# 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?

## 3 Answers

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