A while back I wrote about a method to test whether a number is divisible by seven. I recently ran across another method for testing divisibility by 7 in Martin Gardner’s book The Unexpected Hanging and Other Mathematical Diversions. The method doesn’t save too much effort compared to simply dividing by 7, but it’s interesting. It looks a little mysterious at first, though the explanation of why it works is very simple.

Suppose you want to find whether a number *n* is divisible by 7. Start with the last digit of *n* and write a 1 under the last digit, and a 3 under the next to last digit. The digits under the digits of *n* cycle through 1, 3, 2, 6, 4, 5, repeatedly until there is something under each digit in *n*. Now multiply each digit of *n* by the digit under it and add the results.

For example, suppose n = 1394. Write the digits 1, 3, 2, and 6 underneath, from right to left:

1394

6231

The sum we need to compute is 1*6 + 3*2 + 9*3 + 4*1 = 6 + 6 + 27 + 4 = 43.

This sum, 43 in our example, has the same remainder when divided by 7 as the original number, 1394 in our example. Since 43 is not divisible by 7, neither is 1394. Not only that, the result of our method has the same remainder by 7 as the number we started with. In our example, 43 leaves a remainder of 1 by 7, so 1394 also leaves a remainder of 1.

You could apply this method repeatedly, though in this case 43 is small enough that it’s easy enough to see that it leaves a remainder of 1.

Suppose you started with a 1000-digit number *n*. Each digit is no more than 9, and is being multiplied by a number no more than 6. So the sum would be less than 54000. So you’ve gone from a 1000-digit number to at most a 5-digit number in one step. One or two more steps should be enough for the remainder to be obvious.

Why does this method work? The key is that the multipliers 1, 3, 2, 6, 4, 5 are the remainders when the powers of 10 are divided by 7. Since 10^{6} has a remainder of 1 when divided by 7, the numbers 10^{6a+b} and 10^{b} have the same remainder by 7, and that’s why the multipliers have period 6.

All the trick is doing is expanding the base 10 representation of a number and adding up the remainders when each term is divided by seven. In our example, 1394 = 1000 + 3*100 + 9*10 + 4, and mod 7 this reduces to 1*6 + 3*2 + 9*3 + 4*1, the exact calculation above.

The trick presented here is analogous to casting out nines. But since every power of 10 leaves a remainder of 1 when divided by 9, all the multipliers in casting out nines are 1.

You could follow the pattern of this method to create a divisibility rule for any other divisor, say 13 for example, by letting the multipliers be the remainders when powers of 10 are divided by 13.

**Related post**: Divisibility rules in hex

“You could follow the pattern of this method to create a divisibility rule for any other divisor, say 13 for example, by letting the multipliers be the remainders when powers of 10 are divided by 13.”

True, but a little inconvenient, as the remainders are 1, 10, 9, 12, 3, 4, some of which are 2 digits.

Still, when faced with such a conundrum on a desert isle with no Wolfram Alpha or Raspberry Pi…

QM: You could replace the 2-digit multipliers by negative 1-digit multipliers: change 10 to -3 and 12 to -1.

Casting out 11’s:

A revered high school math teacher adroitly goaded a group of us into discovering “casting out 11’s”:

A number is divisible by 11 if the difference of the sum of odd digits and sum of even digits is divisible by 11. This of course is the same as a pattern of multipliers of 1, -1, 1, -1… in the above general case.

The excitement of a first discovery echoes a long time, I still find myself checking out license plates for div by 11 in idle moments. This brings three zings: fun of doing an obscure thing, warm memory of the discovery process and even warmer memory of the subsequent discovery of indirect teaching skills.