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POHow to check for an integer in Python 3.2?
 

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  1. POHow to check for an integer in Python 3.2?
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    <p>I'm trying to write a program where the user inputs a two-digit integer and the output is the second digit printed the amount of times indicated by the first digit. Here is what I have so far:</p> <pre><code>number = input('Type two-digit integer \n') a = int(number)//10 b = int(number)%10 if len(number) != 2: print(number, 'is not a two-digit integer') else: print(a*str(b)) </code></pre> <p>When I test this out it does what I intend it to do as long as someone types in numbers. If someone were to type in, say, 6r, an error message would pop up saying:</p> <p>a = int(number)//10</p> <p>ValueError: invalid literal for int() with base 10: '6r'</p> <p>So I would assume that something would need to be put in the second line of the code to test if the input is actually an integer, how would I do that? Would I be better off rewriting it in a different way? Please keep in mind that I'm taking an intro course to Python and this is a question on a practice midterm I am taking, so in the case that I would have to answer something like this on the real midterm I can't use a lot of complicated processes.</p> <p>This is something I tried that works if someone types something that isn't an integer, but for some reason that I don't know it gives the same message for non-integers to integers and doesn't function as I intend it to:</p> <pre><code>number = input('Type two-digit integer \n') if (isinstance(number, int)) == False: print(number, 'is not a two-digit integer') elif len(number) != 2: print(number, 'is not a two-digit integer') else: a = int(number)//10 b = int(number)%10 print(a*str(b)) </code></pre> <p>Help would be greatly appreciated!</p>
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