3. The difference of two natural numbers is 2010 and their greatest common divisor is $2014 k$ times smaller than their least common multiple. Determine all such pairs of numbers.
SOLUTION. Let the sought numbers be $a$ and $b (a > b)$ and $d$ their greatest common divisor. Then $a = m d, b = n d$, where $m > n$ are coprime numbers. Since the least common multiple of the numbers $a, b$ is the number $m n d$, substituting into the given relationships gives us the equations

$$
\begin{gathered}
a - b = (m - n) d = 2010 \\
m n d = 2014 d \quad \text { or } \quad m n = 2014 .
\end{gathered}
$$

According to the prime factorization $2014 = 2 \cdot 19 \cdot 53$, we list all possible pairs $(m, n)$ and for each of them, we check whether the number $m - n$ is a divisor of the number 2010.

In the affirmative case, the corresponding quotient gives the number $d$ and the calculation of the unknowns $a = m d$ and $b = n d$ is straightforward:
a) $m = 2014$ and $n = 1: \quad m - n = 2013$ does not divide 2010;
b) $m = 19 \cdot 53 = 1007$ and $n = 2: \quad m - n = 1005 \mid 2010, d = 2, a = 1007 \cdot 2 = 2014$, $b = 2 \cdot 2 = 4$;
c) $m = 2 \cdot 53 = 106$ and $n = 19: \quad m - n = 87$ does not divide 2010;
d) $m = 53$ and $n = 2 \cdot 19 = 38: \quad m - n = 15 \mid 2010, d = 134, a = 53 \cdot 134 = 7102$, $b = 38 \cdot 134 = 5092$.

Conclusion: The sought numbers form one of the pairs $(2014, 4)$ or $(7102, 5092)$.

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