Problem:
n! means n × (n − 1) × ... × 3 × 2 × 1 For example, 10! = 10 × 9 × ... × 3 × 2 × 1 = 3628800, and the sum of the digits in the number 10! is 3 + 6 + 2 + 8 + 8 + 0 + 0 = 27. Find the sum of the digits in the number 100!
n! means n × (n − 1) × ... × 3 × 2 × 1 For example, 10! = 10 × 9 × ... × 3 × 2 × 1 = 3628800, and the sum of the digits in the number 10! is 3 + 6 + 2 + 8 + 8 + 0 + 0 = 27. Find the sum of the digits in the number 100!
My Solution:
Note: You can simplifies the coding :)
- static void Main(string[] args)
- {
- BigInteger multi = 1;
- long result = 0;
- int digits = 100;
- for (int i = digits; i >= 1; i--)
- {
- multi *= i;
- }
- foreach (var item in multi.ToString().ToCharArray())
- {
- result += long.Parse(item.ToString());
- }
- Console.WriteLine(result);
- Console.ReadLine();
- }
Note: You can simplifies the coding :)
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