# A neat problem Dan Anderson shared with us

We’ve spent the last couple of days studying divisors of integers – mainly the number of divisors and the sum of those divisors. This topic came to us via a “things you should know for math contests” list that math team at my older son’s school gave to the kids.

We’ve used Mathematica to help us get a feel for these topics and that computer work (I assume) prompted Dan to share this problem with us:

The problem is: Find the first triangular number with more than 500 divisors.

I asked the boys if they wanted to try to tackle this problem, and they wanted to give it a try. So . . . off we went:

Once the kids understood the problem, I thought it would be useful to spend some time talking about how we could approach the problem. The boys had some pretty good ideas:

The one thing I wanted to spend some extra time on was an alternate way to calculate the triangular numbers. The method that the boys proposed was actually fine, but it seemed like an extra couple of minutes talking about a different approach would be time well spent:

Now we went to the computer to implement our plan. We found that the 12,375th triangular number, 76,576,500, was the first triangular number with more than 500 divisors:

The boys were a little surprised to learn that the first triangular number with more than 500 divisors was smaller than the one with exactly 500 divisors. In fact, we didn’t even find one with 500 divisors yet. In the next part of the project we looked for that number. We did find that number, but it was much larger than we expected – the 1,569,375th triangular number is 1,231,469,730,000, which has exactly 500 divisors!

We wrapped up by looking at 5 of the triangular numbers with exactly 500 factors. They all shared a common factor of 16. We decided to look to see if there was an odd triangular number with exactly 500 factors. As of now (3 hours) after finishing up the project, the computer has not found one.

So, a really fun computer project with the boys. Thanks to Dan Anderson for providing this challenging problem!