Why does adding a suitable multiple of $9$ always lead to the reverse of the number?

For example:

  • $12$ reversed is $21$ and $12$ + $9$ = $21$.

  • $17$ with the two values swapped is $71$, and $17$ + $9$ + $9$ + $9$ + $9$ + $9$ + $9$ = $71$.

  • Take the number $123$ and add $9$ a total of $22$ times you get $321$, which is the reverse.

It seems to work for every number. Why is this the case? Is it just a addition thing?

Answer

Any number is congruent modulo $9$ to the sum of its digits:
$$a_na_{n-1}\ldots a_1a_0\equiv a_n+a_{n-1}+\ldots +a_1+a_0\pmod 9$$
If you reverse the digits, the sum is unchanged. Therefore $$a_0a_1\ldots a_{n-1}a_n\equiv a_na_{n-1}\ldots a_1a_0\pmod 9$$
which is equivalent to your statement.


Note a stronger statement implied by the same reasoning:

By adding or subtracting a suitable amount of $9$’s, you can reach any permutation of the digits.

Attribution
Source : Link , Question Author : fwefwf , Answer Author : Arnaud Mortier

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