Question

Find all $3-digit$ palindromes which are divisible by $11$

Answer 1

A $3-digit$ palindrome must be of the form $\overline{aba}$, where $a\ne 0$ and $b$ are digits.

This is divisible by $11$ if and only if $2a-b$ is divisible by $11$.

This is possible only if $2a-b=0$ or $2a-b=11$ or $2a-b=-11$.

Since $a\ge 1$ and $b\le 9$, we see that $2a-b\ge 2-9=-7>-11$.

Hence, $2a-b=-11$ is not possible.

Suppose, $2a-b=0$.

Then, $2a=b$

Thus $a=1,b=2;a=2,b=4;a=3,b=6$; and $a=4,b=8$ are possible.

We get the numbers $121,242,363,484$.

Similarly, $a=7,b=3,a=8,b=5$; and $a=9,b=7$ give the combinations for which $2a-b$ is divisible by $11$.

We get four more numbers: $616,737,858$, and $979$.

Thus, the required numbers are: $121,242,363,484,616,737,858,979$.