Irreducible Fractions

Write $\frac{A}{600}+\frac{B}{700}=\frac{7A+6B}{4200}$ .

The prime decomposition of $4200$ (the denominator) is $2^{3}.3.5^{2}.7$ , we are attempting to make the numerator share as many similar factors as possible to cancel them out in order to minimize the denominator.

Note that it is impossible for $7A+6B$ to be even since this implies $A$ must be even. However, we know that this is impossible since $A$ is coprime to $600$ . Therefore, $2^{3}$ will remain in the denominator.

Also, it is impossible for $7A+6B$ to be divisible by $3$ since this implies $A$ is divisible by $3$ . However, we know that this is impossible since $A$ is coprime to $600$ . Therefore, $3$ will remain in the denominator.

By the same reasoning, it is impossible for $7A+6B$ to be divisible by $7$ since this implies $B$ is divisible by $7$ . However, we know that this is impossible since $B$ is coprime to $700$ . Therefore, $7$ will remain in the denominator.

However, it is easy to cancel out $25$ in the denominator. Just let $A=1, B=3$ and we get $\frac{25}{4200}=\frac{1}{168}$ .

Therefore, the minimum value of the denominator is $\boxed{168}$ .

We can write this problem in the form of:

$\frac {a} {600}$ $+\frac {b} {700}=$ $\frac {c} {d}$ , where a, b, c, d are positive integers and gcd(a,600) = gcd(b,700) = gcd(c,d) = 1. We need to find the minimum value of d.

It is easy to see, that

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

We can get the minimum value of d by simplifying the fraction on the RHS (which therefore has to be a factor of 4200).

4200= $2^3×3×5^2×7$

Now, we will show, that we can simplify by $5^2=25$ , but not any further:

As 17×7+1×6=125 :

$\frac{17}{600}$ $+ \frac{1}{700}=$ $\frac {125}{4200}=$ $\frac{25×5}{25×168}=$ $\frac{5}{168}$

However, we cannot simplify by any other prime factor of 4200 (2, 3 or 7) than 5, because:

If we could simplify by 2, then there would be an m=2k (positive integer), that

7a + 6b = 2k

If we rearrange this, we get

7a = 2(k-3b)

This can only happen, if a is divisible by 2, which is impossible as gcd(a,600)=1 (the fraction is irreducible according to the question) and 2 is a factor of 600.

Similarly, we cannot simplify by 3 either, as then there would be an m=3k, 7a = 3(k-2b), a cannot be divisible by 3 (as then gcd(a,600) would be at least 3, which contradicts the condition of the question .

And we can't simplify by 7, as then there would be an n=7p, that

7a + 6b = 7p

6b = 7(a-p)

and b cannot be divisible by 7 since gcd(b,700)=1.

For the reasons above, the solution is $\boxed{168}$ .