A grid with 10 rows and some number of columns is made up of unit squares, as shown. A domino $(\square$ ) can be placed horizontally or vertically to exactly cover two unit squares. There are 2004 positions in which the domino could be placed. The number of columns in the grid is
(A) 105
(B) 106
(C) 107
(D) 108
(E) 109

![](https://cdn.mathpix.com/cropped/2024_04_20_46ead6524a8d61e21c51g-082.jpg?height=384&width=444&top_left_y=1952&top_left_x=1255)
Start with a grid with two columns and ten rows. There are 10 ways to place the domino horizontally (one in each row) and 18 ways to place the column vertically (nine in each column), so 28 ways overall. How many more positions are added when a new column is added? When a new column is added, there are 9 new vertical positions (since the column has ten squares) and 10 new horizontal positions (one per row overlapping the new column and the previous column). So there are 19 new positions added.

How many times do we have to add 19 to 28 to get to 2004? In other words, how many times does 19 divide into $2004-28=1976$ ? Well, $1976 \div 19=104$, so we have to add 104 new columns to the original 2 columns, for 106 columns in total.

## Solution 2

Let the number of columns be $n$.

In each column, there are 9 positions for the domino (overlapping squares 1 and 2, 2 and 3, 3 and 4, and so on, down to 9 and 10 ).

In each row, there are $n-1$ positions for the domino (overlapping squares 1 and 2, 2 and 3, 3 and 4, and so on, along to $n-1$ and $n$ ).

Therefore, the total number of positions for the domino equals the number of rows times the number of positions per row plus the number of columns times the number of positions per column, or $n(9)+10(n-1)=19 n-10$. We want this to equal 2004 , so $19 n-10=2004$ or $19 n=2014$ or $n=106$. Thus, there are 106 columns.

Answer: (B)