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    <p>In your script, regardless of if the number is divisble by 2 or not, it breaks the loop immediately. I've reindented the code and this is probably closer to what you were trying to do.</p> <p>In your original code, if the number <em>is divisible by 2</em> (first number in the <code>range(2,number)</code>, then you break the loop and if it <em>is not divisible</em> you also break the loop. So all odd numbers, like 9, looked like primes.</p> <p>The <code>else</code> keyword after a <code>for</code> loop is run <em>iff</em> the loop exits normally. So the "is prime" part will only be printed if no divisor is found.</p> <pre><code>for number in range(0,1000): for x in range(2,number): if(number % x == 0): print number,"divisible by",x break else: print number, "is prime" </code></pre> <p>You can see this is anction here: <a href="http://codepad.org/XdS413LR" rel="nofollow">http://codepad.org/XdS413LR</a></p> <p>Also, this is a naive algorithm (not a critique of the code, exploring simple algorithms is a useful study), but you can make a little more efficient. Technically you only need to check as far as the sqare root of <code>number</code>, as any number larger than the square root must have a complement that is less than the square root, which should have already been encountered. So the logic in the code can be changed to:</p> <pre><code>from math import sqrt for number in range(0,1000): for x in range(2,int(sqrt(number/2))): # Rest of code as above. </code></pre> <p>That said there are many ways that you can optimise the checking or discovery of prime numbers that are worth investigating if you get the chance.</p>
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    1. CO+1 for appropriate use of for/else
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    2. CO@Brian Answer early, and edit often!
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    3. COThanks for the quick answer, and also for the very thorough explanation of what exactly was wrong with the old code. Also for a way to make the code more efficient.
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