A road has houses numbered from 1 to $n$, where $n$ is a three-digit number. Exactly $\frac{1}{k}$ of the numbers start with the digit 2 , where $k$ is a positive integer. Find the possible values of $n$.
SolUTION
We consider the three cases $n<200,200 \leq n<300$ and $300 \leq n$ separately.
Case I
There are eleven houses among the first 99 whose numbers begin with a 2 , house 2 and houses 20 to 29 inclusive.
Thus if $n<200$, we have $\frac{1}{k}=\frac{11}{n}$. This shows that $n$ is a multiple of 11 . There are nine possibilities, namely $n=110,121,132,143,154,165,176,187,198$.
Case II
If $200 \leq n<300$ we may write $n=199+x$ for some integer $1 \leq x \leq 100$.
We have that $\frac{1}{k}=\frac{11+x}{199+x}$ so $k=\frac{199+x}{11+x}=1+\frac{188}{11+x}$. This implies that $k-1=\frac{188}{11+x}$ so $11+x$ must be a factor of 188 (since $k-1$ is an integer).
Thus, we need to find factors of 188 which are between 12 and 111 inclusive. Since, $188=2^{2} \times 47$ its factors are $1,2,4,47,94$ and 188 . Of these we need only consider 47 and 94 which give $(x, n)=(36,235)$ and $(83,282)$ respectively.
Case III
There are 111 houses among the first 299 whose numbers begin with a 2, the eleven already counted and those numbered 200 to 299 inclusive.
Thus if $n \geq 300$, we have $\frac{1}{k}=\frac{111}{n}$. This shows that $n$ is a multiple of 111 . There are seven possibilities, namely $n=333,444,555,666,777,888,999$.

Alternative
The second, most interesting, case can also be tackled by bounding the possible values of $k$ and then checking each in turn. Since $k-1=\frac{188}{11+x}$, and the left hand side decreases as $x$ increases, we see that $\frac{188}{12}+1 \geq k \geq \frac{188}{111}+1$ or $3 \leq k \leq 16$. These fourteen values of $k$ can be tested in turn to see that only two, $k=3$ and $k=5$, give rise to integer values of $x$ and hence $n$. It is important with solutions of this kind to provide enough evidence of the checking to make it clear that the solutions have been found systematically, rather than by lucky guesswork.