1. Rowan is given a square grid whose side length is an odd integer. The square is divided into 1 by 1 squares,. Rowan shades each 1 by 1 square along the two diagonals and does not shade any of the remaining squares. For example, a 5 by 5 grid is shown in which Rowan has shaded the squares along the diagonals.
(a) Rowan is given a square grid that is 7 by 7 . How many 1 by 1 squares does he shade?
(b) Rowan is given a square grid that is 101 by 101. Explain why the number of 1 by 1 squares that he shades is 201 .
(c) Rowan is given another square grid with odd side length, and the number of 1 by 1 squares that he shades is 41 . How many unshaded 1 by 1 squares are there in the grid?
(d) Rowan is given a fourth square grid with odd side length. After shading the squares on the diagonals, there are 196 unshaded 1 by 1 squares. How many 1 by 1 squares does this grid contain in total?
1. (a) A 7 by 7 square has seven columns, and the fourth column is the middle column since it has three columns to the left and three columns to the right.
Rowan shades 2 squares in each of the first, second and third columns, he shades 1 square in the fourth (middle) column, and he shades 2 squares in each of the the fifth, sixth and seventh columns.
In total, he shades $2+2+2+1+2+2+2=13$ squares.
(b) In a 101 by 101 square, each diagonal contains 101 squares. The two diagonals overlap in one square (the middle square), which Rowan does not shade twice.
When Rowan shades the first diagonal, he shades 101 squares.
Therefore, when Rowan shades the second diagonal, he shades $101-1=100$ squares.
In total, he shades $101+100=201$ squares.
(c) Suppose that the grid that Rowan is given in this part is $n$ by $n$, with $n$ odd.
We can use the analysis from (b) to conclude that Rowan shades $n+(n-1)=2 n-1$ squares.
We are told that Rowan shades 41 squares, so $2 n-1=41$ or $2 n=42$, which gives $n=21$. Now the square contains $n^{2}=21^{2}=441$ squares in total, and Rowan shades 41 of them, so there are $441-41=400$ unshaded squares.
(d) Suppose that Rowan is given an $m$ by $m$ square with $m$ odd.
The grid contains $m^{2}$ squares in total, of which he shades $2 m-1$.
This means that there are $m^{2}-(2 m-1)=m^{2}-2 m+1$ unshaded squares in total.
Since we are told that there are 196 unshaded squares in total, then $m^{2}-2 m+1=196$.
Now $(m-1)^{2}=m^{2}-2 m+1$, so $(m-1)^{2}=196$.
Since $m$ is positive, then $m-1=\sqrt{196}=14$ and so $m=15$.
Therefore, the grid that Rowan is given is 15 by 15 , and so contains $15^{2}=225$ squares in total.