The denominators of two irreducible fractions are 600 and 700. What is the smallest possible value of the denominator of their sum when written as an irreducible fraction?

Note: We say that the fraction $p / q$ is irreducible if the integers $p$ and $q$ do not have any prime factors in common in their factorizations. For example, 5/7 is an irreducible fraction.

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Solution

Let $a / 600$ and $b / 700$ be the two irreducible fractions. Thus, $m d c(a, 600)=m d c(b, 700)=1$. The sum of the two fractions can be written as

$$
\begin{aligned}
\frac{a}{600}+\frac{b}{700} & =\frac{7 a+6 b}{6 \cdot 7 \cdot 100} \\
& =\frac{7 a+6 b}{3 \cdot 7 \cdot 2^{3} \cdot 5^{2}}
\end{aligned}
$$

Since $a$ and $b$ are odd, $7 a+6 b$ is odd and thus does not have the prime factor 2 in its factorization. Since 7 does not divide $b$, it follows that the sum $7 a+6 b$ also does not have the prime factor 7 in its factorization. Similarly, since 3 does not divide $a$, we can conclude that this factor is not present in the factorization of $7 a+6 b$. Thus, only the prime factor 5 can be common to the numerator and the denominator of the sum, and consequently, the denominator will be at least $3 \cdot 7 \cdot 2^{3}$. To verify that it is admissible, we just need to find $a$ and $b$ such that 25 is a divisor of $7 a+6 b$. This can be achieved with $a=1$ and $b=3$, for example,

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