Let abc be a 3-digit number.
Find it, if acb + bca + bac + cba + cab = 3961.
(In reply to
re(2): 3 digit number by bernie)
I completely agree with your solution, except that you don't have to
resort to trial and error after realizing that 3961 = 222*(a+b+c) -
abc.
If you consider the original equation mod 9, you know that the sum of
the digits on the left must equal the sum of the digits on the right.
That gives 5 * (a+b+c) = 1 mod 9 or (a+b+c) = 2 mod 9.
With that information, you can immediately go to a+b+c=20 as the only
solution-- the next smallest, a+b+c=11 can never be big enough, and the
next larges, a+b+c = 29 can never occur with a, b, and c being single
digits.
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Posted by Paul
on 2005-10-17 23:05:36 |
re(3): 3 digit number
