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%I #12 Jul 07 2018 19:22:51
%S 124,142,214,241,412,421,1222,2122,2212,2221
%N Numbers such that sum of digits is 7 and product of digits is 8.
%C Sequence is finite.
%C There are no other terms. Proof. From the fact that the product of the digits is 8, we can conclude that each number contains either three '2's or a '2' and a '4' or a single '8' together with arbitrarily many '1's. In the first case there has to be exactly one '1', otherwise the sum wouldn't be 7. This gives the numbers 1222, 2122, 2212, 2221. In the second case, as well, there can be only one '1', this yields the numbers 124, 142, 214, 241, 412, 421. The third case is not possible because 8, by itself, is already bigger than 7. All those terms are listed, hence the sequence is complete. - _Stefan Steinerberger_, Jun 07 2007
%H N. J. A. Sloane et al., <a href="https://oeis.org/wiki/Binary_Quadratic_Forms_and_OEIS">Binary Quadratic Forms and OEIS</a> (Index to related sequences, programs, references)
%K nonn,base,fini,full
%O 1,1
%A _Zak Seidov_, May 13 2005