There's a similar test for divisiblity by 13: cross off the least significant digit, quadruple, and add to what's left.

Example: 325 becomes 32 + 20 = 52 which is divisible by 13. Example: 8632 -> 863 + 8 = 871 -> 87 + 4 = 91 which is divisible by 13.

It works because 39 = 40-1 is divisible by 13 (similarly to why the 7 test works because 21 = 20+1 is divisible by 7).

An equivalent test would be to use a factor of -9 instead of 4.

325 => 32 - 9 * 5 = 32 - 45 = -13

The method itself comes from the fact that what we're doing is really just modular arithmetic. Which has the wonderful property that, as long as all manipulations mod N don't fundamentally change the base we can keep repeating them.

Similar to my mod 7 breakdown:

  AB = 0 (mod 13)
  10*A + B = 0 (mod 13)
  40*A + 4*B = 0 (mod 13)
  A + 4*B = 0 (mod 13)
  A - 9*B = 0 (mod 13) // because 4 = -9 (mod 13)

This is neat. A similar construction should be doable for any number.

EDIT: I have a rule of not questioning down votes, but many of my comments that get down voted have something disagreeable in them (well, I agree with them, but it's typically political/management style/economics/etc.). But how does a break down of modular arithmetic become worth down voting?

Oops. That makes more sense now that you point it out. I thought they were just little scrolling arrows, and never understood why they didn't work right. Sorry about that.

(While it's a good general practice not to question downvotes, there are some cases that just leave you wondering what the other person was thinking. This is one of the explanations I've imagined, although usually I just assume big fingers on a tiny touchscreen.)

Doing it with -9 instead of +4 will terminate faster. But on the other hand one has to think about negative numbers. It's an interesting tradeoff.

"A similar construction should be doable for any number." - yes, except 2 and 5. You just need to find a multiple of the number that's 1 more or less than a multiple of 10.

In fact, we can even come up with a test for divisibility by 3 (or 9) this way: just add the last digit to the rest of the number repeatedly until you can tell if the result is divisible by 3 (or 9):

4752 -> 475+2 = 477 -> 47+7 = 54 -> 5+4 = 9 is divisible by 3 and 9 3752 -> 375+2 = 377 -> 37+7 = 44 -> 4+4 = 8 isn't divisible by 3 or 9 4755 -> 475+5 = 480 -> 48+0 = 48 -> 4+8 = 12 -> 1+2 = 3 is divisible by 3 but not 9.

Of course this ends up being equivalent to the usual "add all the digits" test.

Using the same approach I used to construct the formulation, divisibility by 2 and 5 actually just break down to the normal "is the last digit even" or "is the last digit 5 or 0" tests. We end up with:

  10*A + B = 0 (mod 2)
  B = 0 (mod 2)

and:

  10*A + B = 0 (mod 5)
  B = 0 (mod 5)

Of course, these are trivially small cases and we've all learned the patterns. An interesting case with N = 4, the quick method I learned in grade school was to look at the last two digits. But using the approach I've adopted here, you get something more complex:

  10*A + B = 0 (mod 4)
  2*A + B = 0 or 2*A - 3*B = 0 (mod 4)

If we instead break N into ABC (A * 100 + B * 10 + C = N):

  100*A + BC = 0 (mod 4)
  BC = 0 (mod 4)

And we've got the method I learned in school.

EDIT: Formatting.

So why does this work?

...There was another answer here that seemed close.

I'm not sure the best way to express it, but here's what I've got:

Allow me to use a shorthand, N = AB where AB is shorthand for 10 * A + B, and B is the least significant digit. IF N is divisible by 7, then AB = 0 (mod 7). This allows us to do the following manipulations:

  AB = 0 (mod 7)
  10*A + B = 0 (mod 7)
  3*A + B = 0 (mod 7)
  5*(3*A + B) = 5*0 (mod 7) // EDIT: had -2 on the RHS, that was correct but confusing
  A + 5*B = 0 (mod 7)
  A - 2*B = 0 (mod 7)

So we've arrived at drfuchs' method. You only need to repeat the process until you've proven to yourself that A - 2 * B is divisible by 7 or not.

That's all I've got. I ran a half-marathon yesterday, and I've been up for 16 hours today. My brain might work better tomorrow.

EDIT: changed the steps a bit, had some unneeded ones. Also, this shows that you can use addition instead of subtraction, but you won't be moving towards 0, but you still have to be able to recognize that the new number is divisible by 7.

Thanks for this explanation! I was about to post something similar after working the math out, but I'm sure my answer would have been twice as long and less clear.

50 years from now this will be called "drfuchs last theorem"

with the scribble "but it took decades for me to realize why it obviously works" somewhere in the margins and stating that there wasn't enough room to write that down though..

You are subtracting enough 21s (ie: X * 21) to make the last digit zero. Since 21 is obviously a factor of 7, subtracting those should yield a factor of 7.

You can then divide by zero because if X is a factor of 7 and ends in zero, X / 10 is a factor of 7 as well.

21=3*7

you are subtracting multiples of 21, then dividing by 10, which is coprime with 7

The examples subtracted 10, 16, etc... Those aren't multiples of 21.

Since you scratched off the last digit, 10 is really 100, plus 5 for the digit you scratched off = 105.

105 is a multiple of 21.

Well.. Why does it, then?!