In this sequence:
1 3 6 10 15 . . .
2 5 9 14 20 . . .
4 8 13 19 26 . . .
7 12 18 25 33 . . .
11 17 24 32 41 . . .
16 23 31 40 50 . . .
. . . . .
. . . . .
. . . . .
In what row and column is the number 2012?

Question

In this sequence:
1     3     6    10   15   . . . 
2     5     9    14   20   . . . 
4     8    13   19   26   . . .
7    12   18   25   33   . . . 
11  17   24   32   41   . . . 
16  23   31   40   50   . . .
 .     .      .      .       .
 .     .      .      .       .
 .     .      .      .       .
In what row and column is the number 2012?

btaylor · Answer

@Calcmathlete @Mertsj help please!

mertsj · Answer

Well, I can grind it out, but there is probably some elegant solution I am not seeing.

btaylor · Answer

Does it have to do with triangular numbers? because it is basically the triangular numbers rotated...

mertsj · Answer

I don't know about triangular numbers.

btaylor · Answer

I know that the first row (1,3,6,10,15, ...) are the sums of the first n numbers. Could you just find the item in that row that is greater than 2012, then count backwards?

mertsj · Answer

I would say so

anonymous · Answer

row 5
column 59

anonymous · Answer

The first number of the 63rd row is 1954
The first number of the 64rd row is 2017

anonymous · Answer

notice that the numbers are read|dw:1339365099192:dw|