Problem Solving

kop wrote: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

GMATGuruNY wrote: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:

kop wrote: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.

Mathsbuddy wrote:As x cannot be a multiple of 10^3, then (N+1)(N+2)= k x 10^3 (where k is an integer)

GMATGuruNY wrote:

Mathsbuddy wrote:As x cannot be a multiple of 10^3, then (N+1)(N+2)= k x 10^3 (where k is an integer)

The portion in red overly constrains the value of (N+1)(N+2).
In your solution, N! = x * 10²�.
While x is not a multiple of 10³, it IS a multiple of 2³.
Thus, we can represent N! as follows:
N! = y * 2³ * 10²�.
Thus, if (N+1)(N+2) = z * 5³, then (N+2)! = y * z * 2³ * 5³ * 10²� = yz * 10³¹.
Implication:
(N+1)(N+2) = k * 5³.

Mathsbuddy wrote:To eliminate 123:

to get from (n-1)! to (n+2)! we multiply by n * (n+1) * (n+2)

so if N = 123:

we multiply by 123 * 124 * 125 which ends in a nought

Therefore we would have to change 28 noughts to 29 noughts, which conflicts with the question.

Therefore eliminate N = 123, or indeed any value for N that ends in a 3.

GMATGuruNY wrote:

Mathsbuddy wrote:
I hope that the forgoing explanation clarifies why 123! will yield 28 trailing 0's, while 125! will yield 31 trailing 0's.

Thank you for your time and insight. I see my flaw now:

To not eliminate 123!

to get from (n-1)! to (n+2)! we multiply by M = n * (n+1) * (n+2)

so if N = 123:

we multiply (n-1)! by M = 123 * 124 * 125 = 1906500

At this point I mistakingly thought the 2 noughts at the end meant it would not increase by 3 noughts.

However, as (n-1)! = 122! is even, then (n+2)! could be a multiple of 2*1906500 = 3813000

which would add three noughts to the end of (n-1)!

Nonetheless, it would have to be the last digit before the noughts that needs to be even.

Hence I will concede to GMATGuruNY's method and congratulate him for his patience and logic!

Thank you.