# Find sum of a number and its maximum prime factor

Given an integer N, the task is to find the sum of N and it’s maximum prime factor.

Examples:

Input: 19
Output: 38
Maximum prime factor of 19 is 19.
Hence, 19 + 19 = 38

Input: 8
Output: 10
8 + 2 = 10

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: Find the largest prime factor of the number and store it in maxPrimeFact then print the value of N + maxPrimeFact.

Below is the implementation of the above approach:

 `// C++ program to find sum of n and  ` `// it's largest prime factor ` `#include ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the sum of n and  ` `// it's largest prime factor ` `int` `maxPrimeFactors(``int` `n) ` `{ ` `    ``int` `num = n; ` ` `  `    ``// Initialise maxPrime to -1. ` `    ``int` `maxPrime = -1; ` ` `  `    ``while` `(n % 2 == 0) { ` `        ``maxPrime = 2; ` `        ``n /= 2; ` `    ``} ` ` `  `    ``// n must be odd at this point, thus skip ` `    ``// the even numbers and iterate only odd numbers ` `    ``for` `(``int` `i = 3; i <= ``sqrt``(n); i += 2) { ` `        ``while` `(n % i == 0) { ` `            ``maxPrime = i; ` `            ``n = n / i; ` `        ``} ` `    ``} ` ` `  `    ``// This condition is to handle the case ` `    ``// when n is a prime number greater  than 2 ` `    ``if` `(n > 2) ` `        ``maxPrime = n; ` ` `  `    ``// finally return the sum. ` `    ``int` `sum = maxPrime + num; ` `    ``return` `sum; ` `} ` ` `  `// Driver Program to check the above function. ` `int` `main() ` `{ ` `    ``int` `n = 19; ` ` `  `    ``cout << maxPrimeFactors(n); ` `    ``return` `0; ` `} `

 `// Java program to find sum of n and ` `// it's largest prime factor ` `import` `java.io.*; ` ` `  `class` `GFG ` `{ ` ` `  `// Function to return the sum of n ` `// and it's largest prime factor ` `static` `int` `maxPrimeFactors(``int` `n) ` `{ ` `int` `num = n; ` ` `  `// Initialise maxPrime to -1. ` `int` `maxPrime = -``1``; ` ` `  `while` `(n % ``2` `== ``0``) ` `{ ` `maxPrime = ``2``; ` `n /= ``2``; ` `} ` ` `  `// n must be odd at this point, ` `// thus skip the even numbers and ` `// iterate only odd numbers ` `for` `(``int` `i = ``3``; i <= Math.sqrt(n); i += ``2``) { ` `     `  `    ``while` `(n % i == ``0``) {  ` `        ``maxPrime = i; n = n / i; ` `        ``}  ` `     `  `}  ` `        ``// This condition is to handle the case  ` `        ``// when n is a prime number greater than 2  ` `        ``if` `(n > ``2``) { ` `            ``maxPrime = n; ` `        ``} ` `// finally return the sum. ` `int` `sum = maxPrime + num; ` `return` `sum; ` `} ` ` `  `// Driver Code ` `public` `static` `void` `main (String[] args) ` `{ ` `int` `n = ``19``; ` ` `  `System.out.println(maxPrimeFactors(n)); ` `} ` `} ` `// This code is contributed by anuj_67 `

 `# Python 3 program to find sum of n and  ` `# it's largest prime factor ` `from` `math ``import` `sqrt ` ` `  `# Function to return the sum of n and  ` `# it's largest prime factor ` `def` `maxPrimeFactors(n): ` `    ``num ``=` `n ` ` `  `    ``# Initialise maxPrime to -1. ` `    ``maxPrime ``=` `-``1``; ` ` `  `    ``while` `(n ``%` `2` `=``=` `0``): ` `        ``maxPrime ``=` `2` `        ``n ``=` `n ``/` `2` `     `  `    ``# n must be odd at this point, thus skip ` `    ``# the even numbers and iterate only odd numbers ` `    ``p ``=` `int``(sqrt(n) ``+` `1``) ` `    ``for` `i ``in` `range``(``3``, p, ``2``): ` `        ``while` `(n ``%` `i ``=``=` `0``): ` `            ``maxPrime ``=` `i ` `            ``n ``=` `n ``/` `i ` `         `  `    ``# This condition is to handle the case ` `    ``# when n is a prime number greater than 2 ` `    ``if` `(n > ``2``): ` `        ``maxPrime ``=` `n ` ` `  `    ``# finally return the sum. ` `    ``sum` `=` `maxPrime ``+` `num ` `    ``return` `sum` ` `  `# Driver Code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``n ``=` `19` ` `  `    ``print``(maxPrimeFactors(n)) ` ` `  `# This code is contributed by  ` `# Surendra_Gangwar `

 `// C# program to find sum of n and ` `// it's largest prime factor ` `using` `System; ` ` `  `class` `GFG ` `{ ` `// Function to return the sum of n ` `// and it's largest prime factor ` `static` `int` `maxPrimeFactors(``int` `n) ` `{ ` `int` `num = n; ` ` `  `// Initialise maxPrime to -1. ` `int` `maxPrime = -1; ` ` `  `while` `(n % 2 == 0) ` `{ ` `    ``maxPrime = 2; ` `    ``n /= 2; ` `} ` ` `  `// n must be odd at this point, ` `// thus skip the even numbers and ` `// iterate only odd numbers ` `for` `(``int` `i = 3;  ` `         ``i <= Math.Sqrt(n); i += 2)  ` `{ ` `     `  `    ``while` `(n % i == 0)  ` `    ``{  ` `        ``maxPrime = i; n = n / i; ` `    ``}  ` `     `  `}  ` ` `  `// This condition is to handle the case  ` `// when n is a prime number greater than 2  ` `if` `(n > 2)  ` `{ ` `    ``maxPrime = n; ` `} ` ` `  `// finally return the sum. ` `int` `sum = maxPrime + num; ` `return` `sum; ` `} ` ` `  `// Driver Code ` `static` `void` `Main () ` `{ ` `    ``int` `n = 19; ` `     `  `    ``Console.WriteLine(maxPrimeFactors(n)); ` `} ` `} ` ` `  `// This code is contributed by Ryuga `

 ` 2) ` `        ``\$maxPrime` `= ``\$n``; ` ` `  `    ``// finally return the sum. ` `    ``\$sum` `= ``\$maxPrime` `+ ``\$num``; ` `    ``return` `\$sum``; ` `} ` ` `  `// Driver Code ` `\$n` `= 19; ` ` `  `echo` `maxPrimeFactors(``\$n``); ` ` `  `// This code is contributed  ` `// by inder_verma ` `?> `

Output:
```38
```

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