Given n, find x, y, z such that x, y, z satisfy the equation “2/n = 1/x + 1/y + 1/z”
There are multiple x, y and z that satisfy the equation print anyone of them, if not possible then print -1.
Examples:
Input : 3 Output : 3 4 12 Explanation: here 3 4 and 12 satisfy the given equation Input : 7 Output : 7 8 56
Note that for n = 1 there is no solution, and for n > 1 there is solution x = n, y = n+1, z = n·(n+1).
To come to this solution, represent 2/n as 1/n+1/n and reduce the problem to represent 1/n as a sum of two fractions. Let’s find the difference between 1/n and 1/(n+1) and get a fraction 1/(n*(n+1)), so the solution is
2/n = 1/n + 1/(n+1) + 1/(n*(n+1))
C++
// CPP program to find x y z that// satisfies 2/n = 1/x + 1/y + 1/z...#include <bits/stdc++.h>using namespace std;// function to find x y and z that// satisfy given equation.void printXYZ(int n){ if (n == 1) cout << -1; else cout << "x is " << n << "\ny is " << n + 1 << "\nz is " << n * (n + 1);}// driver program to test the above functionint main(){ int n = 7; printXYZ(n); return 0;} |
Java
// Java program to find x y z that// satisfies 2/n = 1/x + 1/y + 1/z...import java.io.*;class Sums { // function to find x y and z that // satisfy given equation. static void printXYZ(int n){ if (n == 1) System.out.println(-1); else{ System.out.println("x is "+ n); System.out.println("y is "+ (n+1)); System.out.println("z is "+ (n * (n + 1))); } } // Driver program to test the above function public static void main (String[] args) { int n = 7; printXYZ(n); }}// This code is contributed by Chinmoy Lenka |
Python3
# Python3 code to find x y z that# satisfies 2/n = 1/x + 1/y + 1/z...# function to find x y and z that# satisfy given equation.def printXYZ( n ): if n == 1: print(-1) else: print("x is " , n ) print("y is " ,n + 1) print("z is " ,n * (n + 1))# driver code to test the above functionn = 7printXYZ(n)# This code is contributed by "Sharad_Bhardwaj". |
C#
// C# program to find x y z that// satisfies 2/n = 1/x + 1/y + 1/z...using System;class GFG { // function to find x y and z that // satisfy given equation. static void printXYZ(int n) { if (n == 1) Console.WriteLine(-1); else { Console.WriteLine("x is "+ n); Console.WriteLine("y is "+ (n+1)); Console.WriteLine("z is "+ (n * (n + 1))); } } // Driver program public static void Main () { int n = 7; printXYZ(n); }}// This code is contributed by vt_m |
PHP
<?php// PHP program to find x y z that// satisfies 2/n = 1/x + 1/y + 1/z...// function to find x y and z that// satisfy given equation.function printXYZ($n){ if ($n == 1) echo -1; else echo "x is " , $n , "\ny is " , $n + 1 , "\nz is ", $n * ($n + 1);} // Driver Code $n = 7; printXYZ($n);// This code is contributed by anuj_67.?> |
Javascript
<script>// Javascript program to find x y z that// satisfies 2/n = 1/x + 1/y + 1/z...// function to find x y and z that// satisfy given equation.function printXYZ(n){ if (n == 1) document.write(-1); else document.write("x is " + n + "<br>y is " + (n + 1) + "<br>z is " + n * (n + 1));}// driver program to test the above function let n = 7; printXYZ(n);</script> |
Output
x is 7 y is 8 z is 56
Time Complexity: O(1)
Auxiliary Space: O(1)
Alternate Solution
We can write 2/n = 1/n + 1/n. And further as 2/n = 1/n + 1/2n + 1/2n.
so here x = n, y = 2*n and z = 2*n
Below is the code implementation for the alternate solution
C++
// CPP program to find x y z that satisfies 2/n = 1/x + 1/y + 1/z...#include <bits/stdc++.h>using namespace std;// function to find x, y and z that satisfy given equation.void printXYZ(int n){ if (n == 1) cout << -1; else cout << "x is " << n << "\ny is " << 2 * n << "\nz is " << 2 * n;}// driver program to test the above functionint main(){ int n = 7; printXYZ(n); return 0;} |
Java
// Java program to find x y z that satisfies 2/n = 1/x + 1/y + 1/z...import java.io.*;class GFG { // function to find x, y and z that satisfy given equation. public static void printXYZ(int n) { if (n == 1) System.out.println("-1"); else System.out.println( "x is " + n + "\ny is " + 2 * n + "\nz is " + 2 * n); } // driver program to test the above function public static void main(String[] args) { int n = 7; printXYZ(n); }} |
C#
// C# program to find x y z that satisfies 2/n = 1/x + 1/y + 1/z...using System;using System.Linq;using System.Collections.Generic;class GFG { // function to find x, y and z that satisfy given equation. static void printXYZ(int n) { if (n == 1) Console.Write(-1); else Console.Write("x is " + n + "\ny is " + 2 * n + "\nz is " + 2 * n); } // driver program to test the above function static public void Main() { int n = 7; printXYZ(n); } }// This code is contributed by ratiagrawal. |
Javascript
// Javascript program to find x y z that satisfies 2/n = 1/x + 1/y + 1/z...// function to find x, y and z that satisfy given equation.function printXYZ( n){ if (n == 1) console.log(-1); else console.log("x is " + n + "\ny is " + 2 * n + "\nz is " + 2 * n);}// driver program to test the above functionlet n = 7;printXYZ(n); |
Python3
# Python program to find x y z that satisfies 2/n = 1/x + 1/y + 1/z...# function to find x, y and z that satisfy given equation.def printXYZ(n): if n == 1: print("-1") else: print(f"x is {n}\ny is {2 * n}\nz is {2 * n}")# driver program to test the above functionif __name__ == '__main__': n = 7 printXYZ(n)#This code is contributed by chinmaya121221 |
Output
x is 7 y is 14 z is 14
Time Complexity: O(1)
Auxiliary Space: O(1)
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