After the last few essays I’d like to take a moment for a distinct, cute little problem of no practical use but cute.
Write down as many 9’s as you like, and when finished with that place a 6 at the right end. The result is divisible by 6.
That is, whatever number you’ve written, divided by 6, produces a whole number. Divisibility is one of those things which turns up whenever you have a collection of things which can be multiplied, and one thing is divisible by the second if you can find something in your collection so that the second multiplied by your find equals the first. It’s most often used to talk about the integers — the positive counting numbers, their negative counterparts, and zero if we didn’t include that already — and if it isn’t said divisible-with-respect-to-what then integers are what is usually meant. Partly that’s because integers are the first thing where divisibility stands out: if we look at the real numbers, everything is divisible by everything else (as long as that “else” is not zero), and a property that’s (almost) always true is usually too dull to mention. The next topic where divisibility gets mentioned much is usually polynomials, with a few eccentrics holding out for the complex numbers where the real part and the imaginary part are both integers.
There are several ways to prove this string of 9’s followed by 6 is divisible by 6. Here’s a proof which I like.
The first piece: 6 is itself divisible by 6. This is probably not a particularly controversial step — I can’t imagine anyone parting with me over this issue — but I do need it. Proofs often work like that.
The next piece: 96 is divisible by 6. I can tell this by experiment, but I have another way to show it. Another way to write 96 is 6 + 90, and 90 is itself 6 times 15. 6 plus 6 times 15 has to be divisible by 6.
Another piece: 996 is equal to 96 plus 900. 96 is divisible by 6; and 900 is 90 times 10, which is 6 times 15 times 10 and therefore is divisible by 6 all over again.
9,996 will fall next, since the difference between 996 and that is 9000 and that’s again divisible by 6. Then comes 99,996, and 999,996, and so on until we reach whatever number you wrote down originally.
This is an example of an “induction proof”. In an induction proof, you show that some formula or claim or whatnot can be indexed by an integer; in this case, that index is the number of 9’s before the terminal 6. And you show that whatever claim you have is true for some base case, usually when the index is 0 or 1. Typically this is the easy part; often it’s a problem simple enough to just test. 6 is divisible by 6. (I have a nagging memory of some induction proof, I believe in functional analysis, where this base-case is the hard part. But functional analysis is a strange subject, one in which it’s shown that the things which arithmetic does to ordinary numbers can be done to functions themselves. It’s easy enough to imagine adding two functions together or multiplying them, but when one starts taking the square root of a function the math major — this math major, at least — needs to spend some time simply following the rules and seeing where they lead before growing confident they do lead somewhere worth going.)
And here I have to pause a moment, and give in to something: I’ve managed to write coherently referring to some number whose value I don’t care about, just by calling it “the number of 9’s”. For the majority of humanity’s manipulations of numbers, something like that was acceptable; we might speak of the length of something, or the amount to be found, or if all else fails “the thing” or some other abstract semi-name for a quantity we want to talk about without committing ourselves to it being, say, four. Eventually mathematics yielded to using simple letters, such as the famous “x”, for a number whose value we may or may not know, and may or may not care about, and for much the same reason I might in telling an anecdote reduce the label “my brother’s doctor’s office’s receptionist” to “he”. It’s less cluttered, is unambiguous once I’ve established what it does represent, and by being less of a challenge to short-term memory makes understanding clearer. Here I will need a number, whose value I don’t care about, except that it will be one of the counting numbers. Following considerable precedent, I’ll call that “n”.
The second part is where the attempt at induction usually falls apart, as it’s easy to shuffle around equations until one gives up and calls that proved. You must show that if your formula is true for the index n, then it follows the formula must be true for n + 1. Putting this little proof about 9’s and 6 in abstract and less understandable terms: let n represent the count of how many 9’s are in our number. Now suppose we know already that a given number, made of n 9’s followed by a 6, is divisible by 6. The next number in the sequence is n + 1 9’s followed by a 6. This is the “if” part of our statement.
Here is the “then”: the next number minus the given number is 9 followed by n + 1 zeroes. That difference is 90 times the number made by 1 followed by n zeroes. That’s 6 times 15 times the number made by 1 followed by n zeroes, and has to be divisible by 6. And a number divisible by 6 plus another number divisible by 6 gives us a sum which is divisible by 6.
So if the given number, n 9’s followed by a 6, is divisible by 6, then the next number, n + 1 9’s followed by a 6, also must be.
And this is how the induction works. We showed a base case, by experimentation, that the number made by zero 9’s followed by a 6, that is 6, is divisible by 6. Through the induction step, if the number made with zero 9’s and a 6 is divisible by 6, then the number made with one 9 and a 6 (96) must also be. And if the number made with one 9 and a 6 is divisible by 6, then the number with two 9’s and a 6 (996) must be. And therefore also three (9,996), and four (99,996), and so on, up to any length we like, and past that.
This isn’t the only, or maybe even the best, way to prove this. But it brought me a smile when it popped into my head, and that’s of value.
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