14. What is the smallest number of rectangles, each measuring $2 \mathrm{~cm}$ by $3 \mathrm{~cm}$, which are needed to fit together without overlap to form a rectangle whose sides are in the ratio $5: 4$ ?
A 10
B 15
C 20
D 30
E 60
Solution
D

A rectangle whose sides are in the ratio $5: 4$ has dimensions $5 k \mathrm{~cm} \times 4 k \mathrm{~cm}$, for some positive number $k$. Since we aim to cover this rectangle with $2 \mathrm{~cm} \times 3 \mathrm{~cm}$ rectangles without overlap, $k$ needs to be a positive integer.
The area of the $5 k \mathrm{~cm} \times 4 k \mathrm{~cm}$ rectangle is $(5 k \times 4 k) \mathrm{cm}^{2}=20 k^{2} \mathrm{~cm}^{2}$. The $2 \mathrm{~cm} \times 3 \mathrm{~cm}$ rectangles have area $6 \mathrm{~cm}^{2}$.
It follows that $20 k^{2}$ needs to be a multiple of 6 . The least positive integer $k$ for which this is the case is 3 . In this case the larger rectangle has dimensions $15 \mathrm{~cm} \times 12 \mathrm{~cm}$ and area $(15 \times 12) \mathrm{cm}^{2}=180 \mathrm{~cm}^{2}$. Therefore the smallest number of $2 \mathrm{~cm} \times 3 \mathrm{~cm}$ rectangles that are needed would be $\frac{180}{6}=30$, provided that it is possible to cover a $15 \mathrm{~cm} \times 12 \mathrm{~cm}$ with 30 $2 \mathrm{~cm} \times 3 \mathrm{~cm}$ rectangles.
To complete the question we need to show that this is possible.
One way in which this can be done is shown in the diagram on the right.
Therefore the smallest number of $2 \mathrm{~cm} \times 3 \mathrm{~cm}$ rectangles that are needed is 30 .