Divisibility by 3 where each digit is the sum of all prefix digits modulo 10

Given K, number of digits, and d0 and d1 as the two digits to form the k-sized integer. Task is to check whether the k-sized number formed using d0 and d1 is divisible by 3 or not.
For each i, di is the sum of all preceding (more significant) digits, modulo 10 — more formally, the following formula must hold :
\huge{d_{i} = \sum_{j = 0}^{i - 1}d_{j} (mod 10), 2 \leqslant i < k}

Examples :

Input : K = 5 d0 = 3 d1 = 4
Output : NO
Explanation : 
The whole number N is 34748 (Starting from 
third digit, every digit is some of preceding
digits mod 10). Since 34748 is not divisible 
by 3, answer is NO.

Input : K = 13 d0 = 8 d1 = 1
Output : YES
Explanation :
The whole number N is 8198624862486, 
which is divisible by 3, so the answer
is YES.

K may be very long so generating the entire number, calculating the sum of digits and then checking for multiple of 3 is cumbersome. The key idea behind the solution is that the digits start repeating after some time in a cycle of length 4 and then the sum of digits can be determined in O(1) step.
We know d0 and d1,
So, d2 = (d0 + d1) mod 10
d3 = (d2 + d1 + d0) mod 10 = ((d0 + d1) mod 10 + (d0 + d1)) mod 10 = 2 * (d0 + d1) mod 10
d4 = (d3 + d2 + d1 + d0) mod 10 = 4 * (d0 + d1) mod 10
d5 = (d4 + d3 + d2 + d1 + d0) mod 10 = 8 * (d0 + d1) mod 10
d6 = (d5 +…+ d1 + d0) mod 10 = 16 * (d0 + d1) mod 10 = 6 * (d0 + d1) mod 10
d7 = (d6 +…+ d1 + d0) mod 10 = 32 * (d0 + d1) mod 10 = 2 * (d0 + d1) mod 10



If we keep on getting di, we will see that the resultant is just looping through the same values. We can see that (d0 + d1) contributes 1 time(s) to d2, 2 times to d3, 4 times to d4, 8 times to d5, …, 2^(k – 2) times to dk.

But, since the powers of 2 under modulo 10 cycle from 1, 2, 4, 8, 6, 2, 4. Here, the cycle length is of 4, where d2 is not present in the cycle. Let, S = (2 * (d0 + d1)) mod 10 + (4 * (d0 + d1)) mod 10 + (8 * (d0 + d1)) mod 10 + (6 * (d0 + d1)) mod 10, this is the cycle which repeats.

So, the sum of digits = (d0 + d1 + d2) + S * ( (k – 3) / 4 ) + x. Here, the first 3 terms will be covered by d0, d1, d2 and after that the groups of 4 will be covered by S, but this formula will still has not summed some terms at the end. That is the residue that is noted by x.

Below is the implementation of above approach :

C++

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// CPP code to check divisibility by 3
#include <bits/stdc++.h>
using namespace std;
  
// Function to check the divisibility
string check(long int k, int d0, int d1)
  
    // Cycle
    long int s = (2 * (d0 + d1)) % 10 + 
                 (4 * (d0 + d1)) % 10 + 
                 (8 * (d0 + d1)) % 10 + 
                 (6 * (d0 + d1)) % 10;
      
    // no of residual terms
    int a = (k - 3) % 4;
      
    // sum of residual terms
    int x;
      
    switch(a)
    {
      
        // if no of residue term = 0
        case 0:
      
        x = 0;
        break;
      
        // if no of residue term = 1
        case 1:
      
        x = (2 * (d0 + d1)) % 10;
        break
  
        // if no of residue term = 2
        case 2:
      
        x = (2 * (d0 + d1)) % 10 + 
            (4 * (d0 + d1)) % 10;
        break;
      
        // if no of residue term = 3
        case 3:
      
        x = (2 * (d0 + d1)) % 10 +
            (4 * (d0 + d1)) % 10 + 
            (8 * (d0 + d1)) % 10;
          
        break;
    }
      
    // sum of all digits
    long int sum = d0 + d1 + ((k - 3) / 4) * s + x;
      
    // divisibility check
    if(sum % 3 == 0)
        return "YES";
    return "NO";
}
  
// Driver code
int main()
{
  
    long int k, d0, d1;
      
    k = 13;
    d0 = 8;
    d1 = 1;
      
    cout << check(k, d0, d1) << endl;
          
    k = 5; 
    d0 = 3; 
    d1 = 4;
          
    cout << check(k, d0, d1) << endl;
  
    return 0;
}

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Java

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// Java code to check divisibility by 3
  
import java.util.*; 
import java.io.*;
  
class GFG {
    // Function to check the divisibility
static String check( int k, int d0, int d1)
  
    // Cycle
     int s = (2 * (d0 + d1)) % 10
                (4 * (d0 + d1)) % 10
                (8 * (d0 + d1)) % 10
                (6 * (d0 + d1)) % 10;
      
    // no of residual terms
    int a = (k - 3) % 4;
      
    // sum of residual terms
    int x=0;
      
    switch(a)
    {
      
        // if no of residue term = 0
        case 0:
      
        x = 0;
        break;
      
        // if no of residue term = 1
        case 1:
      
        x = (2 * (d0 + d1)) % 10;
        break
  
        // if no of residue term = 2
        case 2:
      
        x = (2 * (d0 + d1)) % 10
            (4 * (d0 + d1)) % 10;
        break;
      
        // if no of residue term = 3
        case 3:
      
        x = (2 * (d0 + d1)) % 10 +
            (4 * (d0 + d1)) % 10
            (8 * (d0 + d1)) % 10;
          
        break;
    }
      
    // sum of all digits
     int sum = d0 + d1 + (((k - 3) / 4) * s + x );
      
