In the most recent episode of our podcast My Favorite Theorem, Kevin Knudson and I talked with aBa Mbirika (who goes by aBa), a math professor at the University of Wisconsin Eau Claire. You can listen to the episode here or at kpknudson.com, where there is also a transcript.

If you'd like to see aBa's talk from the Joint Mathematics Meetings, which he mentioned in the episode, it was recorded, and you can watch it here.

As is the case for many of our guests, aBa had trouble choosing a favorite theorem. He took us on a tour of a few of his favorites before settling on one to tell us more about. The theorem he settled on is not an important theorem. You won’t find it in any textbooks. But it was important to aBa when he was a beginning math student.

Many people are familiar with divisibility tests for 3 and 9: if the (base ten) digits of a number add up to a multiple of 3, the number is divisible by 3. If they add up to a multiple of 9, the number is a multiple of 9. There are also easy ways to tell whether a number is divisible by 2 or 5 or 11. But you probably didn’t learn a divisibility test for 7. (There is one, but it’s not as easy to use as the rules for 3 and 9.)

The proof for divisibility by 9 uses modular arithmetic, where you only look at the remainders of numbers when divided by another number. It is also called clock arithmetic because we use arithmetic mod 12 when we tell time. (Five hours after 10:00 is 3:00 because 10+5=3 mod 12.) The proof that numbers are divisible by 9 if their digits sum to a multiple of 9 uses the fact that 10, the base of the decimal number system, is 1 mod 9. Divisibility rules use modular arithmetic because the basic question is: does this number have a remainder of 0 when divided by another number?

The proof for 9 blew aBa’s mind, so shortly after he learned it, he decided to see what else he could do with the general idea. On a long bus ride, he started in on 7 to see whether the same ideas could apply to a divisibility test for 7. Ten is equal to 3 mod 7, so the test he found uses a lot of threes. In base ten, we write numbers as a sum of multiples of powers of ten. The number 3146 means 6×10^{0}+4×10^{1}+1×10^{2}+3×10^{3}. If we are trying to figure out what this number is mod 7, we can change all those tens to threes. Mod 7, 3846=6×3^{0}+4×3^{1}+1×3^{2}+3×3^{3}. Then, if that number is divisible by 7, 3846 is as well. So we can evaluate that expression: 6+12+9+81=108. And 108 is not a multiple of 7, so 3146 is not as well.

This rule for 7 does not scale up as well as the rules for 3 and 9. The powers of 3 you need to compute for the 7 rule get unwieldy. Using this rule for a number that has 10 digits would require multiplying the leftmost digit by 59,049. (An exercise for the interested reader: figure out how to adapt this rule to be a little more practical by taking advantage of the cyclic nature of powers of 3 mod 7.) But even though the rule is not terribly useful, aBa told us that figuring out this theorem and its proof changed his life. He created something new in mathematics for the first time, and he was hooked.

Enrique Treviño, who was a guest on the Lathisms podcast, told me a similar story. The problem that hooked him was showing that the sequence 2, 5, 8, 11, 14,…, where each successive number is 3 more than the last, has no perfect squares. The proof, which also uses modular arithmetic, is fairly simple, but he describes it as an “eye-opening” experience that helped pave the way for his mathematics career.

I suspect many other mathematicians have similar stories of the simple, often unimpressive theorems and exercises that first made them feel like they had the power to create in mathematics. I certainly felt this way in my very first proof writing class. So today I salute the tiny theorems that helped us become mathematicians.