= Solution
The integer $137$ is prime: none of the primes $2,3,5,7,11$ below $\sqrt{137}$ divides it. Since $43\not\equiv0\pmod{137}$, <Fermat's little theorem> gives $43^{136}\equiv1$. Therefore $43^{135}$ is the <modular inverse> of $43$. The <extended Euclidean algorithm> yields
$$
137=3(43)+8,\quad43=5(8)+3,\quad8=2(3)+2,\quad3=2+1,
$$
and back-substitution gives $1=51(43)-16(137)$. Consequently
$$
\boxed{43^{135}\equiv51\pmod{137}.}
$$
Indeed, $43\cdot51=2193=16\cdot137+1$, directly checking the required <modular inverse>.
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