Minimum Fraction Sum

To mimimize $\frac {p}{q} + \frac {r}{s}$ , $p$ and $r$ should be as small as possible, while $q$ and $s$ should be as large as possible. Either $p$ will be $1$ and $r$ will be $2$ , or vice-versa. Likewise, $q$ and $s$ will each be either $16$ or $17$ . Adding the fractions gives:

$\frac {p}{q} + \frac {r}{s} = \frac {ps + rq}{qs}$

$qs = 16*17=272$

The minimum value of $ps + rq$ given the previous conclusions is: $ps + rq = 1*17 + 16*2 = 49$

So, the minimum value of $\frac {ps + rq}{qs}$ is $\frac{49}{272}$

Because the fraction is already in simplest form, $a + b = 49 + 272 = 321$

Since all we need is to make the value of $\frac{a}{b}$ as small as possible, all we need to do is to choose the number of q and s, which has the largest value between 4 of these variables, and choose the number of p and r, which has the smallest of these variables, from the set of integers. (It's obvious to choose the largest integer as denominator and smallest integer as nominator, since we want to make a smallest fraction number with given set of integers)

Now, we have two candidate for denominator, let's say D = {16,17}, and we have two candidate for nominator, let's say N = {1,2}. now we want to create a pair of fraction, which candidate is F1 = { $\frac {1}{16}$ , $\frac {2}{17}$ } and F2 = { $\frac {1}{17}$ , $\frac {2}{16}$ }. Now, let's calculate the sums of the elements of F1 and F2, and we have : |F1| = $\frac {1}{16}$ + $\frac {2}{17}$ = $\frac {49}{272}$ and |F2| = $\frac {1}{17}$ + $\frac {2}{16}$ = $\frac {50}{272}$

From these two result, |F1| is smaller than |F2|. so the answer of $\frac {a}{b}$ is $\frac {49}{272}$ . since 49 and 272 is already coprime, then we can safely assume that a = 49 and b = 272.

so to answer the value of a + b, we just need to check the result of 49 + 272, which is 321.

To minimize this sum, we want to minimize the numerators, and maximize the denominators. This means our numerators are 1 and 2, and our denominators are 16 and 17. This offers only two possible sums, 1/16+2/17 and 1/17 + 2/16. Because there are only two cases, it is easiest to just check both, and the former is the smaller, so the sum is 49/272.

p/q,r/s=min implies q,s are max ie(17,16)&p,rare min ie(1,2) case-1 (p=1,q=17)&&(r=2,s=16) therefore (p/q)+(r/s)=50/272 case-2 (p=1,q=16)&&(r=2,s=17) therefore (p/q)+(r/s)=49/272 case-3 (p=2,q=17)&&(r=1,s=16) therefore (p/q)+(r/s)=49/272 case-4 (p=2,q=16)&&(r=1,s=17) therefore (p/q)+(r/s)=50/272 therefore case-2,3 are min therefore 49/272=a/b therefore a+b=321

For the sum $\frac {p}{q} + \frac {r}{s}$ to be minimum, $p$ and $r$ must be minimum and $q$ and $s$ must be maximum. Notice that $\frac {1}{17} + \frac {2}{16} > \frac {1}{16} + \frac {2}{17}$ . Hence $\frac {a}{b} = \frac {1}{16} + \frac {2}{17} = \frac {49}{272}$ . Thus, $a+b=321$ .

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

1/17+2/16=50/(16⋅17) and 1/16+2/17=49/(16⋅17). The second one is smaller, the fraction is a/b=49/272 and hence a+b=321. (as in 2n+(n−1) or n+2(n−1) former is smaller) In general case of choosing from {1,2,…,n} you will get a+b=(n+2(n−1))+(n(n−1))=n^2+2n−2.

