Question

A rectangular piece of paper measures $4$ units by $5$ units. Several lines are drawn parallel to the edges of the paper. A rectangle determined by the intersections of some of these lines is called basic if (i) all four sides of the rectangle are segments of drawn line segments, and (ii) no segments of drawn lines lie inside the rectangle.

Given that the total length of all lines drawn is exactly $2007$ units, let $N$ be the maximum possible number of basic rectangles determined. Find the remainder when $N$ is divided by $1000.$

Answer

Answer: 896

Concepts:optimization quadratic modular arithmetic

Difficulty rating: 2840

Solution:

Suppose $h$ of the drawn lines have length $4$ and $v$ have length $5,$ so $4h + 5v = 2007.$ A basic rectangle is bounded by two adjacent lines in each direction, so the lines determine $(h - 1)(v - 1)$ basic rectangles. Setting $x = h - 1$ and $y = v - 1,$ we must maximize $xy$ subject to $4x + 5y = 1998.$

As a function of $x,$ the product $xy = x \cdot \frac{1998 - 4x}{5}$ is a downward parabola with vertex at $x = \frac{999}{4} = 249.75.$ For $y$ to be an integer we need $4x \equiv 1998 \pmod 5,$ i.e. $x \equiv 2 \pmod 5.$ The nearest candidates are $x = 247$ (giving $y = 202$ and product $49894$ ) and $x = 252$ (giving $y = 198$ and product $49896$ ).

So $N = 49896,$ and the remainder upon division by $1000$ is $896.$

2007 AIME II Problem 8

Answer: 896

Difficulty rating: 2840

Solution:

Problem 8 in Other Years