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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?

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3 Answers 3

up vote 2 down vote accepted

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.

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So without calculating both we can't say which one is smaller? –  chndn Dec 4 '12 at 10:27
    
Somehow we can: which is smaller: $2n+(n-1)$ or $n+2(n-1)$? But basically, maybe some complicated constructions would work such that you don't calculate the numbers. However, mathematics is not about making things complicated, only some people think it is. –  tohecz Dec 4 '12 at 10:28
    
Okay I can accept this answer after 8 mins –  chndn Dec 4 '12 at 10:30

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

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

So obviously the first one is smaller.

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p/q + r/s = (p.s + r.q)/q.s

to gain smallest value, denominator q.s must be bigest, there are 16 and 17 for q and s and nominator must be smallest, p.17 + r.16, so p=1 and r=2

(1.17 + 2.16) / 16.17 = 49/272 a+b=321

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