A challenge from the Mensa Puzzle Calendar
Konrad Koller
koko9991 at compuserve.de
Mon Oct 7 05:31:58 EDT 2002
On 4 Oct 2002 10:15:17 -0700, chris.myers at prov.ingenta.com (Chris
Myers) wrote:
>So, my generalized sol'n of Raymond's answer is:
>for a in range(1000):
> for b in range(100):
> digits = list('%.3d%.2d%.5d' % (a, b, a*b))
> digits.sort()
> if digits == list('0123456789'):
> print '\n%5.3d\n%5.2d\n-----\n%5.5d' % (a, b, a*b)
>
Another small changes in Raymond's excellent program:
import string; alldigits=list(string.digits)
for a in range(12,988):
for b in range(1,99):
digits = list('%.3d%.2d%.5d' % (a, b, a*b))
digits.sort()
if digits == alldigits: print '%3u*%2u=%5.5u' % (a, b, a*b)
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