Python3 Program for Count rotations divisible by 8

Namachila October 14, 2025

0 0 2 minutes read

Given a large positive number as string, count all rotations of the given number which are divisible by 8.

Examples:

Input: 8
Output: 1

Input: 40
Output: 1
Rotation: 40 is divisible by 8
          04 is not divisible by 8

Input : 13502
Output : 0
No rotation is divisible by 8

Input : 43262488612
Output : 4

Approach: For large numbers it is difficult to rotate and divide each number by 8. Therefore, ‘divisibility by 8’ property is used which says that a number is divisible by 8 if the last 3 digits of the number is divisible by 8. Here we do not actually rotate the number and check last 8 digits for divisibility, instead we count consecutive sequence of 3 digits (in circular way) which are divisible by 8.

Illustration:

Consider a number 928160
Its rotations are 928160, 092816, 609281, 
160928, 816092, 281609.
Now form consecutive sequence of 3-digits from 
the original number 928160 as mentioned in the 
approach. 
3-digit: (9, 2, 8), (2, 8, 1), (8, 1, 6), 
(1, 6, 0),(6, 0, 9), (0, 9, 2)
We can observe that the 3-digit number formed by 
the these sets, i.e., 928, 281, 816, 160, 609, 092, 
are present in the last 3 digits of some rotation.
Thus, checking divisibility of these 3-digit numbers
gives the required number of rotations.

Python3

# Python3 program to count all 
# rotations divisible by 8
 
# function to count of all 
# rotations divisible by 8
def countRotationsDivBy8(n):
    l = len(n)
    count = 0
 
    # For single digit number
    if (l == 1):
        oneDigit = int(n[0])
        if (oneDigit % 8 == 0):
            return 1
        return 0
 
    # For two-digit numbers 
    # (considering all pairs)
    if (l == 2): 
 
        # first pair
        first = int(n[0]) * 10 + int(n[1])
 
        # second pair
        second = int(n[1]) * 10 + int(n[0])
 
        if (first % 8 == 0):
            count+=1
        if (second % 8 == 0):
            count+=1
        return count
 
    # considering all 
    # three-digit sequences
    threeDigit=0
    for i in range(0,(l - 2)): 
        threeDigit = (int(n[i]) * 100 +
                     int(n[i + 1]) * 10 +
                     int(n[i + 2]))
        if (threeDigit % 8 == 0):
            count+=1
 
    # Considering the number 
    # formed by the last digit
    # and the first two digits
    threeDigit = (int(n[l - 1]) * 100 +
                 int(n[0]) * 10 +
                 int(n[1]))
 
    if (threeDigit % 8 == 0):
        count+=1
 
    # Considering the number 
    # formed by the last two
    # digits and the first digit
    threeDigit = (int(n[l - 2]) * 100 +
                 int(n[l - 1]) * 10 +
                 int(n[0]))
    if (threeDigit % 8 == 0):
        count+=1
 
    # required count 
    # of rotations
    return count
 
 
# Driver Code
if __name__=='__main__':
    n = "43262488612"
    print("Rotations:",countRotationsDivBy8(n))
 
# This code is contributed by mits.

Output:

Rotations: 4

Time Complexity : O(n), where n is the number of digits in input number.
Auxiliary Space: O(1)

Please refer complete article on Count rotations divisible by 8 for more details!

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