How do i show n + k is a integral multiple of 11

Question

(the digits of a positive two digit integer N are interchanged to form a integer)I don't know what this question means so I'm clueless and what do i do to solve? please in steps so i can understand it thanks ahead of time

Answer 1

I believe the question saying the following:

Take a positive integer $AB$ (where $A$ and $B$ are digits), and reverse it to make the two digit number $BA$. Show that $AB+BA$ is a multiple of $11$.

A hint towards solving it: How do you express a two digit positive integer algebraically in terms of its two digits? What happens when you add that to the corresponding expression for the reverse?

Answer 2

Let $n=n_1n_2$ be a two digit integer. Then $n=n_1\cdot10+n_2$ . Let $k=n_2n_1$. Then $k=n_2\cdot10+n_1\cdot$. Then $$n+k=n_1\cdot10+n_2+n_2\cdot10+n_1\cdot=n_1(10+1)+n_2(10+1)=11n_1+11n_2=11(n_1+n_2).$$

Thus $n+k$ is a multiple of 11.