Let's determine all natural numbers $n$ of at most three digits for which $\frac{n+15}{n}$ and its reciprocal can be expressed as finite decimal fractions.

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Determine all natural numbers $n$ of at most three digits for which $\frac{n+15}{n}$ and its reciprocal can be expressed as finite decimal fractions.
Let the number $n+15$ be denoted by $m$, the greatest common divisor of $n$ and $m$ by $d$, and the numbers $n / d, m / d, 15 / d$ by $\nu, \mu, \delta$ respectively. On one hand, $\mu=\nu+\delta$ holds due to $m=n+15$, and on the other hand, neither $\mu$ nor $\nu$ can have prime divisors other than 2 and 5, since they are the denominators of fractions that cannot be further simplified and can be written as finite decimal fractions. Since the possible values of $\delta$ are the divisors of 15, and all of these are odd, one of $\nu$ and $\mu$ is even, and the other is odd. Let the odd one be denoted by $\kappa$ and the even one by $\sigma$, then $\varepsilon=\sigma-\kappa$ is equal to $\delta$ or $(-\delta)$.

We will tabulate the possible values of $\kappa$ and $\varepsilon$ along with the values of the sums $\varepsilon+\kappa$. We will select those sums that are non-negative and have no prime divisors other than 2 and 5; these are the possible values of $\sigma$. Among the obtained $\kappa, \sigma$ pairs, the smaller one is $\nu$, and multiplying it by $d$ gives us the sought $n$ numbers. Since $d$ is equal to the absolute value of the quotient $15 / \varepsilon$, it is expedient to carry out the procedure column by column. Thus, we obtain the numbers $10 ; 60 ; 10 ; 60 ; 15 ; 5$, 25, $625 ; 15,1,5,25,625$, so the problem has seven different solutions for $n$. These are: $1,5,10,15,25,60,625$.

| $\kappa \backslash \varepsilon$ | -15 | -5 | -3 | -1 | 1 | 3 | 5 | 15 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 1 | $-\mathbf{1 4}$ | -4 | -2 | 0 | $\mathbf{2}$ | $\mathbf{4}$ | 6 | $\mathbf{1 6}$ |
| 5 | -10 | 0 | $\mathbf{2}$ | $\mathbf{4}$ | 6 | $\mathbf{8}$ | $\mathbf{1 0}$ | $\mathbf{2 0}$ |
| 25 | $+\mathbf{1 0}$ | $\mathbf{2 0}$ | $\mathbf{2 2}$ | $\mathbf{2 4}$ | 26 | 28 | 30 | $\mathbf{4 0}$ |
| 125 | 110 | 120 | 122 | 124 | 126 | $\mathbf{1 2 8}$ | 130 | 140 |
| 625 | 610 | 620 | 622 | 624 | 626 | 628 | 630 | $\mathbf{6 4 0}$ |