Solve the cryptarithm: ` A B \ X\ A C =\ A C B `

Explanation

Given: `ABxxAB=ACB`
`AB` represent a `2` digit number with `A` at `10s` place and B at units place. We are multiplying `AB` with `AB`.
`A` and `B` can take any values from `0` to `9`.

But, If we multiply `B` by `B` to get `B` itself at the units place of the product,
`B` can be `0`,`1`,`5`,`6`.
Also `A` should be at the `10s` place , therefore middle term should not have any carry over.
Middle term is `AB+AB=C`
Therefore `B` cannot be `5` or `6`.
Also `A` cannot be `5` or `6`.
Therefore `A` and `B` can be `0` or `1`.

If we assume `B` and `C` to be different numbers, `B` cannot be `0`, as this makes `C` also `0`.
Therefore, `B=1`.
Then `A` also `1`.
`C=AB+BA=2`

Therefore `AB=11`, `ABtimesAB=ACB` and `ACB=121`.