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Bunuel
If 100!/x is not an integer,which of the following could be the value of x?

A. 5^24
B. 7^16
C. 11^9
D. 13^6
E. 17^6

Checking the options is the best way to go. The answer will be a number which doesn't completely divide 100!

No of 5's in 100! = 24
No of 7's in 100! = 16
No of 11's in 100! = 9
No of 13's in 100! = 7

No of 17's in 100! = 5

There are 5 seventeens in 100! , however we are dividing by 17^6 , so this can not divide 100!

Hence our answer will be E
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Hi,

Can anyone please help me in regards to finding out 5s, 7s, 11s, 13s , and 17s in 100!. How exactly does one know that 100! has 5 24s, not 25 or has 5 17s not 6?

Thank You!
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Bunuel
If 100!/x is not an integer, which of the following could be the value of x?

A. 5^24
B. 7^16
C. 11^9
D. 13^6
E. 17^6

Let’s start with answer choice E. In 100!, we have the following multiples of 17:

17, 34, 51, 68, and 85. Thus, we have 5 multiples of 17 in 100!.

Therefore, 100!/(17^6) IS NOT an integer.

Answer: E
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If 100!/x is a non-interger, then x must be larger than 100!
Let's look at all the choices to find an option with value greater than 100!, Using concept: power of prime number in n!= n/p + n/p^2 + .......n/p^k, where p^k =<n.

A) Power of 5: 100/5 + 100/5^2 = 20 + 4 = 24. Therfore, 5^24. Same as the option. Hence, incorrect.

B) Power of 7: 100/7 + 100/7^2 = 14 + 2 = 16. Therfore, 7^16. Same as the option. Hence, incorrect.

C) Power of 11: 100/11= 9. Therfore, 11^9. Same as the option. Hence, incorrect.

D) Power of 13: 100/13 = 7. Therfore, 13^7. Smaller than the option. Hence, incorrect.

E) Power of 17: 100/17 = 5. Therfore, 17^5. Larger than the option. Hence, CORRECT
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