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asked 2014-10-18 10:32:22 -0500

anonymous user


updated 2014-10-18 10:46:29 -0500

To find numbers of the form 7aaa that are multiples of 7, I use:

sage:print [a for a in [0..9] if (7000+111*a)%7==0]
[0, 7]

I do not know how to write for 7|7abc.


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I reedit your question. Please double check that it is all right. It was not very clear.

vdelecroix gravatar imagevdelecroix ( 2014-10-18 10:45:49 -0500 )edit

Thank you for edit.

masac gravatar imagemasac ( 2014-10-18 10:48:27 -0500 )edit

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answered 2014-10-18 11:09:42 -0500

masac gravatar image

updated 2014-10-18 11:17:50 -0500

I tried:

sage: a=range(7000,8000)
sage: print[a if a%7==0]
  File "<ipython-input-30-590163f2a51e>", line 1
print[a if a%Integer(7)==Integer(0)]
 SyntaxError: invalid syntax

I find a solution:

print [a for a in [7000..7999] if a%7==0]

Thank you for support.

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answered 2014-10-18 10:51:02 -0500


There are various solutions. To generate the list of numbers of the form 7abc, you can use one of:

sage: numbers0 = range(7000,8000)
sage: numbers1 = [7000 + x for x in range(1000)]
sage: numbers2 = [7000 + a*100 + b*10 + c for a in range(10) for b in range(10) for c in range(10)]

Then you can check divisibility the very same way you did. But note that there is a faster way since x is a multiple of 7 if and only if x+7 is a multiple of 7. So once you find one, you only need to go 7 by 7.

Is that ok ?


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Asked: 2014-10-18 10:32:22 -0500

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Last updated: Oct 18 '14

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