### Summary

Given an integer, sum all of its digits.

### Example

Let's take, for example, the integer 10023.

Its digits are 1, 0, 0, 2, and 3.

Summing them up yields 1 + 0 + 0 + 2 + 3 = 6.

Code
Diff
• ``````using System;
using System.Linq;

namespace Kumite
{
public class Problem
{
public static int SumDigitsOf(long integer)
{
return Math.Abs(integer).ToString().Sum(digit => digit - '0');
}
}
}``````
• def digit_sum(number: int) -> int:
• return(sum([int(num) for num in str(abs(number))]))
• using System;
• using System.Linq;
• namespace Kumite
• {
• public class Problem
• {
• public static int SumDigitsOf(long integer)
• {
• return Math.Abs(integer).ToString().Sum(digit => digit - '0');
• }
• }
• }
Failed Tests

Algorithms
Logic
Algebra
Mathematics

### Summary

Given an integer number - let's call it N -, calculate the sum of the first positive integers up to N (inclusive) that are divisible by 3 or 7.

### Example

Let's take N = 30:

Sum of integers divisible by 3 : 3 + 6 + 9 + 12 + 15 + 18 + 21 + 24 + 27 + 30 + 33 = 165 (#1)

Sum of integers divisible by 7 : 7 + 14 + 21 + 28 + 35 = 70 (#2)

Sum of integers divisible by 3 and 7 : 21 + 42 = 21 (#3)

Total : #1 + #2 - #3 = 165 + 70 - 21 = 214

### Attention

Please be wary that N may be a large number (N larger than 1,000,000). This requires brushing up on optimization.

### Caveat

There is one way of resolving this problem in O(1) complexity. For further details, please refer to this wiki.

``````using System;

namespace Kumite
{
public class Problem
{
public static long Sum(int N)
{