Bunuel 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
Kudos for a correct solution.
VERITAS PREP OFFICIAL SOLUTIONIn any N!, the number of trailing zeroes = number of factors of 2 and 5. For example, for 6! (which is 720), the ending 0 is created by taking the numbers 2 and 5 in (6)(5)(4)(3)(2)(1) and multiplying them together. It's also important to realize that since every second number is divisible by 2 but every fifth number is a multiple of 5, the number of occurrences of 5 is less than 2, hence the limiting factor is 5.
Sine we know we're working with giant numbers, we can start getting an idea of what we're looking at by seeing how many zeroes 100! ends with
Every multiple of 5 between that 1 and 100 provides one factor of 5, and at 25, 50, 75, and 100, you get an extra 5 (for example, 75 = 3 * 5 * 5, so 75 provides two factors of 5). So the number of factors in 100! is 24 as a starting point for us.
Now since N! ends with 28 zeroes, we need to four more 5’s in 100!, so N should be 120. But one thing is to be noticed here is 120!, 121!, 122!, 123! and 124! also have 28 zeroes as from 120! to 124! there is no more extra 5 factor.
Hence Statement 1 is not sufficient.
As per Statement 2 the (N+2)! has 31 zeroes.
So let’s check which numbers can have 30 zeroes. We know from the above that 120!-124! should have 28 zeroes.
But when it comes to 125!, we have 5^3 factor so 125!-129! have 31 zeroes.
So N+2 can be any number between 125 to 129, but the statement also says N-1 has 28 zeroes that means N-1 can be any number from 120! to 124!.
So N can be 125 or 124 or 123 so again Statement 2 is not sufficient.
Let’s combine Statements 1 and 2: in this case, N could be 124 or it could be 123, so E is the correct answer.