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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 digit_count is even, then
- if prod is divisible by k, then
- return True
- return False
Let us see the following implementation to get better understanding −
Example Code
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
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