GRE question which math to use?

[m0bius]

So this isn't a homework problem I suppose, but this seems to be the best place to post. I've been going through the online math subject GRE, trying to first identify what field of math I will need to solve the problem, then figuring out how to solve them. However, the test covers a wide range of fields. About 50% of the test is Calculus and Differential Equations, 25% is Linear Algebra and Abstract Algebra, and the rest could be from any other subject including Analysis (real and complex), Topology, Probability, Combinatorics, etc...

So, there have been a few problems where I cannot even identify which math to use.

Homework Statement

For how many positive integers k does the ordinary decimal representation of the integer k! end in exactly 99 zeros?

Homework Equations

There are 5 answers to choose from : 0, 1, 4, 5, 24.

The Attempt at a Solution

There have been some types of problems I've never seen before, but have been able to figure out using various fields of math. I have never seen a problem like this before and I don't have a clue of what math to use. My guess is that it has to do with Real Analysis, but it also sounds like a Number Theory problem.

If somebody could at least point me to a field of math to use that would be great. I never took a course in Topology, Combinatorics, or Probabilty. I'm about to start learning the basics of those, but if the problem requires any of them it would be great if you could point me towards a theorem or something. Thanks!

 

[tiny-tim]

welcome to pf!

hi m0bius! welcome to pf!

For how many positive integers k does the ordinary decimal representation of the integer k! end in exactly 99 zeros?
…
There are 5 answers to choose from : 0, 1, 4, 5, 24.

it's just common-sense, really …

k! has to be divisible by 2⁹⁹ and 5⁹⁹ …

since 2⁹⁹ isn't going to be a problem, we can just look at 5⁹⁹ …

ok, now just count how large k has to be for k! to include 5⁹⁹ (and the same for 5¹⁰⁰)

 

[jem05]

i passes by this question while studying for gre subject, (my test is in 2 days )
first i thought that was a number theory question but i didnt take such a course, but anyway i solved it this way,
let's say x is a number where x! has 99 zeros, to get a number with 100 zeros we must pass by a multiple of 10, so we need both a 2 and a 5, (to get an extra zero that is) but with any 4 consecutive numbers, you must have at least 2 2's and the fifth at most will get you a 5, so the fifth one will have 100 zeros, so scratch that.
so you are left with the x!, (x+1)!, (x+2)! (x+3)!, and (x+4)!.
the answer is 5

 

[m0bius]

Thank you both for the help! Unfortunately, I am just not understanding this. After looking at your responses and the question for about an hour, I haven't gotten too far.

Ok so we assume k! ends in 99 zeros. We can write k! as n * 10⁹⁹ = n * 2⁹⁹ * 5⁹⁹, where n is some number that comes before the 99 zeros.

since 2⁹⁹ isn't going to be a problem, we can just look at 5⁹⁹ …

I'm not really sure what you mean by "isn't going to be a problem".

ok, now just count how large k has to be for k! to include 5⁹⁹ (and the same for 5¹⁰⁰)

Ok so we know that if k! ends in 99 zeros, then it must have 5⁹⁹ as a factor. The problem I have here is that we don't know how many additional 5's are in n, the number before the zeros.

I did start to count out how many 5's k! has as a factor for various values of k, which is what I think you meant. I must be doing something wrong though, because this has all been taking way too long considering you have about 2.5 minutes to do each question on the test. Only 28% of people got it right so I'm assuming most eliminated 0 as an answer and just guessed from the remaining 4.

 

[vela]

I'm not really sure what you mean by "isn't going to be a problem".

Every even integer between 1 and k will contribute at least one factor of 2 to the product, so there's an abundance of 2s about, many more than the factors of 5.

Ok so we know that if k! ends in 99 zeros, then it must have 5⁹⁹ as a factor. The problem I have here is that we don't know how many additional 5's are in n, the number before the zeros.

There can't be any. If there were, you could combine it with one of the extra 2s to form a 10, which will result in another 0 on the end of k!.

 

[hgfalling]

At some point, k! is going to have 98 zeros at the end of it. Then k is going to increment and then there are going to be 99 zeros. What are the properties of the new number k? What are the properties of the next number k that makes the number of zeros 100?

 

[tiny-tim]

(just got up :zzz: …)

don't forget that the number of zeros doesn't always jump by 1 …

what happens if k is divisible by 25 or 125? ​