The question states:
There are two irreducible rational numbers with denominators $600$ and $700$. Find the minimal possible value of the denominator of their sum.
I tried to rearrange it into $$\frac{x}{600}+\frac{y}{700}=\frac{7x+6y}{4200}$$ where $\gcd(x, 6)=1$ and $\gcd(y, 7)=1$ From that I tried to identify $\gcd(7x+6y, 4200)$ but couldn't. I also tried using Bezout's theorem but failed. I was wondering if I could get some help on this question, maybe a hint. Thank you anyway.
On
Note first that $x$ has no factor in common with $6$ and $y$ is not divisible by $7$ (also neither is divisible by $2$ or $5$).
This means that you will always have the factor $42$ in the denominator, leaving $100$. Since $x$ is not even, no factor $2$ will cancel, so only the factor $25$ in the denominator can actually cancel, leaving a minimum value $4\times 42=168$
Then can $7x+6y$ cancel a factor of $25$ whilst satisfying the conditions $x$ having no factor $2,3,5$ and $y$ having no factor $2,5,7$? And if not, is it possible to cancel $5$?
On
By below $\,d =100,\ a,b = 6,7,\ c = 25,\ $ so least denom $= abd/c = 6\cdot 7\cdot 100/25 = 168$
Theorem $ $ If $\, (x,A)\! =\! 1\! =\! (y,B)\ $ then $\ \dfrac{x}A + \dfrac{y}B = \dfrac{B x \!+\! A y}{AB} = \dfrac{b x \!+\! a y}{abd},\ \ \begin{align} d\, &=\, (A,B)\\ A,B &=\,\ da,db\end{align}$
has least denominator = $abd/c\, $ for all such $\,x,y,\,$ where $\,c = $ max factor of $\,d\,$ coprime to $\,a,b$.
Proof $\quad\ (a,bx\!+\!ay) = (a,bx)=1\,$ by $\,(a,b)=1=(a,x),\,$ since $\,a\mid A\,$ and $(A,x)=1.$
Similarly $\,(b,bx\!+\!ay) = 1.\,$ So $\,a,b\,$ can't cancel from the denominator, nor can any prime factors of $\,d\,$ dividing $\,a\,$ or $\,b.\,$ But we can cancel all the other (co)primes in $\,c\,$ by choosing $\,x,y\,$ such that $\,bx+ay = c,\,$ possible since $(a,b) = 1$.
Hint. Note that $25$ divides $4200$ and by solving $7x+6y=25$ we get $$\frac{1}{600}+\frac{3}{700}=\frac{25}{4200}=\frac{1}{168}.$$ Is $168$ the minimal value of the denominator we can get?