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by vaibhav101 » Fri Aug 10, 2018 9:13 am

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Difficulty—

If 5^a is a factor of n!, and the greatest integer value of a is 6, what is the largest possible value
of b so that 7^b is a factor of n!?
(A) 2
(B) 3
(C) 4
(D) 5
(E) 6
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by [email protected] » Sun Aug 12, 2018 2:20 pm
Hi vaibhav101,

We're told that 5^A is a FACTOR of N! and the greatest integer value of A is 6. We're asked for the largest possible value of B such that 7^B is a factor of N!

This question is based on Prime Factorization, but comes with a 'twist.' We're limited by the 5s that appear in N! 5^6 allows for six 5s - and those 5s would be found in 5, 10, 15, 20 and 25 (note that 25 = 5^2, so that counts as two 5s), so we can't have any larger multiples of 5 in N! The 'twist' is that N! can actually be bigger than 25!, since 26!, 27!, 28! and 29! all have just six 5s (the same as 25!). We cannot use 30! though - since that would create a seventh 5 (which is not allowed).

We would be able to find 7s in 7, 14, 21 AND 28 (assuming that N! was either 28! or 29!), so the largest possible value of B would be 4.

Final Answer: C

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by Jeff@TargetTestPrep » Sat Aug 18, 2018 6:46 pm
vaibhav101 wrote:If 5^a is a factor of n!, and the greatest integer value of a is 6, what is the largest possible value
of b so that 7^b is a factor of n!?
(A) 2
(B) 3
(C) 4
(D) 5
(E) 6
Since 5^6 is the greatest power of 5 that divides n!, we see that we need the factors of 5, 10, 15, 20 and 25 in n! (notice that the first 4 factors each provides a factor of 5 while the last factor, 25, provides two factors of 5). That means, n! can be 25!, 26!, 27!, 28! or 29!. Each of these factorials will have 5^6 as the greatest power of 5 that divides it. If n! Is 25!, 26! or 27!, then b = 3 since each of these contains the factors of 7, 14 and 21. However, if n! is 28! or 29!, then b = 4 since each of these contains the factors of 7, 14, 21 and 28. Therefore, the largest possible value of b is 4.

Answer: C

Jeffrey Miller
Head of GMAT Instruction
[email protected]

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