We want to find [tex]x[/tex] such that
[tex]\begin{cases}x\equiv1\pmod2\\x\equiv2\pmod{25}\end{cases}[/tex]
Note that both 2 and 25 are coprime, so we can apply the Chinese remainder theorem right away.
Let's start with [tex]x=1+2[/tex]. 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 [tex]x=25+2[/tex], then taken mod 25 the 25 will vanish and we're left with 2, as desired. So [tex]x=27[/tex] is the least positive solution.
The whole range of solutions is given by the CRT to be [tex]x=27+(2\cdot25)n=27+50n[/tex] for all integers [tex]n[/tex]. There are only 2 solutions that fall below 100, for [tex]n=0[/tex] and [tex]n=1[/tex], which are 27 and 77.
