$\frac{p}{r}+\frac{q}{s}=2$ with p,r and q,s relatively prime $ps+qr=2rs$ Dividing both sides r we obtain $r|s$, because p and r relatively prime. Dividing both s we obtain $s|r$. Then s=r $p+q=2r=1000 \rightarrow r=500$ solutions are all p and q such that p and q aren't divisible by 5,2. We have 800 pairs of (p,q) and(q,p)

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AdiletR wrote: $\frac{p}{r}+\frac{q}{s}=2$ with p,r and q,s relatively prime $ps+qr=2rs$ Dividing both sides r we obtain $r|s$, because p and r relatively prime. Dividing both s we obtain $s|r$. Then s=r $p+q=2r=1000 \rightarrow r=500$ solutions are all p and q such that p and q aren't divisible by 5,2. We have 800 pairs of (p,q) and(q,p) Note that a number is coprime with $500$ if, and only if, it's not divisible by $2$ and $5$. * The number of multiples of $2$ between $1$ and $1000$ is $\frac{1000}{2}=500$. * The number of multiples of $5$ between $1$ and $1000$ is $\frac{1000}{5}=200$. * The number of multiples of $10$ between $1$ and $1000$ is $\frac{1000}{10}=100$. Therefore, we have $1000-(500+200-100)=400$ ways. But the order doesn't matter, so number of possibilities of Pedro choose is $200$.