Question

Let S (n) denote the number of ordered pairs (x, y) satisfying \[\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{n}\], where n > 1 and x, y \[\in \] N.
(i) Find the value of S (6).
(ii) Show that if n is prime, then S (n) = 3 always.

Answer 1

Hint: Use the condition given to find the simplest possible relation between the n and (x, y). Use general algebra techniques to solve. After getting the simplest relation try to find the number of pairs by substituting n =6. Now, the result will be answered to part (i) of the question. Next take n as a prime number use the prime numbers definition to crack a point which will make the solution simpler.

Complete step-by-step answer:
The three given condition in the question are: -
\[\Rightarrow \dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{n}\] - (1)
\[\Rightarrow n>1\] - (2)
\[\Rightarrow x,y,n\in N\] - (3)
By condition (3), we can say that \[x,y,n>0\].
By general mathematical knowledge we can say that the result obtained when we add 2 positive numbers is greater than the each number added.
As we know, \[x,y,n>0\], we can say \[\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{n}>0\].
From condition (1), we get: Sum of \[\dfrac{1}{x},\dfrac{1}{y}\] is \[\dfrac{1}{n}\].
So, by above statement, we can conclude:
\[\Rightarrow \dfrac{1}{n}>\dfrac{1}{x};\dfrac{1}{n}>\dfrac{1}{y}\]
By general knowledge of fractions, if \[\dfrac{1}{a}>\dfrac{1}{b}\] then b > a, from this we get:
\[\Rightarrow \] x > n; y > n
So, we can write x, y as follows:
\[\Rightarrow \] x = n + a; y = n + b, where a, b > 0
By substituting these x, y values into condition (1), we get:
\[\Rightarrow \dfrac{1}{n+a}+\dfrac{1}{n+b}=\dfrac{1}{n}\]
By taking least common multiple, we can write it as:
\[\Rightarrow \dfrac{n+b+n+a}{\left( n+a \right)\left( n+b \right)}=\dfrac{1}{n}\]
By cross multiplying the terms, equation turns into:
\[\Rightarrow n\left( 2n+a+b \right)=\left( n+a \right)\left( n+b \right)\]
By simplifying the equation, we write it as follows:
\[\Rightarrow n\left( 2n+a+b \right)=\left( n+a \right)\left( n+b \right)\]
By cancelling common terms and subtracting \[{{n}^{2}}\] on both sides we get:
\[\Rightarrow {{n}^{2}}=ab\] - (4)
So, condition (1) is simplified as equation (4).
For a particular n, the number of ordered pairs (a, b) will be the number of ordered pairs (x, y) because n is constant. So, by this we can say S (n) as number of ordered pairs (a, b) such that, \[{{n}^{2}}=ab\].
For solving first part: -
We have n = 6, that is we need the value of S (6).
By substituting this into condition, we get: \[ab={{6}^{2}}=36\].
So, the number 36 can be prime factorized as.
\[\Rightarrow 36=2\times 18\]
The number 18 can be written as \[2\times 9\] by substituting, we get:
\[\Rightarrow 36=2\times 2\times 9\]
The number 9 can be written as \[{{3}^{2}}\]. By substituting this, we get:
\[\Rightarrow 36={{2}^{2}}\times {{3}^{2}}\]
So, the possible values for (a, b) can be written as:
(a, b) = (1, 36); (2, 18); (3, 12); (4, 9); (6, 6); (9, 4); (12, 3); (18, 2); (36, 1)
So, the number of ordered pairs is 9. We did prime factorization. So, we can find factors.
By this we have, S (6) = 9.
For solving second part: -
Given n is the prime number.
Prime number: - A prime number is a natural number greater than 1 that is not a product of 2 other small natural numbers. In other words the number whose factors are only 1 and the number itself is said to be a prime number.
So, for S (n) such that n is prime number we get:
\[\Rightarrow {{n}^{2}}=ab\], where n is a prime number.
So, \[{{n}^{2}}\] can be written as: \[n\times n\].
So, the factors of \[{{n}^{2}}\] as: 1, n, \[{{n}^{2}}\].
So, \[{{n}^{2}}\] can be written as:
\[\Rightarrow {{n}^{2}}=1\times {{n}^{2}};{{n}^{2}}=n\times n;{{n}^{2}}={{n}^{2}}\times 1\]
So, the ordered pair can be written as:
(a, b) = (1, \[{{n}^{2}}\]); (n, n); (\[{{n}^{2}}\], 1).
So, irrespective of n will have only 3 ordered pairs.
So, S (n) = 3, when n is a prime number.
Hence proved the required result.
Therefore we say S (6) = 9 and S (n) = 3 if n is a prime number.

Note: Be careful while calculating S (6) because if you miss one factor the whole answer might change. The idea of converting a given expression to the simplest possible is very crucial because a sum of 2 numbers can have many variations. There is a large chance you miss some but product can be done by taking prime factorization. So, whenever you have a number of ordered pairs try to convert the given sum into product form for easier simplification.