When the positive integers are arranged in order, filling in the successive diagonals of an infinite grid from top to bottom, as shown, the integer 41 is in the (5,5) spot. What integer would we see in the (10,10) spot if the rest of the grid were visible?

ant101
Sep 2, 2018

#1**+1 **

Wow, here's is my try!

In the left column, it's a steadily increasing sum, +2, +3, +4, so 10 down will be 55, we can also use the arithmetic series formula, to get 55!

Now, we just add 10, then 11, and finally end up with \(\boxed{181}\) .

tertre
Sep 3, 2018

#2**+1 **

**When the positive integers are arranged in order, filling in the successive diagonals of an infinite grid from top to bottom, as shown, the integer 41 is in the (5,5) spot. What integer would we see in the (10,10) spot if the rest of the grid were visible?**

\(\text{The red entries in the main diagonal (n,n) form} \\ \text{an arithmetic series of the second order: {$1,5,13,25,41,\ldots$}}\)

\(\begin{array}{|r|r|r|r|} \hline n & (n,n ) & & \text{First difference} & \text{Second difference} \\ \hline 1 & (1,1) & 1 & \\ & & & 4 \\ 2 & (2,2) & 5 & & 4 \\ & & & 8 \\ 3 & (3,3) & 13 & & 4 \\ & & & 12 \\ 4 & (4,4) & 25 & & 4 \\ & & & 16 \\ 5 & (5,5) & 41 & & 4 \\ & & & 20 \\ 6 & (6,6) & 61 & & 4 \\ & & & 24 \\ 7 & (7,7) & 85 & & 4 \\ & & & 28 \\ 8 & (8,8) & 113 & & 4 \\ & & & 32 \\ 9 & (8,9) & 145 & & 4 \\ & & & 36 \\ 10 & (10,10) & \mathbf{181} & & 4 \\ \hline \end{array}\)

heureka
Sep 3, 2018