# Project Euler Problem 1

This is my python solution to the first problem on Project Euler:

n = 1
rn = 0
while n < 1000:
if n%3 == 0 or n%5 == 0:
rn += n
n = n + 1
print(rn)


I would like to find a way to keep everything in this python code to as little number of lines as possible (maybe even a one liner??), and possibly improve the speed (it's currently around 12 ms). By the way, this is 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.

Suggestions?
Thanks.

• sum(n for n in range(1000) if not n%3 or not n%5) – dansalmo Dec 29 '13 at 21:28
• I like your solution to the problem. It is much more efficiently coded than another Project Euler #1 code question I just read. – AdmiralAdama Apr 4 '16 at 8:50

Python hint number 1:

The pythonic way to do :

n = 1
while n < 1000:
# something using n
n = n + 1


is :

for n in range(1,1000):
# something using n


Python hint number 2:

You could make your code a one-liner by using list comprehension/generators :

print sum(n for n in range(1,1000) if (n%3==0 or n%5==0))


Your code works fine but if instead of 1000, it was a much bigger number, the computation would take much longer. A bit of math would make this more more efficient.

Math hint number 1 :

The sum of all the multiples of 3 or 5 below 1000 is really the sum of (the sum of all the multiples of 3 below 1000) plus (the sum of all the multiples of 5 below 1000) minus the numbers you've counted twice.

Math hint number 2 :

The number you've counted twice are the multiple of 15.

Math hint number 3 :

The sum of the multiple of 3 (or 5 or 15) below 1000 is the sum of an arithmetic progression.

• Oh, sorry, my mistake, I inputted 100 instead of 1000 @Josay – LazySloth13 Mar 3 '13 at 13:02
• Math hint #4: every number that is divideable by an odd number is an odd number itself (so you can skip half of the loop iterations). @Lewis – 11684 Mar 3 '13 at 17:04