Rotating multiples of 37

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 = 100a + 10b + c.

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

m = 100c + 10a + 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 + 999c is also a multiple of 37. Now

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

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

k/10 = 100c + 10a + 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.

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