A large rectangle whose perimeter is 300 cm, is divided into a number of identical rectangles, each of perimeter 58 cm. Each side of these rectangles is a whole number of cm. Show that there are exactly two possibilities for the number of smaller rectangles.
My solution so far:
Larger rectangle $2x+2y=300\Rightarrow x+y=150$
Smaller rectangles $2x+\frac{2y}{n}=58\Rightarrow x+\frac{y}{n}=29\Rightarrow x=29-\frac{y}{n}$ for some $n\in Z$.
This gives
$29-\dfrac{y}{n}+y=150$
$y-\dfrac{y}{n}=121$
$y\left(1-\dfrac{1}{n}\right)=121$
Now this means that the pairs of product can only be 1 and 121, 11 and 11, and 121 and 1. However, this can't be. Where did I go wrong? Or have I misunderstood the problem?
Let the small rectangles be $z \times (29-z)$ with $z \lt 15$ and there be $n$ of them. The area of the large rectangle is then $nz(29-z)$. If one dimension of the large rectangle is $w$ we get $$nz(29-z)=w(150-w)\\ 29nz-nz^2=150w-w^2\\ w^2-150w+29nz-nz^2=0\\ w=75\pm\sqrt{nz^2-29nz+5625}$$ I made a spreadsheet to search for solutions and I find $$\begin {array} {r r r r r} n&z&\text{small}&\text{large}&\text{OK}\\ \hline 12&6&6\times 23&12\times 138&\text{yes}\\ 14&4&4\times 25&10\times 140\\ 16&2&2\times 27&6\times 144\\ 18&8&8\times 21&24\times 126&\text{yes}\\ 12&11&11\times 18&18 \times 132&\text{yes}\\ 23&13&13 \times 16&46 \times 104\\ 24&12&12\times 17&48 \times 102&\text{yes}\\ 26&4&4\times 25&20\times 130\\ 28&8&8\times 21&42\times 108\\ 27&13&13 \times 16&72\times78 \end {array}$$ Where yes in the OK column means the small rectangles can tile the large one in an obvious way. I claim the problem is wrong and there are three possibilities for the number of smaller rectangles $$12,18,24$$