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I need to test whether each number from 1 to 1000 is a multiple of 3 or a multiple of 5. The way I thought I'd do this would be to divide the number by 3, and if the result is an integer then it would be a multiple of 3. Same with 5.

How do I test whether the number is an integer?

here is my current code:

n = 0
s = 0

while (n < 1001):
    x = n/3
    if isinstance(x, (int, long)):
        print 'Multiple of 3!'
        s = s + n
    if False:
        y = n/5
        if isinstance(y, (int, long)):
            s = s + n

    print 'Number: '
    print n
    print 'Sum:'
    print s
    n = n + 1
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In Python 2.x, integer division always yields an integer. –  Russell Borogove Nov 3 '11 at 22:19
You should be printing "Fizz" and "Buzz" –  wim Nov 3 '11 at 23:21
Project Euler Problem 1 ? –  Ashutosh Dave Jan 30 '13 at 13:56
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3 Answers

up vote 15 down vote accepted

You do this using the modulus operator, %

n % k == 0

evaluates true if and only if n is an exact multiple of k. In elementary maths this is known as the remainder from a division.

In your current approach you perform a division and the result will be either

  • always an integer if you use integer division, or
  • always a float if you use floating point division.

It's just the wrong way to go about testing divisibility.

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Thanks for the reply, can you explain what exactly is going on there please? –  Taimur Nov 3 '11 at 21:39
@Taimur answered updated –  David Heffernan Nov 3 '11 at 21:40
perfect, thanks a lot! –  Taimur Nov 3 '11 at 21:42
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The simplest way is to test whether a number is an integer is int(x) == x. Otherwise, what David Heffernan said.

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public class Check {

    public void check(int x){
        StringBuilder sb = new StringBuilder();
    public static void main(String[] args) {
        Check test=new Check();     
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