Every so often, I’ll informally teach a class of gifted elementary-school students. I greatly enjoy interacting with them, and I especially enjoy the questions they pose. Often these children pose questions that no one else will think about, and answering these questions requires a surprisingly depth of mathematical knowledge.
Here’s a question I once received:
What is the chance of winning a game of BINGO after only four turns?
When my class posed this question, I was a little concerned that the getting the answer might be beyond the current abilities of a gifted elementary student. As discussed over the past couple of posts, for a non-standard BINGO game with 44 numbers, the answer is
After we got the answer, I then was asked the question that I had fully anticipated but utterly dreaded:
What’s that in decimal?
With these gifted students, I encourage thinking as much as possible without a calculator… and they wanted me to provide the answer to this one in like fashion. For my class, this actually did serve a purpose by illustrating a really complicated long division problem so they could reminded about the number of leading zeroes in such a problem.
Gritting my teeth, I started on the answer:
How long do we have before the digits start repeating?
My stomach immediately started churning.
I told the class that I’d have to figure this one later. But I told them that the answer would definitely be less than 135,751 times. My class was surprised that I could even provide this level of (extremely) modest upper limit on the answer. After some prompting, my class saw the reasoning for this answer: there are only 135,751 possible remainders after performing the subtraction step in the division algorithm, and so a remainder has to be repeated after 135,751 steps. Therefore, the digits will start repeating in 135,751 steps or less.
What I knew — but probably couldn’t explain to these elementary-school students, and so I had to work this out for myself and then get back to them with the answer — is that the length of the repeating block is the least integer so that
which is another way of saying that we’ve used the division algorithm enough times so that a remainder repeats. Written in the language of group theory, is the least integer that satisfies
(A caveat:this rule works because neither 2 nor 5 is a factor of 135,751… otherwise, those factors would have to be taken out first.)
Some elementary group theory can now be used to guess the value of . Let be the multiplicative group of integers modulo which are relatively prime which. The order of this group is denoted by , called the Euler totient function. In general, if is the prime factorization of , then
For the case at hand, the prime factorization of can be recovered by examining the product of the fractions near the top of this post:
Next, there’s a theorem from group theory that says that the order of an element of a group must be a factor of the order of the group. In other words, the number that we’re seeking must be a factor of . This is easy to factor:
Therefore, the number has the form
where , , , and are integers.
So, to summarize, we can say definitively that is at most , and that were have narrowed down the possible values of to only possibilities (the product of one more than all of the exponents). So that’s a definite improvement and reduction from my original answer of possibilities.
At this point, there’s nothing left to do except test all 126 possibilities. Unfortunately, there’s no shortcut to this; it has to be done by trial and error. Thankfully, this can be done with Mathematica:
For more on this, see https://meangreenmath.com/2013/08/23/thoughts-on-17-and-other-rational-numbers-part-6/ and https://meangreenmath.com/2013/08/25/thoughts-on-17-and-other-rational-numbers-part-8/.