Check whether sum of digits at odd places of a number is divisible by K in Python

In this problem, we need to check whether the sum of digits at odd positions (counting from right to left) of a number is divisible by a given value k. The positions are counted starting from 1.

So, if the input is like n = 2416 and k = 5, then the output will be True. The digits at odd positions from right to left are: position 1 (digit 6) and position 3 (digit 4). Their sum is 4 + 6 = 10, which is divisible by 5.

Algorithm

To solve this problem, we will follow these steps ?

• Initialize total = 0 and pos = 1
• While n > 0, do:
  • If pos is odd, then add the last digit (n mod 10) to total
  • Remove the last digit: n = n // 10
  • Increment pos by 1
• If total is divisible by k, return True, otherwise return False

Example

Let us see the following implementation to get better understanding ?

def solve(n, k):
    total = 0
    pos = 1
    while n > 0:
        if pos % 2 == 1:  # Check if position is odd
            total += n % 10  # Add the last digit
        n = n // 10  # Remove the last digit
        pos += 1  # Move to next position
        
    if total % k == 0:
        return True
    return False

n = 2416
k = 5
print(solve(n, k))

True

Step-by-Step Execution

Let's trace through the example with n = 2416 and k = 5:

def solve_with_trace(n, k):
    total = 0
    pos = 1
    original_n = n
    
    print(f"Number: {original_n}, K: {k}")
    print("Position | Digit | Add to sum?")
    print("---------|-------|------------")
    
    while n > 0:
        digit = n % 10
        if pos % 2 == 1:  # Odd position
            total += digit
            print(f"    {pos}    |   {digit}   |    Yes")
        else:
            print(f"    {pos}    |   {digit}   |    No")
        n = n // 10
        pos += 1
    
    print(f"\nSum of digits at odd positions: {total}")
    print(f"Is {total} divisible by {k}? {total % k == 0}")
    return total % k == 0

solve_with_trace(2416, 5)

Number: 2416, K: 5
Position | Digit | Add to sum?
---------|-------|------------
    1    |   6   |    Yes
    2    |   1   |    No
    3    |   4   |    Yes
    4    |   2   |    No

Sum of digits at odd positions: 10
Is 10 divisible by 5? True

Alternative Approach Using String

We can also solve this problem by converting the number to a string and accessing digits by index ?

def solve_string_approach(n, k):
    num_str = str(n)
    total = 0
    
    # Iterate from right to left (reverse the string)
    for i, digit in enumerate(reversed(num_str)):
        if (i + 1) % 2 == 1:  # Odd position (1-indexed)
            total += int(digit)
    
    return total % k == 0

# Test with the same example
n = 2416
k = 5
result = solve_string_approach(n, k)
print(f"Result using string approach: {result}")

Result using string approach: True

Conclusion

Both approaches effectively solve the problem by summing digits at odd positions from right to left. The mathematical approach processes digits by repeatedly dividing by 10, while the string approach reverses the number and uses indexing for position checking.