How to find the sum of all the multiples of 3 or 5 below 1000 in Python?

Question

Not sure if I should\'ve posted this on math.stackexchange instead, but it includes more programming so I posted it here.

The question seems really simple, but I\'ve

Answer 1

You are overcomplicating things. You just need a list of numbers that are multiples of 3 or 5 which you can get easily with a list comprehension:

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

Then use sum to get the total:

>>> sum([i for i in range(1000) if i % 3 == 0 or i % 5 == 0])
<<< 233168

Or even better use a generator expression instead:

>>> sum(i for i in range(1000) if i % 3 == 0 or i % 5 == 0)

Or even better better (courtesy Exelian):

>>> sum(set(list(range(0, 1000, 3)) + list(range(0, 1000, 5))))

Answer 2

here is my solution:

for n in range(100):
    if n % 5==0:
        if n % 3==0:
        print n, "Multiple of both 3 and 5"    #if the number is a multiple of 5, is it a multiple of 3? if yes it has has both.

    elif n % 5==0:
        print n, "Multiple of 5"

    elif n % 3==0:
        print n, "Multiple of 3"

    else:
        print n  "No multiples"

Answer 3

Here is the code:

count = 1000
m = [3, 5, 3*5]
result = 0
Sum = 0
for j in m:
    result = 0
    for i in range(count):
        if i*j < 1000:
            result = result + i*j
        elif i == (count - 1):
            if j < 15:
                Sum = result + Sum
            elif j == 15:
                Sum = Sum - result
                print(Sum)

Answer 4

t = int(input())
for a in range(t):
n = int(input().strip())
sum=0
for i in range(0,n):
       if i%3==0 or i%5==0:
            sum=sum+i
print(sum)

Answer 5

You can also use functional programming tools (filter):

def f(x):
    return x % 3 == 0 or x % 5 == 0
    filter(f, range(1,1000)) 
print(x)

Or use two lists with subtraction of multiples of 15 (which appears in both lists):

sum1 = []
for i in range(0,1000,3):
    sum1.append(i)
sum2 = []
for n in range(0,1000,5):
    sum2.append(n)
del sum2[::3] #delete every 3-rd element in list
print(sum((sum1)+(sum2)))

I like this solution but I guess it needs some improvements...

Answer 6

I know it was 7 years ago but I wanna share my solution to this problem.

x= set()
for i in range(1,1001):
    if (i % 3) == 0:
        x.add(i)

for j in range(1,1001):
    if (j % 5) == 0:
        x.add(j)

print(sum(x))