Large rectangle with perimeter 300 cm divided into smaller identical rectangles. 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
$$12,18,24$$
