Question

I am a number less than 100. When you divide me by 2, my remainder is 1. When you divide me by 25, my remainder is 2. What numbers could I be?

Answer 1

We want to find
`x` such that

`\begin{cases}x\equiv1\pmod2\\x\equiv2\pmod{25}\end{cases}`

Note that both 2 and 25 are coprime, so we can apply the Chinese remainder theorem right away.

Let's start with
`x=1+2`. Taken mod 2, the 2 vanishes and we're left with 1, as desired. But taken mod 25, we're left with 3. If we multiply 1 by 25, so that
`x=25+2`, then taken mod 25 the 25 will vanish and we're left with 2, as desired. So
`x=27` is the least positive solution.

The whole range of solutions is given by the CRT to be
`x=27+(2\cdot25)n=27+50n` for all integers
`n`. There are only 2 solutions that fall below 100, for
`n=0` and
`n=1`, which are 27 and 77.