Check if product of digits of a number at even and odd places is equal in Python

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Suppose we have a number n. We have to check whether the product of odd placed digits and the even placed digits are same or not.

So, if the input is like n = 2364, then the output will be True as product of odd placed numbers are 2 * 6 = 12 and product of even placed numbers are 3 * 4 = 12 which are same.

To solve this, we will follow these steps −

• if num < 10, then
  • return False
• odd_place := 1, even_place := 1
• while num > 0, do
  • d := last digit of num
  • odd_place := odd_place * d
  • num := quotient of (num/10)
  • if num is same as 0, then
    • break
  • d := last digit of num
  • even_place := even_place * d
  • num := quotient of (num/10)
• if odd_place is same as even_place, then
  • return True
• return False

Example

Let us see the following implementation to get better understanding −

 Live Demo

def solve(num):
   if num < 10:
      return False
   odd_place = 1
   even_place = 1
   while num > 0:
      d = num % 10
      odd_place *= d
      num = num//10
      if num == 0:
         break
      d = num % 10
      even_place *= d
      num = num//10
   if odd_place == even_place:
      return True
   return False
num = 2364
print(solve(num))

Input

2364

Output

True

Arnab Chakraborty

Updated on 19-Jan-2021 05:33:59

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