# Let N be a four digit number, and N' be N with its digits reversed. Prove that N-N' is divisble by 9. Prove that N+N' is divisble by 11.

Let $N$ be a four digit number, and $N'$ be $N$ with its digits reversed. Prove that $N-N'$ is divisible by $9$. Prove that $N+N'$ is divisible by $11$.

I let $N=abcd$ and $N'=dcba$

but I dont see how I can proceed to show that either the sum or difference is divisble by 11 or 9.

Any ideas on how to start this problem?

The trick is to look write your number base 10, then use modular arithmetic to make life easy. For example, write $N = a \cdot 10^3 + b \cdot 10^2 + c \cdot 10 + d$. 10 is congruent to 1 modulo 9, so $$N \equiv a (1)^3 + b(1)^2 + c(1) + d \equiv a + b + c + d \mod 9$$ and similarly $N' \equiv d + c + b + a \mod 9$. Then $N - N' \equiv 0 \mod 9$, which just means that 9 divides $N-N'$.
To show that their sum is divisible by 11, do the same thing but look at the sum modulo 11, noting that $10 \equiv -1 \mod 11$.
\begin{align} N &= abcd = 1000a + 100b + 10c + d\\ N' &= dcba = 1000d + 100c + 10b + a\\ N - N' &= 999a + 90b - 90c -999d\\ N + N' &= 1001a + 110b + 110c + 1001d \end{align}
Hint: Analyze the numbers $N$ and $N'$ $\mod 9$ and $\mod 11$. Maybe write them as $$N= 1000a+100b+10c+d$$ and $$N'= 1000d+100c+10b+a$$
If you calculate $N+N' \mod 11$ or $N-N' \mod 9$ this will give you the answer