EndingDigits Basic

Problem - 4159
Determine the last two digits of $312^{123}$.

First, let's simplify the expression as follows: $$312^{123}\equiv 12^{123} = (2^2\times 3)^{123}=2^{246}\times 3^{123}$$

Then, we simplify the two terms separately:

$$\begin{align*} 2^{246}&= (2^{10})^{24}\times 2^6 = (1024)^{24}\times 64\equiv 24^{24}\times 64\equiv (24^2)^{12}\times 64\equiv 76\times 64\pmod{100} \\ 3^{123}&=(3^4)^{30}\times 3^3 = 81^{30} \times 27 \equiv 01 \times 27 = 27\pmod{100} \end{align*}$$

Therefore, the answer we are looking for equals

$$64\times 76\times 27 \equiv \boxed{28} \pmod{100}$$

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