Determine all possible triplets (A,B,C) of positive integers with A ≤ B ≤ C that satisfy this system of equations:
A = P(B)+ P(C), B = P(A) + P(C), and C = P(A) + P(B)
Prove that there are no others.
| No Solution Yet | Submitted by K Sengupta |
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re(2): results so far, but no proof in sight
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 | Comment 3 of 7 |  |
(In reply to re: results so far, but no proof in sight by Charlie)
My results were : 5,5,10, found while thinking of a zero digit and
36,36,36 based a A,A,A assumption and solving for a 2-digit case.
1OP+Q=2*P*Q ==> Q=10P/(2P-1 ) ==>P=3 Q=6.
I was almost sure that no other examples exist but decided
to wait for other solvers.
I still do not see how one can prove that there are no other
solutions.
| Posted by Ady TZIDON on 2013-08-04 15:25:36 |
