 56 factors of 2 within 60!. Since 16 = 2^4, the largest power of 16 that divides 60! is 14.

Doing the same for the prime factor of 3 to determine the number of 9s in 60!, we find that the largest power of 9 that divides 60! is also 14.

Finally, since there are 12 multiples of 5 in 60! and both 25 and 50 contains two 5’s apiece, we see that if we were to break 60! into primes, we’d have 14 factors of 5. Thus, the largest power of 6! that divides 60! is 14.

Answer: D

I solved this using number of 0's in 60!
6! can be written as 2^4*3^2*5 : It will has only 0 which depends on power of 5
So i counted that 60! will have 5^14

so 14

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