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My favorite numbers II (Posted on 2009-12-21) Difficulty: 3 of 5
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.

See The Solution Submitted by K Sengupta    
Rating: 4.0000 (1 votes)

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The start of a proof... Comment 4 of 4 |
Lemma 1) 
    if x, y, z are all >= 2, 
           then xyz > x + y+ z 
           
   This follows because
        xyz >= 4x
        xyz >= 4y
        xyz >= 4z
   Average the 3 together, and we get
        xyz >= (4/3)(x + y + z) 
   Therefore xyz > x + y+ z
   
Step 1) 
   If A >= 2,  
   then D >= 2 
     DEF > D+E+F, (from lemma 1)
     D+E+F = ABC
     ABC > A+B+C  (from lemma 1)
     A+B+C = DEF
     
     implies DEF > DEF, 
     
   which is a contradiction,
   so A = 1

  Posted by Steve Herman on 2009-12-23 10:20:36
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