A young fellow, six years old, recently made the delightful discovery that even numbers are made of pairs of other numbers. In fact, most mathematicians would define even numbers in a similar way: any even integer, divided by two, has a remainder of zero. Adding the quotient to itself is the same as doubling the quotient, and will give the original number. When the remainder is one, the original number is odd.

So how had this young man been taught? He had been taught – as many of us are – that even numbers end with a 0, 2, 4, 6, or 8. This is absolutely true. It is a “math fact” – but it stands in isolation; it has no connection to anything else; it actually obscures the concept of “even” versus “odd.”

Why? It is a lucky accident that our numbers are based on an even number, ten. If we were three-handed space aliens, we might have developed a number system based on fifteens, not tens. Such a species would eventually discover the idea of “even” and “odd” numbers, but they’d probably first develop some concept of the remainder when dividing by three – which would be a natural fit with their system.

Computer programmers often use hexadecimal notation, because it is a good fit with computer systems. I will borrow their idea, but use base 15. The digits in base 15 would be 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E. The letters would represent what we call “10”, “11”, “12”, “13”, and “14”.

Let us take a random number in base 15: 4E33. I can tell immediately that this number is exactly divisible by 3. Why? Because the last digit is divisible by 3, AND the number base is divisible by 3. To illustrate, I shall use a smaller number: E3. This stands for E fifteens plus 3; 14 times 15 plus 3, in our usual decimal system. There is no need to multiply it out. Let us ask “what is 14*15+3 divided by 3?”

To answer that question, divide each part by 3: what is 14*15 divided by 3? Since 15 is divisible by 3, we can divide it directly, “cancelling the factor of 3” – leaving 14*5. In the same way, dividing 3 by 3 gives one. There is no remainder in either case. We put them back together: the answer is exactly 14*5+1, no remainder.

If I know which digits are divisible by 3 in a base 15 system, I can immediately tell if some number is divisible by 3 as a whole. These digits would be 0, 3, 6, 9, and C.

If all that is confusing, let us switch back to base 10. Ten is an “even” base because ten is divisible by 2. Two fives make ten. Two 50s make 100. Two 500s make 1000. And so on and so forth. So, when looking at a ginormous number such as 65342354, we need only look at the one’s digit – the 4 on the right – and we can say that it is divisible by 2.

There’s an obvious place for math facts such as this. If you had to do the complete division by 2, it would take longer to find the answer. But this math fact actually depends on an interesting concept. Knowing both of these things gives you a richer understanding of what an “even number” is; it reveals more interesting information about the decimal number system.

For example, you might discover a test for division by 25: look at the rightmost two digits, the ones and tens; if they are “00”, “25”, “50”, or “75”, the number as a whole is divisible by 25. This works because hundreds, thousands, and higher powers of ten are always divisible by 25.

There are a whole series of similar ideas, for any factor or product of the factors of 10 – that is, 2, 5, 10, 25, 100, and so forth.

What about dividing by 3? There’s a test for that, too, but you have to look at the whole number. Add all the digits together. If this is more than one digit, add the digits of the new sum together. If the result is divisible by 3, then the original number was divisible by 3.

This method is called “casting out nines.” Why? Ten is 9+1. One hundred is 99+1. One thousand is 999+1 – and so forth. If you throw away all the 9s in a number, and keep the matching ones, you are essentially adding the digits as I have described above.

Let’s take an example, the number 27. Two tens and seven – two nines and two ones and seven. Cast out the nines, and you have 2 + 7, which is 9 – which is divisible by 9 – and also by 3.

Another example: 127 – one hundred and twenty and seven. 99 + 1 + 9 + 1 + 9 + 1 + 7. Throw away the 99 and the 9s. 1+1+1+7 is 10. Add the 1 and 0 digits in the same way; the result is 1. If you were to do the long form division, you’d have 14 nines and a remainder of 1. Casting out nines quickly gives you the remainder.

When a child learns concepts this way, they’ll stick better than a collection of random facts. To the mathematical mind, connections and sense and patterns are more valuable than seemingly-random facts.

This was really fascinating. I had never heard about casting nines before, and I’m a high school student taking calculus. That being said, if I had been told this in elementary school, I don’t think I would have been as interesting. I do math competitions semi-regularly (which usually feature more creative problems than school math), and I’ve used divisibility rules countless times. I don’t feel like I could appreciate the proof behind this rule had I not applied it multiple times.

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I forget exactly where I learned about casting out nines, but it was probably a very old – early 1900s – math text. In those days, math was about day-to-day calculation, not so much abstract set theory, etc.. Accountants used casting out nines to check for errors in their ledgers. They’d cast out nines both vertically and horizontally, and check that the figures matched.

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