If a three-digit number is divisible by 37, it remains divisible by 37 if you rotate its digits. For example, 148 is divisible by 37, and so are 814 and 481. This rotation property could make it easier to recognize multiples of 37 or easier to carry out trial division.

Before proving the theorem, I’ll state a couple things the theorem does not say. First of all, it does not say that you can reverse the digits. For example, although 148 is divisible by 37, 841 is not. Second, it does not say that rotating digits preserves the remainder by 37 except when that remainder is 0. For example, 123 = 12 mod 37, but 312 = 16 mod 37.

Now for the proof. Let *a*, *b*, and *c* be integers. Let

*n* = 100*a* + 10*b* + *c*.

and assume that *n* is a multiple of 37. We want to show that

*m* = 100*c* + 10*a* + *b*

is also a multiple of 37. If we can prove this, then we can apply the result to *m* to obtain one more rotation.

Note that 999 = 27 × 37, so *n* + 999*c* is also a multiple of 37. Now

*k = n* + 999*c* = 1000*c* + 100*a* + 10*b*

is clearly divisible by 10, and 10 is relatively prime to 37, so *k*/10 must also be divisible by 37, and

*k*/10 = 100*c* + 10*a* + *b*

which is just the rotation we were looking for.

Note that to prove the theorem at the top of the post it is enough to assume *a*, *b*, and *c* are digits, but the proof is valid for any integer values.

Nice!

It seems to be true of 27 also. There are 34 three digit numbers divisible by 27: 108, 135, 162, 189, 216, 243, 270, 297, 324, 351, 378, 405, 432, 459, 486, 513, 540, 567, 594, 621, 648, 675, 702, 729, 756, 783, 810, 837, 864, 891, 918, 945, 972, and 999. For each of the 34 numbers, its 3-digit rotations are also in the list.

Amateur mathematician here, but shouldn’t this proof also apply to all integer factors of 999? As Bob Lyons points out, 27 is also true by enumeration, but 3, 9, 111, and 333 are valid as well, even though none of them are very interesting.