Find how many solutions the congruence $x^2 \equiv 121 \mod 1800$ has
As $(121,1800)=1,$ $$x^2\equiv121\pmod{1800}\iff(x\cdot 11^{-1})^2\equiv1$$
Now as $1800=3^2\cdot2^3\cdot5^2$
as $y^2\equiv1\pmod{p^n}$ has exactly two in-congruent solutions for odd prime $p$
and $y^2\equiv1\pmod8$ has four
the number of solutions $$(x\cdot 11^{-1})^2\equiv1\pmod{1800}$$ will be $2\cdot4\cdot2$