Number Theory Puzzler: Proving N-S is a Multiple of 9

[sizzlaw]

Given: N is a four digit number. S is the sum of N's digits.

Prove: N minus S is a multiple of 9.

 

[Goongyae]

Call the four digits of N "abcd"
Then N is (a * 1000 + b * 100 + c * 10 + d)
And N - S = N - a - b - c - d = (a * 999 + b * 99 + c * 9)
And (N - S)/9 is (a * 111 + b * 11 + c), which is an integer.
Hence N-S is divisible by 9.

 

[sizzlaw]

Revised puzzler:

Given: N is a four digit number. S is the sum of N's digits.

Prove: The sum of the digits of (N - S) is divisible by 9.

 

[micromass]

This follows easily from Goongyae's post, since a number is divisible by 9 if and only if the sum of it's digits is divisible by 9.

This is real easy to prove: take x₁...x_n be an n-digit number, then

$$x_110^{n-1}+...+x_n10^0=x_1+...+x_n~(mod~9)$$.