Question

Find the largest of three distinct positive integers with the least possible sum such that the sum of the reciprocals of any two integers among them is an integral multiple of the remaining one.

Answer 1

Hint: In this particular question assume any different variables be the three distinct positive integers then first find out the equations according to given condition such as $\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{c}{z},\dfrac{1}{y} + \dfrac{1}{z} = \dfrac{a}{x},\dfrac{1}{z} + \dfrac{1}{x} = \dfrac{b}{y}$, where a, b, and c are the integral multiples i.e. these are also integers so use these concepts to reach the solution of the question.

Complete step-by-step answer:
Let the three distinct positive integers be x, y and z.
Assume, x < y < z
Now according to the given condition that the sum of the reciprocals of any two integers among them is an integral multiple of the remaining one.
$ \Rightarrow \dfrac{1}{x} + \dfrac{1}{y} = \dfrac{c}{z},\dfrac{1}{y} + \dfrac{1}{z} = \dfrac{a}{x},\dfrac{1}{z} + \dfrac{1}{x} = \dfrac{b}{y}$, where a, b, and c are the integral multiples i.e. these are also integers.
Now add both sides by $\dfrac{1}{z},\dfrac{1}{x},\dfrac{1}{y}$ in the above equations respectively we have,
 $ \Rightarrow \dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} = \dfrac{{c + 1}}{z},\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} = \dfrac{{a + 1}}{x},\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} = \dfrac{{b + 1}}{y}$
$ \Rightarrow \dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} = \dfrac{{a + 1}}{x} = \dfrac{{b + 1}}{y} = \dfrac{{c + 1}}{z} = r$..................... (1), where r is some positive quantity.
Since, x < y < z so we can say that from the above equation that a < b < c.
Now from equation (1) we have,
$ \Rightarrow \dfrac{1}{x} = \dfrac{r}{{a + 1}},\dfrac{1}{y} = \dfrac{r}{{b + 1}},\dfrac{1}{z} = \dfrac{r}{{c + 1}}$........................... (2)
Now add all the above equations we have,
$ \Rightarrow \dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} = \dfrac{r}{{a + 1}} + \dfrac{r}{{b + 1}} + \dfrac{r}{{c + 1}}$
Now from equation (1) we have,
$ \Rightarrow r = \dfrac{r}{{a + 1}} + \dfrac{r}{{b + 1}} + \dfrac{r}{{c + 1}}$
$ \Rightarrow 1 = \dfrac{1}{{a + 1}} + \dfrac{1}{{b + 1}} + \dfrac{1}{{c + 1}}$...................... (3)
Now as, a < b < c, so $\dfrac{1}{{a + 1}} + \dfrac{1}{{b + 1}} + \dfrac{1}{{c + 1}} < \dfrac{3}{{a + 1}}$
Therefore, $1 = \dfrac{1}{{a + 1}} + \dfrac{1}{{b + 1}} + \dfrac{1}{{c + 1}} < \dfrac{3}{{a + 1}}$
So we can say that,
$ \Rightarrow 1 < \dfrac{3}{{a + 1}}$
$ \Rightarrow a + 1 < 3$
$ \Rightarrow a < 2$
Now as a, b and c are integers so the only integer less than 2 is 1.
Therefore, a = 1.
Now from equation (3) we have,
$ \Rightarrow 1 = \dfrac{1}{{1 + 1}} + \dfrac{1}{{b + 1}} + \dfrac{1}{{c + 1}}$
$ \Rightarrow 1 - \dfrac{1}{2} = \dfrac{1}{{b + 1}} + \dfrac{1}{{c + 1}}$
$ \Rightarrow \dfrac{1}{2} = \dfrac{1}{{b + 1}} + \dfrac{1}{{c + 1}}$................. (4)
Now as, b < c, so, $\dfrac{1}{{b + 1}} + \dfrac{1}{{c + 1}} < \dfrac{2}{{b + 1}}$............. (5)
So from equation (4) and (5) we have,
$ \Rightarrow \dfrac{1}{2} < \dfrac{2}{{b + 1}}$
$ \Rightarrow b + 1 < 4$
$ \Rightarrow b < 3$
Since, b > a, and a = 1, so the only possible integer for b is 2.
Therefore, b = 2.
Now from equation (3) we have,
$ \Rightarrow 1 = \dfrac{1}{{1 + 1}} + \dfrac{1}{{2 + 1}} + \dfrac{1}{{c + 1}}$
$ \Rightarrow 1 - \dfrac{1}{2} - \dfrac{1}{3} = \dfrac{1}{{c + 1}}$
$ \Rightarrow \dfrac{{6 - 3 - 2}}{6} = \dfrac{1}{{c + 1}}$
$ \Rightarrow \dfrac{1}{6} = \dfrac{1}{{c + 1}}$
$ \Rightarrow c + 1 = 6$
$ \Rightarrow c = 5$
Now from equation (2) we can say that,
$ \Rightarrow x:y:z = \left( {a + 1} \right):\left( {b + 1} \right):\left( {c + 1} \right)$
$ \Rightarrow x:y:z = \left( {1 + 1} \right):\left( {2 + 1} \right):\left( {5 + 1} \right)$
$ \Rightarrow x:y:z = 2:3:6$
Now we have to find out the largest three possible distinct integers with least possible sum, so from the above ratio we can say that,
$ \Rightarrow x = 2, y = 3, z = 6$
So this is the required answer.

Note:Whenever we face such types of questions the key concept we have to remember is that if, a < b < c then the value of $\left( {\dfrac{1}{{a + 1}} + \dfrac{1}{{b + 1}} + \dfrac{1}{{c + 1}}} \right)$ can be written as $\left( { < \dfrac{3}{{a + 1}}} \right)$ and $\left( {\dfrac{1}{{b + 1}} + \dfrac{1}{{c + 1}} < \dfrac{2}{{b + 1}}} \right)$ and if, $\left( {\dfrac{1}{x} = \dfrac{r}{{a + 1}},\dfrac{1}{y} = \dfrac{r}{{b + 1}},\dfrac{1}{z} = \dfrac{r}{{c + 1}}} \right)$, then $x:y:z = \left( {a + 1} \right):\left( {b + 1} \right):\left( {c + 1} \right)$.