This is what I got so far but for some reason I feel like I am wrong. 49 is congruent to 26 mod(23). Therefore we have 49^4 is congruent to 26^4 = 456976 which is congruent to 456923 mod(23)but this doesnt seem like the least residue due to the issue of the theorem stating that 0 < r < m but here r is bigger than m
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1To get started, it's good to note that $49\equiv 3\pmod {23}$. – lulu Oct 11 '18 at 21:26
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1And $26$ is congruent to $3$ modulo $23$. – Angina Seng Oct 11 '18 at 21:26
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It will take a long time to get down from $456976$ by subtracting $23$ at a time. Instead, divide with remainder. You can find that $456976=23\cdot 19868+12$, so your least residue is $12$.
As lulu says, you could also start from the fact that $49 \equiv 3 \pmod {23}$, so $49^4 \equiv 3^4=81 \pmod {23}$. Now it only takes three subtractions of $23$ or you can say $81=23 \cdot 3+12$ and you are there.
Ross Millikan
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To save a bit more work, use $3^3 = 27 \equiv 4 \pmod{23}$. Therefore $$49^4 \equiv 3^4 = 3^3\cdot3 \equiv 4\cdot3 =12 \pmod{23}.$$ – Théophile Oct 11 '18 at 21:35