Let abc be a 3-digit number.
Find it, if acb + bca + bac + cba + cab = 3961.
(In reply to
re: 3 digit number by brianjn)
Since all the permutations of the digits are in the sum except abc then the sum is really 222 times the sum of the digits (i.e., a+b+c) less the number abc. So we're looking for a multiple of 222 that is in excess of 3961. That multiple should be the sum of the digits of the excess.
In practicality, first you try 19 as the multiple because it is the first one that gets the product over 3961, then you try 20, and you get 479.
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Posted by bernie
on 2005-10-17 04:36:57 |
re(2): 3 digit number
