Check whether product of digits at even places is divisible by sum of digits at odd place of a numbers in Python

Suppose we have a number n, we have to check whether the product of digits at even places of n is divisible by sum of digits at odd place of n or not. Places are started counting from right to left. Right most is at place 1.

So, if the input is like n = 59361, then the output will be True as (1*3*5) = (6+9).

To solve this, we will follow these steps −

digit_count := digit count of given number n
total := 0, prod := 1
while n > 0, do
- if digit_count is even, then
  - prod := prod * last digit of n
- otherwise,
  - total := total + last digit of n
- n := quotient of (n / 10)
- digit_count := digit_count - 1
if prod is divisible by total, then
- return True
return False

Let us see the following implementation to get better understanding −

Example Code

from math import log10

def solve(n):
   digit_count = int(log10(n))+1
   total = 0
   prod = 1
   while n > 0 :
      if digit_count % 2 == 0 :
         prod *= n % 10
      else:
         total += n % 10
        
      n = n // 10
      digit_count -= 1
 
   if prod % total == 0:
      return True
   return False
   
n = 59361
print(solve(n))

Input

Output

True