A. Find the modular reciprocal of 1000 in ℤ 1253.
B. Find an integer x ∈ ℤ 1253 that solves 1000 ⊗ x = 13 in ℤ 1253.
I solved the first part by using the extended euclidean algorithm, and I obtained that the multiplicative inverse of 1000 in ℤ 1253 is 104.
how do I solve the second question?
Well, if $1000^{-1}=104$ in ${\Bbb Z}_{1253}$, then multiply $1000\cdot x =13$ with $104$ giving $x = 104\cdot 13$ in ${\Bbb Z}_{1253}$.[\
In general, if you want to solve $ax=b$ in ${\Bbb Z}_n$ and $a$ is a unit, then $x=a^{-1}b$ in ${\Bbb Z}_n$. If $a$ is not a unit, there might be no solution as in $2x=3$ in ${\Bbb Z}_4$.