Project Euler #1: Multiples of 3 and 5

total_sum = 0
for i in range(1000):
    if (i%3 == 0 or i%5 == 0):
        total_sum = total_sum+i
print total_sum  

You should leave your numbers, variables and operators some space to breathe by adding some horizontal space between them which improves readability a lot.

total_sum = 0
for i in range(1000):
    if (i % 3 == 0 or i % 5 == 0):
        total_sum = total_sum + i
print total_sum  

As natural numbers aren't explictly defined containing 0 you could also use the two-parameter form of the range() function and specify the start parameter like so

total_sum = 0
for i in range(1, 1000):
    if (i % 3 == 0 or i % 5 == 0):
        total_sum = total_sum + i
print total_sum  

You don't need iteration at all for this problem.

Consider; the sum of all numbers from 1 to n is equal to n*(n+1)/2. Also the sum of all numbers less than n that divides d equals d times the sum of all numbers less than n/d.

So the sum of all numbers less than 1000 that divides 3 is

3*floor(999/3)*(floor(999/3)+1)/2

Likewise the sum of all numbers less than 1000 that divides 5 is

5*floor(999/5)*(floor(999/5)+1)/2

Adding the two numbers would overcount though. Since the numbers that divides both 3 and 5 would get counted twice. The numbers that divides both 3 and 5 is precisely the numbers that divides 3*5/gcd(3,5)=15/1=15.

The sum of all numbers less than 1000 that divides 15 is

15*floor(999/15)*(floor(999/15)+1)/2

So the final result is that the sum of all numbers less than 1000 that divides either 3 or 5 equals:

  3 * (floor(999/3)  *  (floor(999/3)+1))/2
+ 5 * (floor(999/5)  *  (floor(999/5)+1))/2
-15 * (floor(999/15) * (floor(999/15)+1))/2

Define a function to solve more general problems:

def divisible_by_under(limit, divs):
    return (i for i in  range(limit) if any(i % div == 0 for div in divs))

This works for any limit and any divisor and is inside an easy to test function.

print(sum(divisible_by_under(1000, (3, 5))))

You don't need to type this out fully:

total_sum = total_sum+i

Python has a += operator, basically shorthand for what you have above. Take what's on the left of the operator and add the result of what's on the right.

total_sum += i

Also when in Python2.7 it's recommended you use for i in xrange(1000). range will immediately create a full list of numbers it stores in memory, while xrange is a generator that produces each number as it's needed. The performance difference is helpful for large ranges but it's generally a good habit to keep.

You could use list comprehension, to save a few lines, but it does exactly the same as yours:

print(sum([i for i in range(1000) if i%3==0 or i%5==0]))

This code is extremely inefficient. Using some basic math we can reduce runtime to constant time complexity. For any n (in this case 1000), we can predict the number of numbers < n and divisible by 3 or 5:

• numbers divisible by 3: lowerbound(n / 3)
• numbers divisible by 5: lowerbound(n / 5)

The sum of all numbers divisible by 3 or 5 can then be predicted using eulers formula:
the sum of all numbers from 1 to n is n(n + 1)/2. Thus the sum of all numbers n divisible by 3 is:

int div_3 = (n / 3)
int sum_div_3 = div_3 * (div_3 + 1) / 2 * 3

Now there's only one point left: all numbers that are divisible by 3 and 5 appear twice in the sum (in the sum of all numbers divisible by 3 and the sum of all numbers divisble by 5). Since 3 and 5 are prim, all numbers that are divisible by 3 and 5 are multiples of 15.

int sum_div3_5(int n)
    int div_3 = (n - 1) / 3 , 
        div_5 = (n - 1) / 5 , 
        div_15 = (n - 1) / 15

    int sum = div_3 * (div_3 + 1) * 3 / 2 + //sum of all numbers divisible by 3
              div_5 * (div_5 + 1) * 5 / 2 - //sum of all numbers divisible by 5
              div_15 * (div_15 + 1) * 15 / 2

    return sum

I can't provide python code though.