Last digits of a huge power of eleven mod 50

What remainder does $11^{2360}$ leave on division by $50$?

$1$. By the binomial theorem $11^5=(10+1)^5=1+50+(\text{multiples of }100)$, so $11^5\equiv1\pmod{50}$. Since $2360=5\cdot472$, $11^{2360}=(11^5)^{472}\equiv1$.