### JUNE'S PUZZLE: MEASURING TIME BY BURNING ROPES

You have two ropes. Each rope takes exactly 1 hour to burn from end to end. The rope has different material densities at different points, so there's no guarantee of consistency in the time it takes different sections within the rope to burn (e.g., the first half might burn in 59 minutes, the second half in 1 minute). You need to somehow measure exactly 45 minutes by using only these two ropes (and a match!). This isn't a trick question, so I'm not looking for an answer such as "You look at your watch for 45 minutes and enjoy the light of the burning ropes." The puzzle has a logical solution that relies only on the burning of the ropes.

### SOLUTION TO MAY'S PUZZLE

As a reminder, last month's puzzle asked you to imagine a 4 × 4 square grid drawn on a paper by using square gridlines and identify the number of different rectangles you can draw on the gridlines within the square (including the square itself). As you try to devise the solution to this puzzle, you might be tempted to simply manually count the different rectangles that you can draw within the square. However, this technique is confusing and usually leads to the wrong result. I enjoy this combinatorial puzzle because the solution lies in a methodical approach.

The different types of rectangles you can draw within a 4 × 4 square are different combinations of rectangles that have 1 to 4 rows and 1 to 4 columns. You can create the matrix that Table A shows to accommodate all possible rectangle row and column lengths. Populate each cell with the number of rectangles of a size (rows × columns) that you can draw within a 4 × 4 square. The number of different rectangles of a size *r *× *c* that you can draw within an *n *× *n* square can generally be expressed as (*n *− *r* + 1) × (*n *− *c* + 1). The logic behind this formula is that within *n* rows, a rectangle with *r *rows has *n *− *r* additional vertical locations to move to besides its current location. Similarly, it has *n *− *c* + 1 possible horizontal locations. So, in total, we're talking about (*n *− *r* + 1) × (*n *− *c* + 1) different rectangles of *r *× *c* size. When you populate all cells in the matrix according to this formula, you get the matrix that Table B shows.

Next, simplify the expression that just adds all values, and you get the solution for this puzzle:

4 × 4 + 4 × 3 + 4 × 2 + 4 × 1 +

3 × 4 + 3 × 3 + 3 × 2 + 3 × 1 +

2 × 4 + 2 × 3 + 2 × 2 + 2 × 1 +

1 × 4 + 1 × 3 + 1 × 2 + 1 × 1 =

4 × (4 + 3 + 2 + 1) +

3 × (4 + 3 + 2 + 1) +

2 × (4 + 3 + 2 + 1) +

1 × (4 + 3 + 2 + 1) =

(4 + 3 + 2 + 1) × (4 + 3 + 2 + 1) =

(1 + 2 + 3 + 4)^{2} = 100

The solution for a more generic case with an *n *× *n* square is (1 + 2 + 3 +...+ *n*)^{2 }.