As shown in the figure, a rectangular table has 8 columns. 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 with the number 1 black, then skips 1 cell, and colors the cell with the number 3 black; then skips 2 cells, and colors the cell with the number 6 black; then skips 3 cells, and colors the cell with the number 10 black. This continues until every column contains at least one black cell (no more coloring after that). Therefore, the number in the last black cell he colored is $\qquad$
(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 $10=1+2+3+4$;
$\qquad$
And so on, the $n$-th blackened cell is $1+2+\cdots+n$.
Thus, we can get the blackened cells in sequence as $1 \equiv 1(\bmod 8), 1+2=3 \equiv 3(\bmod 8)$,
$$
\begin{array}{l}
1+2+3=6 \equiv 6(\bmod 8), 1+2+3+4=10 \equiv 2(\bmod 8) \\
1+2+\cdots+5=15 \equiv 7(\bmod 8), \quad 1+2+\cdots+6=21 \equiv 5(\bmod 8) \\
1+2+\cdots+7=28 \equiv 4(\bmod 8), \quad 1+2+\cdots+8=36 \equiv 4(\bmod 8) \\
1+2+\cdots+9=45 \equiv 5(\bmod 8), \quad 1+2+\cdots+10=55 \equiv 7(\bmod 8) \\
1+2+\cdots+11=66 \equiv 2(\bmod 8), \quad 1+2+\cdots+12=78 \equiv 6(\bmod 8) \\
1+2+\cdots+13=91 \equiv 3(\bmod 8), \quad 1+2+\cdots+14=105 \equiv 1(\bmod 8) \\
1+2+\cdots+15=120 \equiv 0(\bmod 8)
\end{array}
$$

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 $10=1+2+3+4$;
$\qquad$
And so on, the $n$-th blackened cell is $1+2+\cdots+n$.
The blackened cells $\bmod 8$ have remainders in sequence as $1,3,6,2, 7, 5, 4, \ldots$
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 $1+2+\cdots+n \equiv 0(\bmod 8)$;
$$
\begin{array}{l}
8 \left\lvert\, 1+2+\cdots+n=\frac{1}{2} n(n+1)\right. \\
16 \mid n(n+1)
\end{array}
$$

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 $n+1=16$;
$$
n=15
$$

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