closetheloopkid wrote:
Understood that the first four columns get canceled out. Why do we sum horizontally as opposed to vertically?
You can also sum it vertically as the second last column = \(3(1+2+3+4+5+6)\)
Similarly the last column = \(4(1+2+3+4+5+6)\)
Combining those two, we get \(3(1+2+3+4+5+6) + 4(1+2+3+4+5+6) = (3+4)(1+2+3+4+5+6) = 7(1+2+3+4+5+6)\)
Now it becomes easy, you just need to calculate the sum of first 6 integers.
the answer becomes \(7 \times 21 = 147\)