Math is cool — Explained

Not only are there neat mathematical facts out there, but you can also show why they work!

Be sure to read the article “Math is cool” before reading the explanations below

Multiplying by 11

Let’s start with a two digit number. A two digit number can be represented by 10a+b, where a is the first digit and b is the second digit. When we multiply by 11 we are doing (10a+b)(11).

The good ol’ distributive property comes in handy here to show why our little trick works.





This would make the hundreds column a, the tens column (a+b) and the units column b. So we get our trick!

The proof is the same for three + digits. A three digit number can be represent by 100a+10b+c.





This would make the thousands column a, hundreds column (a+b), the tens column (b+c) and the units column c.

Squares of numbers ending in 5

The proof for this one is similar to the one above. Any number ending with 5 can be represented as (10n+5), where n represents the number formed from the digits preceding the 5.

If we square this we get:







This would make the last two digits 25 and the preceding digits n(n+1) because 100(n(n+1)) is just n(n+1) followed by two zeros. Pretty cool.

Dividing by 7

This one won’t be quite as algebraically rigorous.

Let’s go back to long division. If we are doing 1/7 through long division, we would go through the process below or something similar. The main thing to note is that when we get to the last line, we basically end up back at the first line and will end up cycling through the lines in between again, so this confirms that 1/7=0.142857142857…repeat.

Here is the important thing to note: The only possible remainders when dividing by 7 are 3, 2, 6, 4, 5, 1 or more nicely: 1, 2, 3, 4, 5, 6.

Hopefully you can see that if we were doing 2/7, we would just start from where the remainder of 2 is written in bright blue in our long division and then we would end up back in our cycle. Since our possible remainders are 1–6, this covers why the pattern works for 1/7, 2/7, 3/7, 4/7, 5/7, and 6/7.

What about 9/7 or 153/7? How did Mr. Brown do those? Well, we can think of these as n + k/7 where n is an integer and k is in the set {1, 2, 3, 4, 5, 6}. All we need to do is calculate the integer n, using our multiplication tables or otherwise, and then the decimal after it comes from the remainder, k. Well done Mr. Brown!