# Check whether product of digits at even places of a number is divisible by K in Python

Suppose we have a number n, and another number k, we have to check whether the product of digits at even places of n is divisible by k 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) is divisible by 3.

To solve this, we will follow these steps −

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

Let us see the following implementation to get better understanding −

## Example Code

Live Demo

from math import log10

def solve(n, k):
digit_count = int(log10(n))+1
prod = 1
while n > 0 :
if digit_count % 2 == 0 :
prod *= n % 10

n = n // 10
digit_count -= 1

if prod % k == 0:
return True
return False

n = 59361
k = 3
print(solve(n, k))

## Input

59361, 3


## Output

True