Question

Why does this not find the sum of multiples of 3 and 5 that go up to 1000?
sum, three_multiple, five_multiple = 0, 1, 1
while True:
if 3 * three_multiple >= 1000:
break
sum += 3 * three_multiple
three_multiple += 1
while True:
if 5 * five_multiple >= 1000:
break
sum += 5 * five_multiple
five_multiple += 1
print sum

Answer 1

You're counting every multiple of 15 (3*5) twice.

Answer 2

Thank you!

Answer 3

This gets the correct answer, but I don't know why I should add the 1 in the for loop.
sum, three_multiple, five_multiple = 0, 1, 1
for three_multiple in range(1, 1000/3 + 1): #decimal is truncated
sum += 3 * three_multiple
for five_multiple in range(1, 1000/5):
if (5 * five_multiple) % 3 != 0:
sum += 5 * five_multiple
print sum
print 233168

Answer 4

1000 / 3 = 333, and the range function goes up to *but not including* the second number. So since you want to include 333, you actually need that number to be 334.

For the record, you can do this in one loop.

total = 0
for x in range(1,1001):
if x%3==0 or x%5==0:
total += x
print total

Answer 5

Wow, much nicer. Why does it not work when I make it be 201 for my second for loop?

Answer 6

Do you want to include 1000 or not? My code actually includes 1000 as a possibility, yours does not. If you changed it to 201, then it would count 1000 and give 234168 as an answer instead of 233168.

If you don't want to include 1000, then my code should say "for x in range(1,1000)" instead.

Answer 7

IC, silly me.

Answer 8

Thanks; and your solution is really elegant.

Answer 9

Gracias, but for beginning comp sci students, you should feel proud that you got something up and running that works. "Short and sweet" comes later, and Python is full of short and sweet functions.

Answer 10

Like this one: print sum( [ i for i in range( 1000 ) if not i % 3 or not i % 5 ] )

Answer 11

This problem is from Project Euler by the way.