When 2^{301} is divided by 5,the least positi | vedclass

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(C) We know that $2^2 = 4 \equiv -1 \pmod{5}$.Squaring both sides,we get $(2^2)^2 \equiv (-1)^2 \pmod{5}$,which implies $2^4 \equiv 1 \pmod{5}$.We can

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$2$

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(C) We know that $2^2 = 4 \equiv -1 \pmod{5}$.Squaring both sides,we get $(2^2)^2 \equiv (-1)^2 \pmod{5}$,which implies $2^4 \equiv 1 \pmod{5}$.We can write $2^{301}$ as $2^{300} 	imes 2 = (2^4)^{75} 	imes 2$.Since $2^4 \equiv 1 \pmod{5}$,then $(2^4)^{75} \equiv 1^{75} \equiv 1 \pmod{5}$.Therefore,$2^{301} \equiv 1 	imes 2 \equiv 2 \pmod{5}$.The least positive remainder is $2$.

When $2^{301}$ is divided by $5$,the least positive remainder is

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