A checkerboard of 13 rows and 17 columns has a number written in each

Imagine that each block has x-y coordinates, where x is the value across and y is the value going down.

So when the the numbers increase by rows, the first block which is 1 has the coordinates of (1, 1), 2 has the coordinates of (2, 1), 18 (1, 2) etc. One notices that the columns increase by 17, while the rows increase by 1. Because of this one can find the value of each block by plugging in the x-y coordinates into \(17(y-1)+x\).

When the numbers increase by columns, the rows increase by 13 and the columns by 1. For this one can find the value of each block by plugging in the coordinates into \(13(x-1)+y\) [Eg. (2,1) = 14].

The blocks which do not change in value will obviously have the same coordinates in both equations, thus making the two equations equal to one another will give an equation to solve for these values.

\(17(y-1)+x = 13(x-1)+y\)

\(17y - 17 + x = 13x - 13 + y\)

\(16y = 12x + 4\)

\(y = \frac{3x+1}{4}\)

One could go about this by plugging in values for x which will yield an integer value for x, but because we know that the first block and last block will not change [(1,1) = 1 & (17,13) = 221], we can use this together with the fact that \(y = \frac{3x+1}{4}\), we know that \(3x+1\) must be divisible by 4. Therefore, \(x = 4*(n-1)+1\) where \(x≤17\) will yield all of the x-coordinates of the blocks that don't change in value.

1)\((4*0)+1 = 1\) y-coordinate: 1
2)\((4*1)+1 = 5\) y-coordinate: 4
3)\((4*2)+1 = 9\) y-coordinate: 7
4)\((4*3)+1 = 13\) y-coordinate: 10
5)\((4*4)+1 = 17\) y-coordinate: 13

Plugging in the coordinates into either \(17(y-1)+x\) or \(13(x-1)+y\) will give the values of each block:

(1,1) will have a value of 1
(5,4) will have a value of 56
(9,7) will have a value of 111
(13,10) will have a value of 166
(17, 13) will have the value of 221

\(1+56+111+166+221 = 555\)

Answer D