Find last two digits of $302^{46}$

$$302^{46} = 2^{46} = (2^{12})^3 \times 2^{10} = (-4)^3 \times 24 = -64 \times 24 = 64 \quad [100]$$


So you wanna calculate $2^{46}$ modulo $100$. For that note that $$2^{46}=(2^{20}\times 2^{3})^2=((2^{10})^2\times 8)^2=(24^2\times 8)^2=(76\times 8)^2=(8)^2=64$$in $\mathbb Z/100\mathbb Z$. Thus, $2^{46}\equiv 64\pmod{100}$.


$2^{10} = 1024$, so $2^{40} = (2^{10})^4 = 24^4$(mod 100).

Hence, $2^{46} = 24^4 \times 2^6 $ = $21233664$ (mod $100$) = $64$

Tags:

Abstract Algebra

Elementary Number Theory

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