DDSA Solutions

3622. Check Divisibility by Digit Sum and Product

EasyView on LeetCode
Math

Problem Overview

You only need two numbers from the digits of n: their sum and their product.

Intuition

You only need two numbers from the digits of n: their sum and their product. Add those together and ask whether n is a multiple of that total. Walking digits with modulo 10 is enough - no extra data structure.

Algorithm

  1. 1Copy n. Start sum = 0 and product = 1.
  2. 2While the copy is non-zero: take the last digit (copy % 10), add it to the sum, multiply it into the product, then drop it (copy / 10).
  3. 3Let total = sum + product.
  4. 4Return whether n % total == 0.

Example Walkthrough

Input: n = 99

  1. 1.Digits 9 and 9: sum = 18, product = 81, total = 99.
  2. 2.99 % 99 == 0, so the answer is true.

Output: true

Common Pitfalls

  • •A zero digit zeros the product. The divisor then equals the digit sum, which is still valid.
  • •Start the product at 1, not 0, or every answer becomes n % sum.
  • •Do not mutate n while peeling digits if you still need the original for the modulo check.
  • •Constraints keep n small, so digit product will not overflow a 32-bit int.
3622.cs
C#
// Approach: Peel digits with n % 10. Accumulate sum and product, then test
// whether n is divisible by (sum + product). A zero digit makes the product
// 0, which is fine: the divisor becomes the digit sum alone.
// Complexity: O(log n) time and O(1) space.
public class Solution
{
    public bool CheckDivisibility(int n)
    {
        int digitSum = 0;
        int digitProduct = 1;
        int number = n;

        while (number != 0)
        {
            int currentDigit = number % 10;
            number /= 10;
            digitSum += currentDigit;
            digitProduct *= currentDigit;
        }

        int total = digitSum + digitProduct;
        return n % total == 0;
    }
}
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