Round 3

[Nadim Ul Abrar]

Easy solution of $\boxed 8$
Let $p > q \geq r$
Then $p > \frac {q+r} 2$
But the highest divisor of $n$ may be $\frac n2$
Do the rest

Let $p \geq q \geq r$
Then $p \geq \frac {q+r} 2$
But the highest divisor of $n$ may be $\frac n2$
so $p= \frac {q+r} 2$ cz $p$ is not equal to$ q+r$

$r=2p-q $
now $q|3p+q$
or $q|3p $
or$ q=3,p $
if $q=p $then$ r=p$ too

and if $q=3$ then$ 2p-3|p+3$ . so $p\leq5$ .
$p=5$ doesnt work . if$ p=3$ then $r$ is also $3$ and if $p=2$ then $r=1$ .

hance all solution is $p=q=r $.. right ?

[*Mahi*]

The solution is a lot easier... we have just proved $p > \frac {q+r} 2$ is not possible, right? Then $p=q \geq r$.