#1 - Multiples of 3 and 5
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.
Nothing fancy to do here. We can simply look through each natural number and add it to a rolling sum if it’s a multiple of 3 or 5:
# file: "problem001.py"
s = 0
for n in range(1, 1000):
if n % 3 == 0 or n % 5 == 0:
s += n
print(n)
Running the code results in
233168
0.00019780000000002573 seconds.
Therefore, 233168 is the answer to the first problem.
Bonus
We can solve this problem completely analytically. For example, $\sum_{i=1}^{333}3i$ will get us all multiples of 3. We can do the same summation for 5. However, we have to be a bit careful. Adding these two summations will lead us to double count all numbers that are both multiples of 3 and 5 i.e. 15. Thus, we need to subtract these multiples. The full sum is therefore: \[\begin{aligned} S &= \sum_{i=1}^{333}3i + \sum_{i=1}^{199}5i - \sum_{i=1}^{66}15i \\ &= 3\left(\frac{333(334)}{2}\right) + 5\left(\frac{199(200)}{2}\right)-15\left(\frac{66(67)}{2}\right) \\ &= 3(55611) + 5(19900) - 15(2211) \\ &= \boxed{233168} \end{aligned}\]
