by showmyiq » Thu Jun 20, 2013 5:57 am
Correct answer!
You are the one who solved the puzzle first!
I will add some further explanation for the other users, who are interested in the approach of solving such problems in future.
\begin{equation}
(207^{19}-41)^{10}\equiv r \pmod {100} \\
\text{We have that: } 207\equiv 7 \pmod {100} \\
\text{Therefore: } 207^3\equiv 7^3 = 343 \equiv 43 \pmod {100} \\
207^4 = 207^3*207 \equiv 43*7 = 301 \equiv 1 \pmod {100} \\
207^{16} \equiv 1 \pmod {100} \\
207^{19} \equiv 43 \pmod {100} \\
207^{19} - 41 \equiv 2 \pmod {100} \\
(207^{19} - 41)^{10} \equiv 2^{10} = 1024 \equiv 24 \pmod {100} \\
\end{equation}