I am getting a problem with the pattern. I am assuming that the pattern appears after the 1st 3 digits, as is the case in all given test cases. And the pattern is always made up of 4 digits. With the digits being 2,4,6,8. Now the problem is, I was using only the first 3 digits, the pattern and the extra numbers to check divisibility.

Example, for the test case,**(13 8 1)**

The first 3 digits are 819, the pattern is 8624, and the extra digits are 86. So the final number becomes **8198624862486**.

Now, with my solution, I was trying to check divisibility by just using the pattern once and then the extra digits. For the above example, I would only check for **819862486**. The 1st 3 digits, the pattern only once and the extra digits. But it is not working. It’s printing NO for all the given test cases. Maybe I am checking for the wrong number.I am pasting the code for reference. There are some print statements that I used for checking. Ignore them.

Please tell me some suggestions. Thank you.

#include<bits/stdc++.h>

using namespace std;`int main() { int t; cin>>t; while(t!=0) { t--; long long int k,d1,d2; cin>>k>>d1>>d2; long long int d3=(d1+d2)%10; long long int sum=d1+d2+d3; long long int first_three=d3+d2*10+d1*100; long long int n=4; vector<long long int> v; while(n!=0) { n--; long long int d_temp=sum%10; v.push_back(d_temp); sum=sum+d_temp; } long long int pattern=v[3]+v[2]*10+v[1]*100+v[0]*1000; long long int final_num=first_three*10000+pattern; //cout<<final_num<<" "; if(k>7) { k=(k-3)%4; //cout<<k<<endl; //cout<<final_num; if(k==1)//1 digit { final_num=final_num*10+v[0]; } else if(k==2)//2 digit { final_num=final_num*100+v[0]*10+v[1]; } else if(k==3)//3 digit { final_num=final_num*1000+v[0]*100+v[1]*10+v[2]; } else if(k==0)//4 digit,no need to append { continue; } cout<<k<<endl<<final_num; if(final_num%3==0) { cout<<"YES"<<endl; } else { cout<<"NO"<<endl; } } else { final_num=final_num/(int(pow(10,7-k))); //cout<<final_num<<endl; if(final_num%3==0) { cout<<"YES"<<endl; } else { cout<<"NO"<<endl; } } } return 0; }`