Decoding divisibility tricks

Mukta's turn. Choose a two-digit number with two different digits and keep it secret, say 47. Reverse the digits: 74. Subtract the smaller from the larger: 74 − 47 = 27. Divide by 9: exactly 3, no remainder. She promises this always happens.

To see why, write the number using its digits. The number ab means 10a + b (a tens and b ones). Reversed, ba means 10b + a.

If b > a, the difference is (10b + a) − (10a + b) = 10b − b − 10a + a = 9b − 9a = 9(b − a).

There is the 9, out in the open. And the answer after dividing by 9 is not random either: it is b − a, the difference of the two digits. For 47 and 74 that is 7 − 4 = 3, which is what we got. If a > b instead, the same working gives 9(a − b), so the rule still holds.

Change subtraction to addition and a different number appears. 31 + 13 = 44. 28 + 82 = 110. 12 + 21 = 33. All are multiples of 11. Why? (10a + b) + (10b + a) = 11a + 11b = 11(a + b).

Three digits give a nicer one. Take abc and move the digits round to get bca and cab. Add all three:

(100a + 10b + c) + (100b + 10c + a) + (100c + 10a + b) = 111a + 111b + 111c = 111(a + b + c).

And 111 = 3 × 37. So the sum is always divisible by 37, and by 3 as well.

Last, the prettiest. Take abc and repeat it to make the six-digit number abcabc. Divide it by 7, then by 11, then by 13, and you get abc back. Why? abcabc = abc × 1001 (for example 123123 = 123 × 1001), and 7 × 11 × 13 = 1001. Every one of these 'magic' tricks is really a hidden factor.