Okay, so, I was previously familiar with the ‘tricks’ for the digits apart from 7, but I found myself curious about the math behind the one for 7. Join me as I explore!
Let’s start with looking at exactly what we are told to do for the trick to work. We have a number—in the first example, it is 294—and we are being told to separate the last digit (29 separated from 4), to multiply that last digit by 2 (4x2=8), and then to subtract that product from the remaining original digits (29-8=21). Then, if the number is small enough to check against our 7s table (21=7x3), we have our answer (yay, it’s divisible!). If not, as was the case with our second example, we repeat the process.
Okay, that’s all well and good, but let’s look at what was *actually* happening. What did we accomplish? There is a bit of sleight of hand within this trick. The trick turned 294 into 21. How? Let’s run it in slow motion and find out.
What is lost in the “separate the last digit, double it, and subtract it from the rest” bit is that you are taking the last digit and multiplying it by 21 (a number divisible by 7) and subtracting *that* product from the original number.
In our example, 4x21=84, and 294-84=210
Then, because 210 (and in fact every possible number we would get at this step) is divisible by 10, we can drop the 0 (210 becomes 21) and carry on with a more mentally manageable number.
The trick simplifies the process of getting there by giving us a handy roadmap, and it utilizes the fact that 21 is divisible by 7, and the fact that subtracting 21(X), where X is the last digit, will always return a round (divisible by 10) number. It’s very clever!
You can check every number in the set {21, 42, 63, 84, 105, 126, 147, 168, 189}. Within the context of our trick, these are all the possible values we would subtract from the original number (or the new number we are checking within subsequent iterations of the trick).
Our goal is to see if our original number (294) is divisible by 7, and we have subtracted a number we know is divisible by 7 (84). By the distributive property (which states that a(b+c)=ab+ac), we can check the new number (210, but because that number is divisible by 10, we are actually checking 21), and if it is also divisible (which it is), it confirms that the original number is (yay!).