Any Number Minus its Reverse is Divisible by 9

Assume that we have the number x = a10^n + b.10^(n-1) + c10^(n-2) + ... + m10^1 + n10^0
where a, b, c...m, n members of the set {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}.
The number made up from the reverse of these digits y = n10^n + m10^(n-1) + ... + c10^2 + b10^1 + a10^0.
Without loss of generality we can assume that x > y, therefore
x - y = (a10^n + b.10^(n-1) + c10^(n-2) + ... + m10^1 + n10^0) - (n10^n + m10^(n-1) + ... + c10^2 + b10^1 + a10^0)

x - y = a10^n + b.10^(n-1) + c10^(n-2) + ... + m10^1 + n10^0 - n10^n - m10^(n-1) - ... - c10^2 - b10^1 - a10^0

x - y = a10^n - a10^0 + b10^(n-1) - b10^1 + c10^(n-2) - c10^2 + ... + m10^1 - m10^(n-1) + n10^0 - n10^n

x - y =  a(10^n - 10^0) + b(10^(n-1) - b10^1) + c(10^(n-2) - 10^2) + ... + m(10^1 - 10^(n-1)) + n(10^0 - 10^n)

However, we know from before that any power of 10 minus a power of 10 is divisible by 9, thus each set of brackets of the form (10^a - 10^b) can be replace by 9ki for some ki member of the integers.
Therefore,
 x - y = a9k1 + b9k2 + c9k3 + ... + m9kn-1 + n9kn


x - y = 9.(ak1 + bk2 + ck3 + ... + mkn-1 + nkn)

QED

Note
Without much effort it can easily be shown that any number minus a number made from a permutation of its digits must also be divisible by 9.