# How To Recognize Multiples Of 100 From Not So Far Away

MJ Howard last week answered my little demonstration that it was easy to tell multiples of two, five, and ten by looking at just the last digit of a whole number, but that there weren’t any ways to tell from just the last digit whether it was divisible by four. He pointed out we could look at the last two digits, and if those were divisible by four, then the entire number would be. This is perfectly true, and it’s only by asserting that I was looking for a rule based on the last digit alone that my forecast of doom about an instant check for divisibility-by-four could be sustained.

Remember the reasoning by which we wrote out a whole number as some string of digits which I call R followed by whatever goes in the units column, which I call a. (I had been thinking of R as in the “rest” of the number, but it struck me over the week that R is also the symbol used in organic chemistry to denote a chain of carbon atoms when one doesn’t really care how many of them are lined up. This interests me as I got on this thread with a set of numbers I called “alcoholic” due to their structural resemblance to organic chemistry’s idea of alcohols.) Since we’re writing in base ten, then, the number written as Ra is ten times R plus a. Ten times R can’t help being divisible by ten, or by any of the factors of ten, which are two and five (and one, which nobody cares about).

If we want to look at tests using the last two digits, we need a convenient way to write this all down. Let us again use R to mean the rest of the number from wherever it starts all the way up to the hundreds column. Let b be whatever is in the tens column, and a in the units column. The number Rba is therefore one hundred times R plus ten time b plus a. Just as we saw above, a hundred times R can’t help being divisible by a hundred or by any factor of 100, which is a not bad set: 2, 4, 5, 10, 20, 25, and 50 make the cut. Also 100 and 1, if we were interested in those.

We’re still stuck looking for anything to do for divisibility by eight, but the pattern looks pretty clear. We can’t fit eight into 100 a whole number of times, but if we go out to 1,000 we’re back on solid grounds. So we need to look at the number Rcba, with c whatever was in the hundreds column. It perhaps doesn’t seem like any great step to check for the divisibility of a three-digit number, but it’s surely easier than if we had to do something using all nine digits, which is where we seem to be stuck if we want to check whether a number can be divided by some needlessly difficult number such as seven or fifteen. With a little work you can prove — well, make a believable argument, at least — that we can eventually form a look-at-the-last-couple-digits rule for multiples of any number made up of factors of two and five, but the rest look hopeless.

Yet some of these rules exist. We know them for three and nine, since they’re the same test, and my Dearly Beloved had wanted me to explain how they work. For three or nine, we add up all the digits in a number, and if that sum’s divisible by three or nine, so is the original number. And if we come up with a sum that’s so large it isn’t obvious whether that’s divisible by three or nine, we can just add up the digits of that number, and we can keep repeating this process until we get to a number we’re comfortable with. It’s beautiful, simple, and even lets us add a simple (if not very rigorous) error-catching mechanism to arithmetic, the process of “casting out nines”.

So why can’t a similar rule work to check whether something can be divided by thirteen?

It can, and as soon as I show why all the obvious stuff above is actually not at all obvious, we can not just write a rule but know why they work.

## Author: Joseph Nebus

I was born 198 years to the day after Johnny Appleseed. The differences between us do not end there. He/him.

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