Solution 1: The minimum is achieved when the numerators are as small as possible, and the denominators are as large as possible. Hence, we have $\{p, r\} = \{1, 2\}$ and $\{ q, s \} = \{16, 17 \}$ . It remains to check that $\frac {1}{16} + \frac {2} {17} < \frac {1}{17} + \frac {2}{16} \Leftrightarrow \frac {1}{17} < \frac {1}{16}$ . Hence, the minimum value is $\frac{1}{16} + \frac{2}{17} = \frac{49}{272}$ . Thus $a + b = 49+272 = 321$ .

Solution 2: The minimum is achieved when the numerators are as small as possible, and the denominators are as large as possible. Hence, we have $\{p, r\} = \{1, 2\}$ and $\{ q, s \} = \{16, 17 \}$ . By the rearrangement inequality, we know that $\frac {p}{q} + \frac {r}{s}$ is minimized when $\{ p, r\}$ and $\{ \frac {1}{q}, \frac {1}{s} \}$ are arranged in opposite order. Hence, the minimum value is $\frac {1}{16} + \frac {2}{17} = \frac{49}{272}$ . Thus $a + b = 49+272 = 321$ .

The answers would probably involve the numbers 1,2,16,17. If we choose 1/17, then the other would be 2/16 or 1/8, but this would not result to the minimum sum. If we try 1/16 and 2/17, then this would result to the minimum. 17+2(16)/16*17=49/272, a+b=321

Let p=1,q=16,r=2 and s=17 then (p/q)+(r/s)=(1/16)+(2/17)=49/272 =a/b. a+b=49+272=321.

Select small p,r as possible, and as large q,s as possible. There are 2 choices: 1/17+2/16=50/(16⋅17) (Discard it) & 1/16+2/17=49/(16⋅17). so let Nos. be 1,2,16 & 17 from the given set: and suppose: p=1, q=16, r=2 & s=17 so that: p/q+r/s= 1/16+2/17 =17+32/16.17=49/272=a/b so a=49 and b=272 which shows a+b=49+272=321

since p/q + r/s is added up to form (ps+qr)/qs .

Therefore , we must choose 16 and 17 for q or p , whereas 1 and 2 must be either p or r .

First case : 1/17 + 2/16 = 50/272

the value for ps+qr = 50 , since qs is a constant which is 272 .

For second case : 2/17+1/16 = 49/272

It has a smaller value of ps+qr = 49 .

Therefore , the answer will be 49/272 .

Minimum value of \frac {p}{q } + \frac {r}{s} = .... (1) To make (1) minimum, p and r must be less than q and s. First possibility : \frac {1}{17}+\frac {2}{16} ...(2) Second possibility : \frac {1}{16}+\frac {2}{17} ...(3) Just take 2 combined number. After both are counted, the equation (3) < (2), the equation (3) is \frac {49}{272} = \frac {a}{b} whereas a=49 and b=272. So a+b = 49+272=321

For the minimum possible value of p/q+r/s, we have to choose two fractions with the smallest possible value from the nos. from 1 to 17 the smallest possible fraction is 1/17 and the next smallest is 1/16 but the values should be distinct and hence we cant chose these two fractions and so we will have to choose a fraction with smallest possible numerator from the given nos. except 1 (as we have already taken 1) i.e. 2 so now the smallest fraction with numerator 2 is 2/17 Hence from the given nos. the two fractions with smallest possible values are 1/16 and 2/17 (we cant take 1/17 because we want four distinct nos.) hence p/q=1/16 & r/s=2/17 Now, p/q+r/s=1/16+2/17=49/272 Therefore, a/b=49/272 So, a = 49 & b = 272 Hence, a + b = 49 + 272 = 321 The answer is 321

\frac{1}{16}+\frac{2}{17}=\frac{49}{272} thus 49+272=321

p/q is the smallest when p is minimum and q is maximum. So, here p/q = 1/17 and r/s = 2/16 OR p/q = 1/16 and r/s = 2/17. The second situation makes their sum minimum. Hence, p/q + r/s = 1/16 + 2/17 = 49/272 = a/b. Hence, a + b = 321

hence let p =1, q = 16; r = 2 and s = 17

a/b = 49/272

thus a+b = 321