    // divisibility check
    if(sum % 3 == 0)
        return "YES";
    return "NO";
}
  
    //Code driven
      
    public static void main (String[] args) {
          
     int k, d0, d1;
      
    k = 13;
    d0 = 8;
    d1 = 1;
      
    System.out.println (check(k, d0, d1));
          
    k = 5
    d0 = 3
    d1 = 4;
          
        System.out.println (check(k, d0, d1));
          
    }
}

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Python3

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# Python3 code to check divisibility by 3
  
  
# Function to check the divisibility
def check(k, d0, d1):
  
    # Cycle
    s = ((2 * (d0 + d1)) % 10 + 
        (4 * (d0 + d1)) % 10 + 
        (8 * (d0 + d1)) % 10 + 
        (6 * (d0 + d1)) % 10)
  
    # no of residual terms
    a = (k - 3) % 4
  
    # if no of residue term = 0
    if(a == 0):
        x = 0
  
    # if no of residue term = 1
    elif(a == 1):
        x = (2 * (d0 + d1)) % 10
  
    # if no of residue term = 2
    elif(a == 2):
        x = ((2 * (d0 + d1)) % 10 + 
            (4 * (d0 + d1)) % 10)
  
    # if no of residue term = 3
    elif(a == 3):
        x = ((2 * (d0 + d1)) % 10 + 
            (4 * (d0 + d1)) % 10 + 
            (8 * (d0 + d1)) % 10)
  
    # sum of all digits
    sum = d0 + d1 + ((k - 3) // 4) * s + x
  
    # divisibility check
    if(sum % 3 == 0):
        return "YES"
    else:
        return "NO"
  
  
# Driver code
if __name__=='__main__':
    k = 13
    d0 = 8
    d1 = 1
  
    print(check(k, d0, d1))
  
    k = 5
    d0 = 3
    d1 = 4
  
    print(check(k, d0, d1))
  
# This code is contributed by
# Sanjit_Prasad

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C#

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// C# code to check divisibility by 3
using System;
  
class GFG
{
// Function to check the divisibility
static String check(int k, int d0, int d1)
  
    // Cycle
    int s = (2 * (d0 + d1)) % 10 + 
            (4 * (d0 + d1)) % 10 + 
            (8 * (d0 + d1)) % 10 + 
            (6 * (d0 + d1)) % 10;
      
    // no of residual terms
    int a = (k - 3) % 4;
      
    // sum of residual terms
    int x = 0;
      
    switch(a)
    {
      
        // if no of residue term = 0
        case 0:
      
        x = 0;
        break;
      
        // if no of residue term = 1
        case 1:
      
        x = (2 * (d0 + d1)) % 10;
        break
  
        // if no of residue term = 2
        case 2:
      
        x = (2 * (d0 + d1)) % 10 + 
            (4 * (d0 + d1)) % 10;
        break;
      
        // if no of residue term = 3
        case 3:
      
        x = (2 * (d0 + d1)) % 10 +
            (4 * (d0 + d1)) % 10 + 
            (8 * (d0 + d1)) % 10;
          
        break;
    }
      
    // sum of all digits
    int sum = d0 + d1 + (((k - 3) / 4) * s + x );
      
    // divisibility check
    if(sum % 3 == 0)
        return "YES";
    return "NO";
}
  
// Driver Code
static public void Main ()
{
      
    int k, d0, d1;
    k = 13;
    d0 = 8;
    d1 = 1;
      
    Console.WriteLine (check(k, d0, d1));
          
    k = 5; 
    d0 = 3; 
    d1 = 4;
          
    Console.WriteLine(check(k, d0, d1));
}
}
  
// This code is contributed by Sach_Code

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PHP

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<?php
// PHP code to check
// divisibility by 3
  
// Function to check 
// the divisibility
function check($k, $d0,$d1)
  
    // Cycle
    $s = (2 * ($d0 + $d1)) % 10 + 
         (4 * ($d0 + $d1)) % 10 + 
         (8 * ($d0 + $d1)) % 10 + 
         (6 * ($d0 + $d1)) % 10;
      
    // no of residual terms
    $a = ($k - 3) % 4;
      
    // sum of residual
    // terms
    $x;
      
    switch($a)
    {
      
        // if no of residue 
        // term = 0
        case 0:
      
        $x = 0;
        break;
      
        // if no of residue
        // term = 1
        case 1:
      
        $x = (2 * ($d0
                   $d1)) % 10;
        break
  
        // if no of residue 
        // term = 2
        case 2:
      
        $x = (2 * ($d0 + $d1)) % 10 + 
             (4 * ($d0 + $d1)) % 10;
        break;
      
        // if no of 
        // residue term = 3
        case 3:
      
        $x = (2 * ($d0 + $d1)) % 10 +
             (4 * ($d0 + $d1)) % 10 + 
             (8 * ($d0 + $d1)) % 10;
          
        break;
    }
      
    // sum of all digits
    $sum = $d0 + $d1
           (int)(($k - 3) / 
              4) * $s + $x;
      
    // divisibility check
    if($sum % 3 == 0)
        return "YES";
    return "NO";
}
  
// Driver code
$k; $d0; $d1;
  
$k = 13;
$d0 = 8;
$d1 = 1;
  
echo check($k, $d0, $d1), "\n";
      
$k = 5; 
$d0 = 3; 
$d1 = 4;
      
echo check($k, $d0, $d1), "\n";
  
// This code is contributed by ajit
?>

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Output:

YES
NO

Time Complexity : O(1)



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Improved By : jit_t, Sanjit_Prasad, Sach_Code