# The Center of Math Blog

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## Monday, April 27, 2015

### Problem of the Week

That weekend was too short, wasn't it? But anyways, we're back at the Center with a new Problem of the Week. I've been feeling logic problems more and more since Cheryl's Birthday a few weeks ago. I particularly like the problems that lean more mathematical. The problem below, which I posted on our Facebook, Twitter, and Google+ accounts, looks so short and simple compared to some past problems of the week. But looks can be decieving! I found this problem on this Reddit comment. The user posted a quick solution in a follow-up comment,but I've expanded it and made it more detailed below.
 Click on any picture to expand!
This problem can be solved using simple mathematics. In theory, a young high schooler has all of the math skills required to solve the problem, but it requires a few very logical steps. Here's my solution:

So at this point, it's important to look at the problem critically. We're looking for the number of trailing zeros, and actually expanding the number is not feasible or necessary. So what makes a trailing zero? Every time a number is multiplied by 10, it gets a zero. So, how many times is 1005! multiplied by 10?

A more reasonable question, though it may not look like it at first, is to ask how many times 1005! is mutiplied by the pair 2x5. These are the prime numbers that make up 10, and we can figure this out by imagining the prime factorization of 1005!. It is simple to realize that there are more 2s in the prime factorization, so we need to find how many 5s exist in the prime factorization of our quantity.

I started below by identifying a set N of each of the numbers that make up 1005!.

So I've identified that when a number n in the set of N is divisible by 5, there will be a 2x5 pair in 1005!. Any time there is a number divisible by another factor of 5 (25, 125, or 625), it will add another pair to the prime factorization. And from this point, when you've identified what makes a factor of 5 in the expansion, it's easy to find how many will exist.

There are 201 n in the set of N that are divisible by 5, so that contributes 201 pairs. Continue this logic, and you'll find that there are 250 instances of (2 x 5) in the prime factorization of 1005!. Thus, there are 250 trailing zeros in the number.

If you have any questions about my solution, or if you'vee discovered another way to solve it, please let me know in the comments! Have a great Monday.
-Tori