Problem (Number Theory):
Let $T=49$. Compute the last digit, in base 10, of the sum

$$
T^{2}+(2 T)^{2}+(3 T)^{2}+\ldots+\left(T^{2}\right)^{2}
$$

Solution:
Let $S$ be the required sum. Factoring $T^{2}$ from the sum yields

$$
\begin{aligned}
S & =T^{2}\left(1+4+9+\ldots+T^{2}\right) \\
& =T^{2}\left(\frac{T(T+1)(2 T+1)}{6}\right) \\
& =\frac{T^{3}(T+1)(2 T+1)}{6} .
\end{aligned}
$$

Further analysis makes the final computation simpler. If $T \equiv 0,2$, or $3 \bmod 4$, then $S$ is even. Otherwise, $S$ is odd. And if $T \equiv 0,2$, or $4 \bmod 5$, then $S \equiv 0 \bmod 5$; otherwise, $S \equiv 1 \bmod 5$. These observations yield the following table:

| $T \bmod 4$ | $T \bmod 5$ | $S \bmod 10$ |
| :---: | :---: | :---: |
| $0,2,3$ | $0,2,4$ | 0 |
| $0,2,3$ | 1,3 | 6 |
| 1 | $0,2,4$ | 5 |
| 1 | 1,3 | 1 |

Because $T=49$, the value corresponds to the third case above; the last digit is $\mathbf{5}$.

Final answer: 5