Take a whole number, for example 1746, and reverse it, obtaining for example 6471. Three non-obvious things:

- The difference of the number and its reverse (or vice versa) is divisible by 9.
- If the original number has an odd number of digits, the difference of the number and its reverse is divisible by 11 (and by 9, hence by 99).
- If the original number has an even number of digits, the sum of the number and its reverse is divisible by 11.

## Divisibility by 9

The number 1746 can be written as 1×10^{3} + 7×10^{2} + 4×10 + 6; in general an *n–*digit number can be written sum(*i* = 0..*n*-1) [*d _{i}* x 10

^{i}], where the

*d*are the digits. Then the number minus its reverse, or vice versa if the reverse is larger, is |sum(

_{i}*i*= 0..

*n*-1) [

*d*

_{i}–d_{(n–i)}x 10

^{i}]| = |sum(

*i*= 0..

*n*-1) [

*d*x (10

_{i}^{i}–10

^{(n–1–i)})]|. For instance, 6471–1746 = (6–1)x10

^{3}+ (4–7)x10

^{2}+ (7–4)*10 + (1–6) = 6 x (10

^{3}–1) + 4 x (10

^{2}–10) + 7 x (10–10

^{2}) + 1 x (1–10

^{3}) = 6 x 999 + 4 x 90 – 7 x 90 – 1 x 999. That last number is obviously divisible by 9, and in the general case the result is divisible by 9 if (10

^{i}–10

^{(n–1–i)}) is. But 10

^{m}mod 9 = 1 for all

*m*, so (10

^{i}–10

^{(n–1–i)}) mod 9 = 0.

## Divisibility by 11

As before, the difference is divisible by 11 if (10^{i}–10^{(n–1–i)}) is, where in this case *n* is odd; let *n* = 2*k*+1; then (10^{i}–10^{(n–1–i)}) = (10^{i}–10^{(2k–i)}). But 10 mod 11 = 10 = –1 mod 11, therefore 10^{m} mod 11 = 1 (10) if *m* is even (odd). *i* and (2*k*–*i*) have the same parity, so 10^{i} mod 11 = 10^{(2k–i)} mod 11, and (10^{i}–10^{(2k–i)}) mod 11 = 0.

By similar reasoning, when *n* = 2*k *is even, the sum is divisible by 11 if (10^{i}+10^{(2k–1–i)}) is. *i* and 2*k*–1–*i* have opposite parity, so 10^{i} mod 11 = –10^{(2k–1–i)} mod 11, and (10^{i}+10^{(2k–1–i)}) mod 11 = 0.