Revisiting Factorials

Last week I did a fun project with my older son about this problem from the 2015 AMC 10b:

Problem 23 from the 2015 AMC 10b

Here’s the problem:

Screen Shot 2015-12-10 at 8.35.55 PM

I wrote about the project here:

A Challenging Factorial Problem

and wrote about somethings the project make me think about here:

Mostly I don’t miss academic math, but sometimes I do

Today I wanted to revisit the project with the boys so that we could answer one question that my older son had – are there infinitely many positive integers with the property mentioned in the AMC 10 problem. So, are there infinitely many positive integers n with the property that (2n)! ends in three times as many zeros as n! does?

Let’s see . . . we began by discussing the problem and discussing some basic ideas about factorials. Mostly it is my younger son speaking in this and the next video since my older son already worked through the problem:

At the end of the last video my younger son hit on the main idea you need to find the number of zeros at the end of n! – you just have to count the number of 5’s in the multiplication. He expands on this idea here:

Now we started to tackle the AMC 10 problem. Since this part wasn’t the main point of today’s project, we went through this section fairly quickly:

Next we dug in a little deeper to see if we could find other numbers where the number of zeros tripled as you moved from n! to (2n)!.

The boys had some great ideas here and began to think that we’d never see 3x again. In fact, their ideas was that we’d see roughly 2x the number of zeros most of the time.

To investigate the number of zeros in n! and (2n)! more carefully we moved to Mathematica. This part ran a little long (~7 min) but we were able to refine the ideas about the number of zeros roughly doubling.

Finally, just for a fun wrap up, I showed them some other fun factorial facts – what is (1/2)! for example:

So, a fun slightly deeper than usual dive into factorials. I’m happy that my son was wondering how many integers satisfied the conditions of the original problem. It was neat showing him that you could actually answer that question 🙂


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