Two irreducible fractions have their denominators equal to 600 and 700. Find the minimum value for the denominator of the sum of the fractions.
Suppose the fractions are a/600 and b/700. Since they are irreducible, a and 600 have no common factor greater than 1, and the same applies to b and 700.

Adding the two fractions, we get

$$
\frac{a}{600}+\frac{b}{700}=\frac{7 a+6 b}{4200}=\frac{7 a+6 b}{2^{3} \times 3 \times 5^{2} \times 7}
$$

Observe that the numerator is not divisible by 2 or 3, because a has no common factor with 6. The numerator is also not divisible by 7 because b and 7 have no common factor.

Thus, the only factor of the denominator that we can possibly simplify is $5^{2}=25$. For this, we just need to take, for example, $a=1$ and $\mathrm{b}=3$.

$$
\frac{1}{600}+\frac{3}{700}=\frac{1}{168}
$$

Therefore, the minimum denominator of the sum is 168.