7.5 Applications: Divisibility and Approximations

We need $11^{12} \pmod{100}$. $11^{12} = (1 + 10)^{12} = 1 + ^{12}C_1(10) + ^{12}C_2(10^2) + ...$ Terms with $10^3$ and higher will end in 000, so they don't affect last two digits. We consider: $1 + 12(10) + \frac{12 \cdot 11}{2}(100)$ $= 1 + 120 + 6600$ $= 6721$. Last two digits are 21.