A rectangular board of 8 columns has squares numbered beginning in the upper left corner and moving left to right so row one is numbered 1 through 8, row two is 9 through 16, and so on. A student shades square 1, then skips one square and shades square 3, skips two squares and shades square 6, skips 3 squares and shades square 10, and continues in this way until there is at least one shaded square in each column. What is the number of the shaded square that first achieves this result?

$\text{(A)}\ 36\qquad\text{(B)}\ 64\qquad\text{(C)}\ 78\qquad\text{(D)}\ 91\qquad\text{(E)}\ 120$
The numbers that are shaded are the triangular numbers, which are numbers in the form $\frac{(n)(n+1)}{2}$ for positive integers.  They can also be generated by starting with $1$, and adding $1, 2, 3, 4...$ as in the description of the problem.
Squares that have the same remainder after being divided by $8$ will be in the same column.  Thus, we want to find when the last remainder, from $0$ to $7$, is found.
So, instead of adding $1, 2, 3, 4, 5, 6, 7, 8, 9, 10...$, we can effectively either add $1, 2, 3, 4, 5, 6, 7, 0, 1, 2, 3...$ or subtract $7, 6, 5, 4, 3, 2, 1, 0, 7, 6, 5...$ if we are only concerned about remainders when divided by $8$.  We will pick the number that keeps the terms on the list between $1$ and $8$.  We get:
$1$
$1 + 2 = 3$
$3 + 3 = 6$
$6 - 4 = 2$
$2 + 5 = 7$
$7 - 2 = 5$
$5 - 1 = 4$
$4 + 0 = 4$
$4 + 1 = 5$
$5 + 2 = 7$
$7 - 5 = 2$
$2 + 4 = 6$
$6 - 3 = 3$
$3 - 2 = 1$
$1 - 1 = 0$
Finally, a term with $0$ is found, and checking, all numbers $1$ through $7$ are also on the right side of the list.  This means the last term in our sequence is the first time that column $8$ is shaded.  There are $15$ terms in the sequence, leading to an answer of $\frac{15\cdot 16}{2} = 120$, which is choice $\boxed{E}$.