# C++ find four factors of N with maximum product and sum equal to N .

## Concept

With respect of a given integer N, our task is to determine all factors of N print the product of four factors of N so that −

• The sum of the four factors is equal to N.
• The product of the four factors is largest.

It has been seen that if it is impossible to find 4 such factors then print “Not possible”.

It should be noted that all the four factors can be equal to each other to maximize the product.

## Input

24

## Output

All the factors are -> 1 2 4 5 8 10 16 20 40 80
Product is -> 160000

Select the factor 20 four times,

Therefore, 20+20+20+20 = 24 and product is maximum.

## Method

Following is the step by step algorithm to solve this problem −

• At first determine the factors of a number ‘N’ by visiting from 1 to square root of ‘N’ and verify if ‘i’ and ‘n/i’ divides N and store them in a vector.
• Now we sort the vector and print every element.
• Determine three numbers to maximize the product with the fourth number, implementing three loops.
• Finally we replace the next maximum product with the previous product.
• Print the product when you’ll find the four factors.

## Example

Live Demo

// C++ program to find four factors of N
// with maximum product and sum equal to N
#include <bits/stdc++.h>
using namespace std;
// Shows function to find factors
// and to print those four factors
void findfactors2(int n1){
vector<int> vec2;
// Now inserting all the factors in a vector s
for (int i = 1; i * i <= n1; i++) {
if (n1 % i == 0) {
vec2.push_back(i);
vec2.push_back(n1 / i);
}
}
// Used to sort the vector
sort(vec2.begin(), vec2.end());
// Used to print all the factors
cout << "All the factors are -> ";
for (int i = 0; i < vec2.size(); i++)
cout << vec2[i] << " ";
cout << endl;
// Now any elements is divisible by 1
int maxProduct2 = 1;
bool flag2 = 1;
// implementing three loop we'll find
// the three maximum factors
for (int i = 0; i < vec2.size(); i++) {
for (int j = i; j < vec2.size(); j++) {
for (int k = j; k < vec2.size(); k++) {
// Now storing the fourth factor in y
int y = n1 - vec2[i] - vec2[j] - vec2[k];
// It has been seen that if the fouth factor become negative
// then break
if (y <= 0)
break;
// Now we will replace more optimum number
// than the previous one
if (n1 % y == 0) {
flag2 = 0;
maxProduct2 = max(vec2[i] * vec2[j] * vec2[k] *y,maxProduct2);
}
}
}
}
// Used to print the product if the numbers exist
if (flag2 == 0)
cout << "Product is -> " << maxProduct2 << endl;
else
cout << "Not possible" << endl;
}
// Driver code
int main(){
int n1;
n1 = 80;
findfactors2(n1);
return 0;
}

## Output

All the factors are -> 1 2 4 5 8 10 16 20 40 80
Product is -> 160000