More Sums

By adding up the $15$ numbers, we added up each letter five times, so $p+q+r+s+t+u=\frac{25+30+38+41+49+52+54+63+68+76+79+90+95+103+117}{5}=196$ . Since $25$ is the least and $117$ is the largest, $25=p+q$ and $117=t+u$ .

So $r+s=p+q+r+s+t+u-(p+q)-(t+u)=196-25-117=\boxed{54}$

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Now we will see what numbers can be $p, q, r, s, t, u$ be. We can notice that $p+q=25, p+r=30, t+u=117, s+u=103$ . So $r=q+5$ and $t=s+14$ . It is easy to see that $q+r$ is is $38, 41, 49 or 52$ . Since $q+r=2q+5$ , $q+r$ can't be $38$ or $52$ , because they are positive integers, so $2q+5=41$ or $49$ , $2q=36$ or $44$ , $q=18$ or $22$ , $r=23$ or $27$ . Similarly we get: $s+t$ is $95, 90, 79$ or $76$ , but $s+t=2s+14$ , so $s+t=2s+14=90$ or $76$ , $s=38$ or $31$ , $t=52$ or $45$ . We know $q$ and $t$ , so we know that $p=7$ or $3$ , $u=65$ or $72$ .

Finally we have:

About $p$ , $q$ and $r$ : $\large \begin{cases} p=7, q=18, r=23 \\ p=3, q=22, r=27\end{cases}$

About $s, t$ and $u$ : $\large \begin{cases} s=31, t=45, u=72 \\ s=38, t=52, u=65\end{cases}$

Since $r+s=54$ , the only possible solution is:

$p=7<q=18<r=23<s=31<t=45<u=72$

Note : In the beginning we proved that if there exist six positive integers that satisfy the given condition, then $r+s=54$ . But it isn't enough!!! We have to give an example, because if there isn't exist six positive integers like this, then the answer is not 54. Be careful!

If we take each letter and add it to the others we are going to have $5$ sums. Take for example $p$ and add it to every other letter, we will obtain: $p+q, p+r, p+s, p+t, p+u$ . Similarly, the same method can be applied to the other letters.

Adding all these sums, we get:

$5p+5q+5r+5t+5u +5s = 5( p+q+r+s+t+u ) = 25 + 30 + 38 + 41 + ... + 103+ 117 = 980$

Thus, $p+q+r+s+t+u = \frac{980}{5} = 196$ .

Since $p+q$ is the smallest sum, implies $p+q= 25$ , and $t+u$ is the greatest sum, then $t+u = 117$ .

Now, $p+q+r+s+t+u = (p+q) + (r+s) + (t+ u) = 25 + (r+s) + 117 = 196 \implies (r+s)= 54$