Solution 5c28c08a-9fad-4e1e-adb1-6421a9e520ff

Points: 

By modular arithmetic, we know
\[
\begin{aligned}
272760 &\equiv 15 \pmod{N},\\[2mm]
273436 &\equiv 16 \pmod{N}.
\end{aligned}
\]
In other words,
\[
\begin{aligned}
272745 &\equiv 0 \pmod{N},\\[2mm]
273420 &\equiv 0 \pmod{N}.
\end{aligned}
\]
Thus, $N$ divides both numbers, so the greatest possible value of $N$ is
\[
N_{\max}=\gcd(272745,273420)=\gcd(273420,675).
\]
We then see that
\[
675=27\cdot25.
\]
Applying divisibility rules, we see that the largest powers of $5$ and $3$
dividing $273420$ are $5$ and $9$, respectively:
\[
\begin{aligned}
5 &\mid 273420, & 25 &\nmid 273420,\\[2mm]
9 &\mid 273420, & 27 &\nmid 273420.
\end{aligned}
\]
This means
\[
\gcd(273420,675)=5\cdot9=45.
\]
But we also know $N>16$. Since every proper positive divisor of $45$ is
at most $15$, the maximum is the only possible value. Hence $N=45$,
whose tens digit is $4$.

Final answer:
4\quad\text{(E)}