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# 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

// 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

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