 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  My favorite numbers II (Posted on 2009-12-21) Determine all possible sextuplets (A, B, C, D, E, F) of positive integers, with A ≤ B ≤ C, and, D ≤ E ≤ F and, A ≤ D, that satisfy both the equations: A+B+C = D*E*F and, A*B*C = D+E+F.

Prove that these are the only sextuplets that exist.

 No Solution Yet Submitted by K Sengupta Rating: 4.0000 (1 votes) Comments: ( Back to comment list | You must be logged in to post comments.) An attempt at a proof... | Comment 3 of 4 | The difference between the product and sum increases as a multiplicand-addend is increased. An increase in a multiplicand-addend increases both the sum and the product, but the product increases by a multiple.
Where S is the sum and P is the product, the difference, S - P, is 2 where the first two multiplicands-addends are each the lowest positive integer, 1. Therefore, in order for S1=P2 and S2=P1, the difference, P - S, can not be greater than 2 and the sum and product of the first two multiplicands-addends can not be greater than 4. The first two multiplicands-addends can then be of the following: 1 and 1, 1 and 2, 1 and 3, or 2 and 2.
Iterating through the small number of positive integers from 1 to n for the third multiplicand until the difference, P - S, is greater than 2 for each of the first two multiplicands-addends results in the six possible sextuplets:

(1,1,6,1,2,3)
(1,1,7,1,3,3)
(1,1,8,1,2,5)
(1,2,3,1,2,3)
(1,2,5,1,1,8)
(1,3,3,1,1,7)

 Posted by Dej Mar on 2009-12-22 19:28:57 Please log in:
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