# A simple problem on congruences.

This is a problem on congruences given in the book of Andy Liu, titled "Arithmetical Wonderland".

The teacher asked Ace, Beatrice and Cecil to write down some $4$-digit number, write down a second $4$-digit number obtained from the first by reversing the order of the digits, and then add the two numbers. Ace’s answer was $5985$, Beatrice’s $2212$ and Cecil’s $4983$.
(a) Without looking at their work, the teacher said that both Ace and Beatrice had made mistakes. How could she tell?
(b) If Cecil had not made mistakes, what was his initial number, given that it was not divisible by $10$ and was greater than the number obtained by digit reversal?

My attempt:
If in base $10$ arithmetic, there is a $4$ digit number, let : 'abcd', then its reverse would be 'dcba'. Adding the two leads to (a+d)(b+c)(b+c)(a+d). This means that first and last digits are the same, and similarly for the middle digits.