Okay, I know divisibility rules are kind of child's play, especially since we all have supercomputers in our pockets nowadays, so there's no real practical use for this. But I've always heard that there's no quicker way to test for divisibility by 7 than just doing the long division, and using this method, I've found that with a just a little bit of practice I can do large numbers (12 digits or more) in seconds. Just wondering what others think.

Let's say the number is:

571,243,814,354

Starting from the right, if we group the digits in blocks of 3 and do the alternating sum:

354 - 814 243 - 571

the result will be congruent mod 7 to the original number.

Of course, whatever that number is, it is going to be difficult to calculate mentally. But the beauty is that we don't have to. Instead, if we can get good at finding the remainders mod 7 of 3-digit numbers, we can simplify this calculation to:

4 - 2 5 - 4 = 3

Therefore, the original number is congruent to 3 mod 7.

As for getting good at finding remainders mod 7 of 3-digit numbers that's actually really easy. There are lots of ways to do it really, but what I like to do is this:

• double the hundreds digit
• add result to the remaining 2 digits
• the result of that will be congruent mod 7 to the original number

Example: 571

5 * 2 = 10

10 71 = 81 = 77 4

Therefore 571 is congruent to 4 mod 7.

In addition, there are quite a few of what I call 'low-hanging fruit' numbers that are super easy.

For example, for all numbers in the range of 700-799 we can just ignore the 7 and treat it like a number in the 0-99 range, over which most "mathy" people already have the multiples of 7 memorized anyway (it's pretty much a requirement for this method).

Also, if the first 2 digits of the number is a multiple of 7, then the remainder is just the last (1s) digit. E.g. for 354 in my example above, 35 is a multiple of 7, so the remainder is just 4.

Further, if the last 2 digits of the number are a multiple of 7, then the remainder is just double the hundreds digit. E.g. for 814 in my example above, 14 is a multiple of 7, and 8*2 = 16 = 14 2, so the remainder is 2.

One other thing. This is not required, but I find it easier to, instead of actually doing the doubling of the hundreds digit, just memorize the the following:

Given a 3-digit number abc:

• If a is 0 or 7 -> add 0 to bc
• If a is 1 or 8 -> add 2 to bc
• If a is 2 or 9 -> add 4 to bc
• If a is 3 -> add 6 to bc
• If a is 4 -> add 1 to bc
• If a is 5 -> add 3 to bc
• If a is 6 -> add 5 to bc

I find it cuts down a little bit on the mental math on numbers like say, 895. If we double the 8, we're doing the calculation 16 95 = 111. Not difficult or anything but it does require a tiny bit more mental math because of the carrying. But more importantly we're left with another 3-digit number to figure out. Instead, I find it easier to just do, 2 95 = 97 = 91 6.

Then you just do the alternating sum calculation mentally by keeping a running total. My inner monologue goes something like:

354 -> cool, starts with 35 so remainder is 4

so 4 minus...

814 -> cool, ends with 14 and hundreds digit is 8 so remainder is 2

4 minus 2 is 2

so 2 plus...

243 -> 4 43 = 47 = 42 5 so remainder is 5

2 plus 5 is 7

so 7 minus...

571 -> 3 71 = 74 = 70 4 so remainder is 4

7 minus 4 is 3

Like I said, it does take a little bit of practice but honestly after just an hour of tooling around with this method, I was doing large numbers like this in seconds, and I'm definitely no mental math savant. Would be curious what y'all think.