An artist wants to completely cover a rectangle with identically sized squares which do not overlap and do not extend beyond the edges of the rectangle. If the rectangle is $60 \frac{1}{2} \mathrm{~cm}$ long and $47 \frac{2}{3} \mathrm{~cm}$ wide, what is the minimum number of squares required?
(A) 429
(B) 858
(C) 1573
(D) 1716
(E) 5148
Solution 1

Suppose that the artist uses squares of side length $s$, and that he uses $M$ of the squares along the length, and $N$ along the width. (Because of the constraints of the problem, $M$ and $N$ must both be integers.)

Then we must have $M s=60 \frac{1}{2} \mathrm{~cm}$ and $N s=47 \frac{2}{3} \mathrm{~cm}$.

The total number of squares covering the rectangle is $M N$, and so we don't need to know $s$. Dividing the two equations will cancel $s$ to obtain

$$
\frac{M s}{N s}=\frac{60 \frac{1}{2}}{47 \frac{2}{3}}
$$

or

$$
\frac{M}{N}=\frac{\left[\frac{121}{2}\right]}{\left[\frac{143}{3}\right]}=\frac{363}{286}=\frac{11(33)}{11(26)}=\frac{33}{26}
$$

So we want to determine positive integers $M$ and $N$ as small as possible so that $\frac{M}{N}=\frac{33}{26}$.

Since the fraction $\frac{33}{26}$ is in lowest terms, then the smallest integers $M$ and $N$ that will work are $M=33$ and $N=26$, which gives a total of $M N=(33)(26)=858$ squares.

## Solution 2

The area of the rectangle is $\frac{121}{2} \times \frac{143}{3}=\frac{11 \times 11}{2} \times \frac{11 \times 13}{3}$.

We would like to write this expression as the number of small squares times the area of the square, that is as an integer times the square of some number.

We can do this by rearranging the above to give

$$
\frac{11 \times 11}{2} \times \frac{11 \times 13}{3}=\frac{11 \times 13}{6} \times 11^{2}
$$

At this point, we need to make the first factor into an integer, so we want to incorporate the 6 into the "squared" part, which we can do by rewriting again as

$$
\frac{11 \times 13}{6} \times 11^{2}=\frac{6 \times 11 \times 13}{6^{2}} \times 11^{2}=[6 \times 11 \times 13] \times\left(\frac{11}{6}\right)^{2}=858 \times\left(\frac{11}{6}\right)^{2}
$$

Thus there are 858 squares which are $\frac{11}{6}$ by $\frac{11}{6}$.

ANSWER: (B)