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Thanks for @MartinR for the hint, and @BorisTheSpider to point out the problem with my original answer that didn't subtract.

Thanks for @MartinR for the hint, and @BorisTheSpider to point out the problem with my original answer that didn't subtract.

Use the Math, @Heslacher ...

Instead of iterating over the numbers in the range and performing modulo, you can:

The code becomes:

int sum = 0;
foreach (int multiple in multiples) {
    int count = (target - 1) / multiple;
    sum += multiple * count * (count + 1) / 2;
}
return sum;

... unfortunately ...

As @BorisTheSpider pointed out in a comment, and @Edward demonstrated in his answer, this algorithm results in double-counting the common multiples. So in case of 3 and 5, you would have to subtract multiples of 3 * 5.

Use The Math, @Heslacher ...

... well ok, in the case of one or two arguments, there's a simple math solution. Instead of iterating over the numbers in the range and performing modulo, you could:

With a helper function:

private int SumOfMultiples(int target, int multiple) {
    int count = (target - 1) / multiple;
    return multiple * count * (count + 1) / 2;
}

This gives a straight \$O(1)\$ solution for one multiple. For two multiples a and b, you cannot simply sum them as that would include the multiples of their multiples twice. So after summing, you would have to subtract the multiples of a * b:

return SumOfMultiples(target, a)
    + SumOfMultiples(target, b)
    - SumOfMultiples(target, a * b);

Thanks for @MartinR for the hint, and @BorisTheSpider to point out the problem with my original answer that didn't subtract.

... unfortunately ...

As @BorisTheSpider pointed out in a comment, and @Edward demonstrated in his answer, this algorithm results in double-counting the common multiples. So in case of 3 and 5, you would have to subtract multiples of 3 * 5.

It's easy to handle two parameters like 3 and 5 in this example. But handling arbitrary number of parameters, the math gets a lot trickier.

In conclusion, when multiples.Length <= 2, a fairly simple math solution exists. When multiples.Length > 2 the implementation needs more work, and more math.

Your solution with iteration may be a bit inefficient, but it's simple and it works.