# Find the largest multiple of 2, 3 and 5 in C++

In this problem, we are given an array arr[] of size N consisting of single digits only. Our task is to find the largest multiple of 2, 3 and 5.

Let's take an example to understand the problem,

Input : arr[] = {1, 0, 5, 2}
Output : 510

Explanation

The number 510 is divisible by all 2, 3, 5.


## Solution Approach

A simple solution to the problem is by checking for basic divisibility of the number created.

So, if the number needs to be divisible by 2 and 5 i.e. it is divisible by 10. For creating a number divisible by 10, the array must have zero.

If it has a zero then, we will create the largest possible number with zero at the end which is divisible by 3.

The method is shown here. Largest Multiple of Three in C++

## Example

Program to illustrate the working of our solution

#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
string largestMultipleOfThree(vector<int>& digits) {
vector<vector<int>> d(3);
int sum = 0;
for (int i = 0; i < digits.size(); i++) {
int x = digits[i];
d[x % 3].push_back(digits[i]);
sum += x;
sum %= 3;
}
if (sum) {
if (!d[sum].size()) {
int rem = 3 - sum;
if (d[rem].size() < 2)
return "";
d[rem].pop_back();
d[rem].pop_back();
}
else {
d[sum].pop_back();
}
}
string ret = "";
for (int i = 0; i < 3; i++) {
for (int j = 0; j < d[i].size(); j++) {
ret += to_string(d[i][j]);
}
}
sort(ret.begin(), ret.end(), greater<int>());
if (ret.size() && ret[0] == '0')
return "0";
return ret;
}
};
int main(){
Solution ob;
vector<int> v = {7, 2, 0, 8};
sort(v.begin(), v.end(), greater<int>());
if(v[v.size() - 1 ] != 0){
cout<<"Not Possible!";
}
else{
cout<<"The largest number is "<<(ob.largestMultipleOfThree(v));
}
}

## Output

The largest number is 870