The figures from $1$ to $120$ were made in $15$ rows. Which column has the largest sum of the numbers? 

Question: The figures from $1$ to $120$ were made in $15$ rows. Which column has the largest sum of the numbers? (Starting from the left)

I tried only with calculating. I could not find an elegant way. I'm looking for a simple solution that does not require calculation. Because this is the exam question. We have up to maximum 2-3 minutes.
 A: Hint. There are $15$ columns and the $n$-column start with $n(n+1)/2$ and has $16-n$ numbers. 
Note that the difference between the sum of the numbers in the $n$-th column and the sum of the numbers in the $(n+1)$-th column is
$$\frac{n(n+1)}{2}\underbrace{-1-1\dots-1}_{16-(n+1)}=\frac{(n+1)(n+2)}{2}-16.$$
The change of sign of the above difference will tell you where the maximum is. 
P.S. Note that the number $\frac{(n+1)(n+2)}{2}$ appears at the top of the $(n+1)$-th column, so you don't have to calculate it...
A: Say the sum of the 15 numbers in the first column is $S_1$. Then notice that in the second column, you have 14 numbers so you remove $1$ from the sum and you add $14$ to the sum (since these numbers are $1$ more than the numbers in the same row). So you have $S_2 = S_1-1+14 = S_1+13$. Then in the third column, you remove $3$ from the sum $S_2$ and add $13$ to the sum, which is $S_3 = S_2-3+13 = S_2+10$. In the fourth column, you remove $6$ from the sum $S_3$ and you add $12$ to the sum so $S_4 = S_3-6+12 = S_3+6$. Finally, in the fifth column, you remove $10$ from the sum $S_4$ and you add $11$ to the sum so $S_5 = S_4-10+11 = S_4+1$. Notice that after this step, your sum will be less than the sum in the previous column so the answer must be the fifth column.
A: Let $S_n$ denote the sum of the $n^{th}$ column.
Then the sum of the immediate left column is $S_{n-1}= S_n + (15 - n) + {n(n+1) \over 2}$
The maximum will occur the first time when ${n(n+1) \over 2}$ exceeds $(15-n)$ and this happens on the $5^{th}$ column.
