【Question 5】
As shown in the figure, a rectangular table has 8 columns, and numbers $1, 2, \cdots$ are filled into the table in a certain order (filled from left to right, and when a row is full, move to the next row, still filling from left to right). A student first colors the cell containing the number 1, then skips 1 cell, and colors the cell containing the number 3; then skips 2 cells, and colors the cell containing the number 6; then skips 3 cells, and colors the cell containing the number 10. This continues until every column contains at least one black cell (no more coloring after that). Therefore, the number in the last cell he colored is $\qquad$
【Analysis and Solution】
(Method One)
The 1st blackened cell is 1;
The 2nd blackened cell is $3=1+2$;
The 3rd blackened cell is $6=1+2+3$;
The 4th blackened cell is $10=1+2+3+4$;
$\qquad$
By analogy, the $n$-th blackened cell is $1+2+\cdots+n$. Thus, we can see that the blackened cells are $1 \equiv 1(\bmod 8), 1+2=3 \equiv 3(\bmod 8)$,
$$
\begin{array}{l}
1+2+3=6 \equiv 6(\bmod 8), 1+2+3+4=10 \equiv 2(\bmod 8) \\
1+2+\cdots+5=15 \equiv 7(\bmod 8), \quad 1+2+\cdots+6=21 \equiv 5(\bmod 8) \\
1+2+\cdots+7=28 \equiv 4(\bmod 8), \quad 1+2+\cdots+8=36 \equiv 4(\bmod 8) \\
1+2+\cdots+9=45 \equiv 5(\bmod 8), \quad 1+2+\cdots+10=55 \equiv 7(\bmod 8) \\
1+2+\cdots+11=66 \equiv 2(\bmod 8), \quad 1+2+\cdots+12=78 \equiv 6(\bmod 8) \\
1+2+\cdots+13=91 \equiv 3(\bmod 8), \quad 1+2+\cdots+14=105 \equiv 1(\bmod 8) \\
1+2+\cdots+15=120 \equiv 0(\bmod 8)
\end{array}
$$

Up to this point, the numbers that are $\bmod 8$ with remainders $0 \sim 7$ have all appeared;
That is, every column contains at least one black cell;
The number in the last blackened cell is 120.
(Method Two)
The 1st blackened cell is 1;
The 2nd blackened cell is $3=1+2$;
The 3rd blackened cell is $6=1+2+3$;
The 4th blackened cell is $10=1+2+3+4$;
$\qquad$
By analogy, the $n$-th blackened cell is $1+2+\cdots+n$.
The remainders of the blackened cells $\bmod 8$ are $1,3,6,2, 7, 5, 4, \ldots$
Up to this point, except for $\bmod 8$ with remainder 0, all other $\bmod 8$ with remainders $1 \sim 7$ have already appeared;
Therefore, we only need to consider $1+2+\cdots+n \equiv 0(\bmod 8)$;
$8 \left\lvert\, 1+2+\cdots+n=\frac{1}{2} n(n+1)\right.$
$16 \mid n(n+1)$
Since $n$ and $n+1$ are coprime;
Thus, $16 \mid n$ or $16 \mid n+1$;
We hope that $n$ is as small as possible;
Then take $n+1=16$;
$$
n=15 \text {; }
$$

Thus, the number in the last blackened cell is $1+2+\cdots+15=(1+15) \times 15 \div 2=120$.