Cutting out squares

Since the rectangle can be cut into squares, suppose that the side of the largest possible square is $x$ . Then the height and width of the rectangle must be $h = mx$ and $w = nx$ for positive integers $m,n$ . If $d > 1$ was a common factor of $m$ and $n$ , then the rectangle could be cut into squares of side $dx$ , which would contradict the maximality of $x$ . Thus $m$ and $n$ are coprime.

Suppose that the rectangle can be cut into squares of side length $y$ . Then we can find positive integers $M,N$ such that $h = My$ and $w = Ny$ , and hence $mN = nM$ . Since $m,n$ are coprime, this means that $m$ divides $M$ and $n$ divides $N$ . Cutting the rectangle into $y \times y$ square yields $MN$ squares, so always yields a multiple of $mn$ squares. Thus the total number of squares obtained by the class must be a multiple of $mn$ . Since this total number is prime, we deduce that $m=n=1$ , and hence $h = w$ , and the rectangle is in fact a square.

A total of $30$ $x \times x$ squares can be cut into

• a total of $28$ sets of $9$ squares of side $\tfrac13x$ ,
• one set of $4$ squares of side $\tfrac12x$ ,
• one $x \times x$ square,

which makes a total of $(28 \times 9) + 4 + 1 = 257$ squares, which is a prime number. Thus it is possible to do the cutting as described when the rectangle is in fact a square, and so the ratio of the height to the width of the rectangle is $\boxed{1}$ .