# What is the greatest number of rectangles with integer side lengths and perimeter 10 that can be cut from a piece of paper with width 24 and length 60?

Aug 22, 2016

$360$

#### Explanation:

If a rectangle has perimeter $10$ then the sum of its length and width is $5$, giving two choices with integer sides:

• $2 \times 3$ rectangle of area $6$
• $1 \times 4$ rectangle of area $4$

The piece of paper has area $24 \times 60 = 1440$

This can be divided into $12 \times 20 = 240$ rectangles with sides $2 \times 3$.

It can be divided into $24 \times 15 = 360$ rectangles with sides $1 \times 4$

So the greatest number of rectangles is $360$.

Aug 23, 2016

$360$

#### Explanation:

Calling $S = 60 \times 24 = {2}^{5} \times {3}^{2} \times 5 \times 1$ the problem can be stated as

Determine

$\max n \in {\mathbb{N}}^{+}$

such that

$n \le \frac{S}{a \cdot b}$
$a + b = 5$
$\left\{a , b\right\} \in \left\{1 , 2 , 3 , 4\right\}$

giving the feasible pairs

$\left\{1 , 4\right\} , \left\{2 , 3\right\}$ and the desired result is

$n = \frac{1440}{4} = 360$