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# Find two numbers with sum and product both same as N

• Last Updated : 10 May, 2021

Given an integer N, the task is to find two numbers a and b such that a * b = N and a + b = N. Print “NO” if no such numbers are possible.
Examples:

Input: N = 69
Output: a = 67.9851
b = 1.01493
Input: N = 1
Output: NO

Approach: If observed carefully, we are given with sum and product of roots of a quadratic equation.
If N2 – 4*N < 0 then only imaginary roots are possible for the equation, hence “NO” will be the answer. Else a and b will be:

a = ( N + sqrt( N2 – 4*N ) ) / 2
b = ( N – sqrt( N2 – 4*N ) ) / 2

Below is the implementation of the above approach:

## C++

 `// C++ program to find a and b``// such that a*b=N and a+b=N``#include ``using` `namespace` `std;` `// Function to return the smallest string``void` `findAandB(``double` `N)``{``    ``double` `val = N * N - 4.0 * N;` `    ``// Not possible``    ``if` `(val < 0) {``        ``cout << ``"NO"``;``        ``return``;``    ``}` `    ``// find a and b``    ``double` `a = (N + ``sqrt``(val)) / 2.0;``    ``double` `b = (N - ``sqrt``(val)) / 2.0;` `    ``cout << ``"a = "` `<< a << endl;``    ``cout << ``"b = "` `<< b << endl;``}` `// Driver Code``int` `main()``{``    ``double` `N = 69.0;``    ``findAandB(N);``    ``return` `0;``}`

## Java

 `// Java program to find a and b``// such that a*b=N and a+b=N` `class` `GFG{``// Function to return the smallest string``static` `void` `findAandB(``double` `N)``{``    ``double` `val = N * N - ``4.0` `* N;` `    ``// Not possible``    ``if` `(val < ``0``) {``        ``System.out.println(``"NO"``);``        ``return``;``    ``}` `    ``// find a and b``    ``double` `a = (N + Math.sqrt(val)) / ``2.0``;``    ``double` `b = (N - Math.sqrt(val)) / ``2.0``;` `    ``System.out.println(``"a = "``+a);``    ``System.out.println(``"b = "``+b);``}` `// Driver Code``public` `static` `void` `main(String[] args)``{``    ``double` `N = ``69.0``;``    ``findAandB(N);``}``}``// This Code is contributed by mits`

## Python3

 `# Python 3 program to find a and b``# such that a*b=N and a+b=N``from` `math ``import` `sqrt` `# Function to return the``# smallest string``def` `findAandB(N):``    ``val ``=` `N ``*` `N ``-` `4.0` `*` `N` `    ``# Not possible``    ``if` `(val < ``0``):``        ``print``(``"NO"``)``        ``return``    ` `    ``# find a and b``    ``a ``=` `(N ``+` `sqrt(val)) ``/` `2.0``    ``b ``=` `(N ``-` `sqrt(val)) ``/` `2.0` `    ``print``(``"a ="``, ``'{0:.6}'` `. ``format``(a))``    ``print``(``"b ="``, ``'{0:.6}'` `. ``format``(b))` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``    ``N ``=` `69.0``    ``findAandB(N)``    ` `# This code is contributed``# by SURENDRA_GANGWAR   `

## C#

 `// C# program to find a and b``// such that a*b=N and a+b=N` `using` `System;``class` `GFG``{``// Function to return the smallest string``static` `void` `findAandB(``double` `N)``{``double` `val = N * N - 4.0 * N;` `// Not possible``if` `(val < 0) {``Console.WriteLine(``"NO"``);``return``;``}` `// find a and b``double` `a = (N + Math.Sqrt(val)) / 2.0;``double` `b = (N - Math.Sqrt(val)) / 2.0;` `Console.WriteLine(``"a = "``+a);``Console.WriteLine(``"b = "``+b);``}` `// Driver Code``static` `void` `Main()``{``double` `N = 69.0;``findAandB(N);``}``// This code is contributed by ANKITRAI1``}`

## PHP

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## Javascript

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Output:
```a = 67.9851
b = 1.01493```

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