Three positive integers sum to 15, and the sum of their reciprocals is $\frac{71}{105}$. Determine the numbers!
Let $a, b$ and $c$ denote the three numbers, then according to the conditions,

$$
\begin{aligned}
a+b+c & =15 \\
\frac{1}{a}+\frac{1}{b}+\frac{1}{c} & =\frac{71}{105}
\end{aligned}
$$

By bringing the left side of (2) to a common denominator and multiplying, we get

$$
105(a b+b c+c a)=71 a b c
$$

From this, $105 \mid 71 a b c$ and since $(105,71)=1$, it follows that $105 \mid a b c$. Since $105=3 \cdot 5 \cdot 7$, among the three numbers $a, b$, and $c$, there is one divisible by 3, one by 5, and one by 7, because a prime number - 3, 5, or 7 - can only divide a product - $a b c$ - if it divides one of its factors. The product of any two of 3, 5, and 7 is at least 15, so neither $a$, nor $b$, nor $c$ can be a multiple of any two of them, as in that case their sum would be greater than 15.

Therefore, among the three numbers $a, b$, and $c$, one is a multiple of 3, another of 5, and the third of 7. Since $3+5+7=15$, and the numbers are positive, this can only be true if one number is 3, another is 5, and the third is 7. Then (1) is clearly satisfied, and (2) is immediately apparent.

The three numbers sought are therefore 3, 5, and 7.

Remarks. 1. The condition $105|abc$ can also lead us to the goal in another way. According to the inequality between the arithmetic and geometric means,

$$
\frac{a+b+c}{3} \geqq \sqrt[3]{a b c}
$$

After cubing, we get $a b c \leqq\left(\frac{a+b+c}{3}\right)^{3}=125$. From this, only $a b c=105$ is possible. All three numbers are greater than 1, since the sum of their reciprocals is less than 1, so $105=3 \cdot 5 \cdot 7$ implies the numbers are 3, 5, and 7.

2. From $\frac{a b+a c+b c}{a b c}=\frac{71}{105}$, many concluded that $a b+a c+b c=71$ and $a b c=105$, i.e., the numerators and denominators of the two fractions are equal separately. Although in this case this is indeed the case, generally it is possible that $\frac{71}{105}$ is the simplified form of the fraction on the left side, so the conclusion is incorrect. Similarly, it is not true that 105 is the least common multiple of $a, b$, and $c$.

Both statements are contradicted by the equality $\frac{1}{2}+\frac{1}{6}+\frac{1}{7}=\frac{17}{21}$, since $2 \cdot 6 \cdot 7 \neq 21$ and $[2,6,7] \neq 21$.