# Largest number with the given set of N digits that is divisible by 2, 3 and 5 in C++

In this tutorial, we are going to write a program that finds the largest number formed from the array that is divisible by 2, 3, and 5.

Let's see the steps to solve the problem.

• Initialise the array.
• The number must end with 0 and the sum of all the numbers should be divisible by 3 to be divisible by 2, 3, and 5.
• Check for the 0 in the array and print not possible if it's not present in the array.
• Sort the array in descending order.
• Find the remainder for sum % 3.
• If the remainder is not 1, then delete all the digits from the end whose remainder for digit % 3 is equal to the above remainder.
• If there are no digits with the same remainder as above, then subtract 3 from the above remainder and delete the last two digits whose remainder is the same as above.
• Print all the digits from the array.

## Example

Let's see the code.

Live Demo

#include <bits/stdc++.h>
using namespace std;
void findLargestDivibleNumber(int n, vector<int>& v){
int flag = 0;
long long sum = 0;
for (int i = 0; i < n; i++) {
if (v[i] == 0) {
flag = 1;
}
sum += v[i];
}
if (!flag) {
cout << "Not possible" << endl;
}else {
sort(v.begin(), v.end(), greater<int>());
if (v[0] == 0) {
cout << "0" << endl;
}else {
int flag = 0;
int remainder = sum % 3;
if (remainder != 0) {
for (int i = n - 1; i >= 0; i--) {
if (v[i] % 3 == remainder) {
v.erase(v.begin() + i);
flag = 1;
break;
}
}
if (flag == 0) {
remainder = 3 - remainder;
int count = 0;
for (int i = n - 1; i >= 0; i--) {
if (v[i] % 3 == remainder) {
v.erase(v.begin() + i);
count++;
if (count >= 2) {
break;
}
}
}
}
}
if (*v.begin() == 0) {
cout << "0" << endl;
}else {
for (int i : v) {
cout << i;
}
}
}
}
}
int main() {
int n = 9;
vector<int> v{ 4, 5, 0, 3, 2, 4, 5, 6, 7 };
findLargestDivibleNumber(n, v);
return 0;
}

## Output

If you run the above code, then you will get the following result.

765544320

## Conclusion

If you have any queries in the tutorial, mention them in the comment section.