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I'm supposed to use Fermat's Little Theorem $($ If $p$ is a prime and $p$ doesn't divide $a$, then $a^{p-1} \equiv1 \mod p)$ and to find the least residue of the following:

$24^{(54n + 1)} +11 \mod 19$, where $n \in \mathbb{Z}^+.$

The 11 is trivial, so I'm just concerned with reducing the term with $24^{(54n + 1)}$.

(24, 19) = 1, so the Theorem applies. $24\equiv 5\mod19$, so

$24^{(54n + 1)} \equiv 24^{54n}(5)\mod19$

At this point, I'm not really sure what to do next because I can't really see how to bring the theorem into play!

mrtaurho
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3 Answers3

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$24^{54n}=((24)^{54})^n$, since $54=3*18$, $24^{54}=1$ mod $19$ since $24^{18}=24^{19-1}$, so the answer is $16$ mod $19$

Tsemo Aristide
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Since $54n+1\equiv 1\pmod{18}$ you have \begin{align} 24^{54n + 1} +11 &\equiv 24+11\\ &\equiv 35\\ &\equiv 16\pmod{19} \end{align}

Fabio Lucchini
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We have $24 = 5\mod 19$. Hence, $$ 24^{54 n + 1} + 11 = 5^{54 n + 1} + 11\mod 19. $$ As you noted, the difficult part is in the first one of course. You need to apply Fermat's theorem which states that $$ a^p = a^1\mod p$$ if the gcd of $a$ and $p$ is $1$. In our case, you are going to apply it where $a = 5$ and $p = 19$. The key part is that you note that Fermat's theorem implies that $$ a^n = a^{n\mod (p-1)} \mod p$$ holds if $a$ and $p$ are relatively prime. Once you see this, it is easy to find the actual result since $54 n + 1 = 1\mod (19-1)$ and you are left with $$ 5^{54 n + 1} + 11 = 5^{54 n + 1\mod 18} + 11 = 5^1 + 11 = 16\mod 19. $$

Stan Tendijck
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