Find the least residue of $7^5$ modulo $50$ without using a calculator, and using an efficient method. So far I have that $7=1\pmod{50}$, $7^5=1^5 \pmod{50}$ Really stuck what to do from here :(
2026-03-27 17:57:30.1774634250
Least residue of $7^5$ modulo $50$
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$$7^2\equiv-1\pmod{50}$$
$$7^{2n+1}=7(7^2)^n\equiv7(-1)^n\pmod{50}$$