## Exercise 4

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)?
## Solution to Exercise 4

Answer: 168. Let $a$ and $b$ be the numerators of these two fractions, so that the sum we are looking for is $\frac{a}{600}+\frac{b}{700}$.

First, let's show that the value 168 is indeed achieved. Indeed, by setting $a=1$ and $b=3$, we get:

$$
\frac{1}{600}+\frac{3}{700}=\frac{7 \times 1}{7 \times 600}+\frac{6 \times 3}{6 \times 700}=\frac{7+18}{4200}=\frac{25}{4200}=\frac{1}{168}
$$

The fractions $\frac{1}{600}$ and $\frac{3}{700}$ are indeed irreducible, and the fraction $\frac{1}{168}$ is also irreducible.

Now, let's show that the denominator cannot be less than 168. The sum we are interested in can be written as:

$$
\frac{a}{600}+\frac{b}{700}=\frac{7 a}{7 \times 600}+\frac{6 b}{6 \times 700}=\frac{7 a+6 b}{4200}
$$

Since the fraction $\frac{a}{600}$ is irreducible, $a$ is coprime with 600 (i.e., there is no number that divides both $a$ and 600). Since $600=2^{3} \times 3 \times 5^{2}$, we know that $a$ is not divisible by 2, 3, or 5. Similarly, $b$ is coprime with 700 and $700=2^{2} \times 5^{2} \times 7$, so $b$ is not divisible by 2, 5, or 7.

We deduce that the product $7a$ is divisible by 7, but not by 2, 3, or 5. Similarly, $6b$ is divisible by 2 and 3, but not by 7. Since the sum of a number divisible by $p$ and a number not divisible by $p$ is never divisible by $p$, we deduce that the sum $7a + 6b$ is not divisible by 2, 3, or 7.

We have $4200=2^{3} \times 3 \times 5^{2} \times 7$. According to the above, we can never simplify the fraction $\frac{7a + 6b}{4200}$ by 2, 3, or 7. Therefore, the denominator of the reduced fraction will still be divisible by $2^{3} \times 3 \times 7$, which is 168. It will be at least 168, Q.E.D.