Math Notes

Let’s go back to divisibility today. We’ve done 2, 4, and 8. And we’ve done 3, 6, and 9. I’m going to leave 5 & 10 for you. Everybody knows those, anyway. Today, I want to talk about the rules for 7 and 11.

You can get a summary of the divisibility rules here: Divisibility Rules.

Divisibility by Seven

There are lots of rules for seven. Some of them are so complicated that you may as well go ahead and divide by seven. Here’s one I like that is not so complicated, and I can finally explain why it works, too!

Take the last digit off the number. Double it and subtract it from the remaining digits. If the result is divisible by seven, then the original number is divisible by seven. If the number is still too large, repeat the process.

Here’s an example: Take 1523. We will remove the 3 and double it. That makes 6. Now we find 152 - 6 = 146. Still too big (for me, anyway) so we will repeat. Remove the 6 and double it. That makes 12. 14 - 12 = 2, so this number is not divisible by 7.

What makes this rule so cool to me is that it seems completely improbable. How in the world would anyone come up with it? Does it work every time? The answer is that someone discovered it by using modular arithmetic, so it’s not as random as you might think, and yes, it works every time. (We use modular arithmetic when we are concerned more with the remainder than with the quotient.)

Here’s why it works. I’m indebted to jimloy.com for the explanation.

First of all, any number with 2 or more digits can be written as 10 times all the digits from the ten’s place forward plus the digit in the one’s place. Let’s use 196 as an example:

196 = 190 + 6 = (19 x 10) + 6

Second, since 2 and 7 have no common factors, then a number is divisible by 7 if and only if 2 times that number is also divisible by 7. Therefore, if the twice the number is not divisible by 7, then the original number is not divisible by 7. (For example, 50 is not divisible by 7, so neither is 25.)

In our example:
196 = (19 x 10) + 6 so
2(196) = 2[(19 x 10) + 6] = 20(19) + 2(6) <– This number will not be divisible by 7
unless 196 is divisible by 7.

Now for the twist. (I love this part!) If we have a multiple of 7 and we subtract a different multiple of 7, the result is still a multiple of 7. (Try it mentally. Say 63 - 14 or 35 - 7.) On the other hand, if our number is not a multiple of 7, then the result of the subtraction will still not be a multiple of 7. The trick is to choose our multiple of 7 very carefully.

In our example:
If 20(19) + 2(6) is divisible by 7, then
20(19) + 2(6) - 21(19) is also divisible by 7. (21 x 19 is our special multiple of 7.)
That simplifies to -19 + 2(6) or
19 - 2(6) if we multiply by -1.

Now notice that we have produced our rule. Double the last digit (the 6) and subtract it from the remaining digits. You see, 21 x 19 was specially chosen because 21 - 20 = 1. There are many variations on this rule, but to me, this is the easiest to remember and use.

Divisibility by 11

So how about 11? Again, there are a lot of rules for 11, but the one I like best is to start with the digit in the one’s place, subtract the digit in the ten’s place, add the digit in the hundred’s place, . . . etc. If the result is a multiple of 11, then the original number is a multiple of 11.

Why does that work? Here’s one explanation: Since 10 + 1 = 11, then 10 times a number plus that number is a multiple of 11. For example 10 x 2 + 2 = 22, which is a multiple of 11. 10 x 5 + 5 = 55, which is a multiple of 11. In other words, we could say that 10 times any number “lacks” that number to be a multiple of 11. (20 lacks 2 to be a multiple of 11, and 50 lacks 5.)

In the hundreds place, we have the opposite problem. 100 - 1 = 99, which is a multiple of 11. So 100 is 1 too many to be a multiple of 11. 200 is 2 too many, 300 is 3 too many. You get the idea.

And it turns out that every power of ten alternates in this way. The thousands are one too small, the 10 thousands are one too big…. etc. Again, we are letting modular arithmetic do the work for us.

Now we know that if we are dividing by 11 and we have a remainder of 11, then we actually have no remainder at all. And that’s what is making this rule work. The one’s place is contributing remainders, while the ten’s place is using them up, and the hundred’s place is contributing remainders, and so on. The neat thing is that the remainders that each column is contributing or using are exactly equal to the digits we see in the number.

Let’s take an example of 418. This is really 400 + 10 + 8.
400 / 11 has a remainder of 4
10 will use 1 of those remainders to make a multiple of 11.
8 / 11 has a remainder of 8.
8 - 1 + 4 = 11, which is no remainder at all. So 418 is a multiple of 11.

Let’s try again. Is 528 divisible by 11? 8 - 2 + 5 = 11, so the rule says yes. Let’s see why:

528 = 500 + 20 + 8.
500 / 11 has a remainder of 5.
20 needs 2 added to it to make a multiple of 11.
8 / 11 has a remainder of 8.
So our remainder of 5, plus our remainder of 8, minus the two remainders that we gave the 20, make 11, which is like having no remainder at all.

You don’t have to think through that every time, but I wanted to show you why it works. All you have to remember is to add & subtract the digits, starting with the one’s place.

If this kind of reasoning is interesting to you, you might like to take a course in college called number theory. That’s where you get to learn all about modular arithmetic and a lot of other fun stuff, too.