Bogdanov I.I.

The denominators of two irreducible fractions are 600 and 700. Find the smallest possible value of the denominator of their sum (in irreducible form).

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Let our fractions be ${ }^{a} / 600$ and ${ }^{b} / 700$. Then $a$ is coprime with 6, and $b$ is coprime with 7. Therefore, the numerator of their sum ${ }^{7 a+6 b / 4200}$ is coprime with both $6=2 \cdot 3$ and 7. Since $4200=2^{3 \cdot 3} \cdot 7 \cdot 5^{2}$, this means that the denominator after simplification will be no less than $2^{3} \cdot 3 \cdot 7=168$. Such a denominator can indeed be obtained; for example, $1 / 600+3 / 700=1 / 168$.