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Modified Product = Sum (Posted on 2010-05-24) Difficulty: 3 of 5
Determine all possible quintuplet(s) (A,B,C,D,E) of positive integers, with A ≤ B ≤ C ≤ D ≤ E, that satisfy this equation:

(A-1)*(B-2)*(C-3)*(D-4)*(E-5) = A+B+C+D+E

Prove that these are the only quintuplet(s) that exist.

No Solution Yet Submitted by K Sengupta    
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Solution computer solution + proof | Comment 1 of 2

FOR a = 1 TO 25
FOR b = a TO 26
FOR c = b TO 27
FOR d = c TO 28
FOR e = d TO 29
  lhs = (a - 1) * (b - 2) * (c - 3) * (d - 4) * (e - 5)
  rhs = a + b + c + d + e
  IF lhs = rhs THEN PRINT a; b; c; d; e, lhs; rhs
NEXT
NEXT
NEXT
NEXT
NEXT

produces

 A  B  C  D   E             LHS  RHS    
 2  3  4  6  25              40  40
 2  3  5  5  25              40  40
 2  3  5  6  12              28  28
 2  4  4  5  25              40  40
 2  4  4  6  12              28  28
 2  4  5  5  12              28  28
 3  3  4  5  25              40  40
 3  3  4  6  12              28  28
 3  3  5  5  12              28  28
 3  4  4  5  12              28  28
 
In (A-1)*(B-2)*(C-3)*(D-4)*(E-5) = A+B+C+D+E
 
The RHS <= 5*E.

If D > 20 and therefore E > 20, LHS > 16*(E-5), so

LHS > 16*E - 80 = 5*E + 11*E - 80 > 5*E + 220 - 80 = 5*E + 140 > RHS

So the cases covered by the program exhaust the range over which the LHS could equal the RHS.

The limits and argument could be made tighter, but the numbers chosen are sufficient for the proof.


  Posted by Charlie on 2010-05-24 13:44:42
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