The problem:

If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23. Find the sum of all the multiples of 3 or 5 below 1000.

Solutions:

Solve this problem using the computer way is easy. Just loop through 1000 numbers, find numbers that are either multiples of 3 or multiples of 5 then add them up.

for i in range(0, 1000):
    if i % 3 == 0 or i % 5 == 0:
        total += i
 
#result: 233168
print(total)
 
# or much shorter statement
total = sum([i for i in range(1000) if (i % 3 == 0 or i % 5 == 0)])

However this isn't a efficient way to do it as the loop runs 1000 times in this case. Although it still runs very fast in any computer, there is a better way to do it: the mathematical way.

Notice the sequence of all numbers divisible by 3 from 1 to 1000 looks like this

total = 3 + 6 + 9 + ... + 999
      = 1*3 + 2*3 + 3*3 + ... 333*3
      = 3 * (1 + 2 + 3 + ... + 333) 

The sequence for 5 is quite similar

total = 5 * (1 + 2 + 3 + ... + 199)

So how the sequence 1 + 2 + 3 + ... + n is calculated. The formula is quite simple:

total = (first_numer + last_number) * length_of_sequence / 2

Here is the solution written in Python:

# Sum of sequence like 1 2 3 4 5 ... n
def sum_of_sequence(start=1, end=1000):
    return int((start + end) * (end - start + 1) / 2)
 
total = 3 * sum_of_sequence(1, 333) + 5 * sum_of_sequence(1, 199)
 
# The reason for this part is when summing up the sequence of numbers divisible by 3 and 5
# the numbers that is divisible by both 3 and 5 is calculated twice 
total = total - 15 * sum_of_sequence(1, 66)

I'm gonna make this code a bit better. Continued from the above code:

def sum_of_sequence_divisible_by(x, start=1, end=1000):
    # y is smallest multiply of x
    # if the start number of the sequence less than x
    # then the smallest multiply of x in the sequence is x
    y = x
    if start > x:
        # if start number of the sequence greater than x
        # find the nearest multiply of x
        for i in range(start, end):
            if i % x == 0:
                y = i
                break
            
    return x * sum_of_sequence(y / x, int(end / x))
 
total = sum_of_sequence_divisible_by(3) + sum_of_sequence_divisible_by(5)
total = total - sum_of_sequence_divisible_by(15)
 
print(total)