[sorry if this isn’t edited well, limited editing time due to lots of kid activities today]

Saw this post on twitter earlier in the week:

It made for a great follow up to a previous Family Math which also happened to arise from a problem we saw on twitter:

The first thing that we did this morning was talk through the problem to be clear about a couple of math terms – proper divisors, for example. We also looked at a few other integers to help make sure that we really did have our arms around the problem. One neat thing is that the kids were able to see that when we listed the factors of a number, the solution to Dave’s problem required exactly four factors (or exactly two rows of divisors the way were were listing them).

The next step was to try to understand how you could sort out which numbers had exactly two rows of divisors. My older son noticed that numbers that were the product of two primes (to the first power) would have exactly four divisors. We worked through a couple of examples of integers of this form and found that they did indeed have that property. My younger son was able to explain why this was the case – thanks to Art of Problem Solving’s *Introduction to Number Theory* book! The last thing we talked about in this part was whether or not numbers of this form comprised all of the solutions to Dave’s problem:

Next up we started searching for other potential solutions. We started off down the path of looking at powers of primes – it turns out this was a lucky road to head down. We looked at squares and fourth powers, but neither of these types of numbers seemed to solve the problem. It did give us the idea to look at cubes of primes, though, and that showed us one more set of solutions:

The last task was to see why the two types of numbers that we’d found – products of exactly two primes, and a single prime cubed – formed a complete set of solutions to the problem. This part of the talk has a little more theory in it, but I think the lesson here is important – how do we know that our solution is complete? We talk about how you count the number of divisors, and then how that counting process could arrive at exactly 4 divisors. Funny enough, the divisors of 4 play an important role in answering that question. The fact that there are only two ways to multiply integers together to arrive at 4 (2×2 and 4×1) tells us that we have a complete solution to the problem. Yay!

So, in a couple of weeks we went from a neat problem shared by Tracy Johnston Zager about the sum of the divisors to a neat problem shared by David Radcliffe about products of proper divisors of integers. Of course this journey was helped by the fact that I’m going through a number theory book with my younger son (and have gone through the same book with my older son previously). These are challenging problems for kids to think through for sure, but I think that kids will enjoy the challenge. These problems also do a great job of building up number sense because as you work through them you are constantly thinking about factoring, multiplying, and integers that share certain types of properties. I forget the exact phrase, but in my mind problems like these definitely belong in the “low entry point / high exit point” problems that the people who study math education seem to really like.

As always, I’m glad that people are sharing problems like these ones on Twitter!

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