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Need to use Fermat's Little Theorem (Let $p$ be a prime number and let $a$ be an integer. Then $a^p = a \mod p$. If $p$ does not divide $a$ then $a^p-1 \equiv 1 \mod p$.) $154$ is not prime, but $154 = 22\cdot 7$ and $23$ is prime, so $a^{22} \equiv 1 \mod 23$. $a^{154} \equiv a^{22} \cdot a^7 \equiv 1 \cdot 6 \equiv 6 \mod 23$.

Not sure how to proceed further. Would be grateful for any help.

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You solved it! apart from an arithmetic error.

$\displaystyle a^{mn} = (a^{m})^n$, not $\displaystyle a^{m}* a^{n}$ (which is actually $\displaystyle a^{m+n}$)

Also, not sure how you got $a^7 = 6 \mod 23$.

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    @Maths: Saying $x = y \mod 23$ is equivalent to saying that $x-y$ is divisible by $23$.2010-10-17
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HINT $\rm\ \ a^{22}\ \equiv 1\ \Rightarrow\ a^{22\: N}\ \equiv\ 1^N\ \equiv\ 1\:.\ $ Now put $\rm\ N = 7\:$.