Finding the minimum of $\frac pq + \frac rs$ for distinct integers $p, q, r, s$ from $\{1,2,3,4,5,\ldots,16,17\}$

Question

Here is the question:

Four distinct integers $p$, $q$, $r$ and $s$ are chosen from the set $\{1, 2, 3, 4, 5, \ldots, 16, 17\}$. The minimum possible value of $\frac pq + \frac rs$ can be written as $\frac ab$, where $a$ and $b$ are positive, co-prime integers. What is the value of $a+b$?

I got the answer as $321$. I took the numbers as $\frac pq$ as $\frac 1{16}$ and $\frac rs$ as $\frac 2{17}$ and added them to get $\frac ab = \frac{49}{272}$. Well I first took $\frac pq$ and $\frac rs$ as $\frac 1{17}$ and $\frac 2{16}$ respectively.

So without calculating both the results I wouldn't have got the answer. So can you help me by giving the way to get the answer?

Answer 1

Obviously, we want the minimum, so as small $p,r$ as possible, and as large $q,s$ as possible. There are 2 choices:

$1/17+2/16=50/(16\cdot17)$ and $1/16+2/17=49/(16\cdot17)$. The second one is smaller, the fraction is $a/b=49/272$ and hence $a+b=321$.

I don't know what other answer you want. In general case of choosing from $\{1,2,\dotsc,n\}$ you will get $a+b=(n+2(n-1))+(n(n-1))=n^2+2n-2$, the reasoning is the same.

Answer 2

$\frac{1}{16} + \frac{2}{17} = \frac{1}{16} + \frac{1}{17} + \frac{1}{17}$

$\frac{2}{16} + \frac{1}{17} = \frac{1}{16} + \frac{1}{16} + \frac{1}{17}$

So obviously the first one is smaller.