This problem is about TRAILING 0's: the number of 0's at the end of a large product.

If N = 120, then N! = 120*119*118*....*3*2*1.

Since 10=2*5, EVERY COMBINATION OF 2*5 contained within the prime-factorization of 120! will yield a 0 at the end of the integer representation of 120!.
The prime-factorization of 120! is composed of FAR MORE 2'S than 5's.
Thus, the number of 0's depends on the NUMBER OF 5's contained within 120!.

To count the number of 5's, simply divide increasing POWERS OF 5 into 120.

Every multiple of 5 within 120! provides at least one 5:
120/5 = 24 --> 24 5's.
Every multiple of 5² provides a SECOND 5:
120/5² = 4 --> 4 more 5's.
Thus, the total number of 5's contained within 120! = 24+4 = 28.

Note that 121!, 122!, 123!, and 124! all contain the same number of 5's as 120!.

For more than 28 5's to be contained within N!, N must be AT LEAST 125, the next greatest multiple of 5 after 120.
Every multiple of 5 within 125! provides at least one 5:
125/5 = 25 --> 25 5's.
Every multiple of 5² provides a SECOND 5:
125/5² = 5 --> 5 more 5's.
Every multiple of 5³ provides a THIRD 5:
125/5³ = 1 --> 1 more 5.
Thus, the total number of 5's contained within 125! = 25+5+1 = 31.

Note that 126!, 127!, 128!, and 129! all contain the same number of 5's as 125!.

Onto the problem:

What is the value of N?

(1) N! ends with 28 zeroes

(2) (N+2)! ends with 31 zeroes and (N-1)! ends with 28 zeroes

The following cases satisfy both statements:

Case 1: N=123
(123+2)! = 125!, which has 31 5's.
(123-1)! = 122!, which has 28 5's.

Case 2: N=124
(124+2)! = 126!, which has 31 5's.
(124-1)! = 123!, which has 28 5's.

Thus, the two statements combined are INSUFFICIENT.

The correct answer is E.

GMATGuruNY wrote: Since 10=2*5, EVERY COMBINATION OF 2*5 contained within the prime-factorization of 120! will yield a 0 at the end of the integer representation of 120!.
The prime-factorization of 120! is composed of FAR MORE 2'S than 5's.
Thus, the number of 0's depends on the NUMBER OF 5's contained within 120!.

To count the number of 5's, simply divide increasing POWERS OF 5 into 120.

Every multiple of 5 within 120! provides at least one 5:
120/5 = 24 --> 24 5's.
Every multiple of 5² provides a SECOND 5:
120/5² = 4 --> 4 more 5's.
Thus, the total number of 5's contained within 120! = 24+4 = 28.

Note that 121!, 122!, 123!, and 124! all contain the same number of 5's as 120!.

For more than 28 5's to be contained within N!, N must be AT LEAST 125, the next greatest multiple of 5 after 120.
Every multiple of 5 within 125! provides at least one 5:
125/5 = 25 --> 25 5's.
Every multiple of 5² provides a SECOND 5:
125/5² = 5 --> 5 more 5's.
Every multiple of 5³ provides a THIRD 5:
125/5³ = 1 --> 1 more 5.
Thus, the total number of 5's contained within 125! = 25+5+1 = 31.

Note that 126!, 127!, 128!, and 129! all contain the same number of 5's as 125!.

Thanks for the amazing explanation Mitch. I have two follow up questions:
1. You said that number of zeroes in a number depends upon the combination of 2's and 5's. So, would just counting the number of 5's suffice?
2. Applying your approach, if in a situation I have to count the number of zeroes in (for eg. 500!)
then
Multiple of 5 in 500! = 500/5 = 100
Multiple of 5^2 in 500! = 500/25 = 25
Multiple of 5^3 in 500! = 500/125 = 4

Can I conclude that 500! will have (100+25+4) = 129 trailing zeroes?

Thanks
Manik

manik11 wrote:Thanks for the amazing explanation Mitch. I have two follow up questions:
1. You said that number of zeroes in a number depends upon the combination of 2's and 5's. So, would just counting the number of 5's suffice?

Correct.

2. Applying your approach, if in a situation I have to count the number of zeroes in (for eg. 500!)
then
Multiple of 5 in 500! = 500/5 = 100
Multiple of 5^2 in 500! = 500/25 = 25
Multiple of 5^3 in 500! = 500/125 = 4

Can I conclude that 500! will have (100+25+4) = 129 trailing zeroes?

Your approach is great, but the values in red are incorrect.

Every multiple of 5 within 500! provides at least one 5:
500/5 = 100 --> 100 5's.
Every multiple of 5² provides a SECOND 5:
500/5² = 20 --> 20 more 5's.
Every multiple of 5³ provides a THIRD 5:
500/5³ = 4 --> 4 more 5's.
Thus:
The total number of trailing zeros = 100+20+4 = 124.