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edited Sep 3, 2016 at 21:02

I used a slightly simpler method, but essentially did the same thing:

total = 0

for n in range(3,1000):
    if n % 3 == 0:
        total += n
    elif n % 5 == 0:
        total += n


print total

The elifelif makes sure that you only count any factor/divisor once.

I used a slightly simpler method, but essentially did the same thing:

total = 0

for n in range(3,1000):
    if n % 3 == 0:
        total += n
    elif n % 5 == 0:
        total += n


print total

The elif makes sure that you only count any factor/divisor once.

I used a slightly simpler method, but essentially did the same thing:

total = 0

for n in range(3,1000):
    if n % 3 == 0:
        total += n
    elif n % 5 == 0:
        total += n


print total

The elif makes sure that you only count any factor/divisor once.

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created Sep 3, 2016 at 19:15

I used a slightly simpler method, but essentially did the same thing:

total = 0

for n in range(3,1000):
    if n % 3 == 0:
        total += n
    elif n % 5 == 0:
        total += n


print total

The elif makes sure that you only count any factor/divisor once.