Determine all possible sextuplet(s) (A, B, C, D, E, F), with B < C, E < F and, A < D, that satisfy this system of equations:
A/(B*C) = D - E - F, and:
D/(E*F) = A - B - C
Prove that these are the only sextuplet(s) that exist.
(In reply to
One Solution by Steve Herman)
OK, Now I am claiming that the only solution is (4,1,2,8,2,4).
a) If A/(B*C) = D/(E*F), then it follows that
A/(B*C) = A - B - C and that D/(E*F) = D - E - F
A/(B*C) = (A - B - C) for only three values of (A,B,C):
(6,1,2), (6,1,3), and (6,2,3).
Since all of these start with 6 and A cannot be equal to D, it cannot be the case that one if these is (A,B,C) and the other is (D,E,F).
Therefore, A/(B*C) is not equal to D/(E*F)
b) Since A/(B*C) is not equal to D/(E*F), one must be less than the other.
Assume, for the moment, that A/(B*C) > D/(E*F).
Then A/(B*C) > A - B - C. There are not many triplets for which this is true; as B and C get larger, the left hand side (lhs) gets smaller faster than the rhs. And if A > 4, then for (B,C) = (1,2) we already have the case that A/(B+C) > (A - B - C).
In fact, A/(B*C) > A - B - C only for (A,B,C) = (4,1,2).
Then, D/(E*F) = A -B - C = 1, and D - E - F = A/(B*C) = 2
This is only true for (D,E,F) = (8,2,4).
So, the only solution is (4,1,2,8,2,4).
c) And yes, an algebraic proof would be better, but I am not there yet. Going back to real work.
Edited on November 23, 2010, 12:25 am
re: One Solution (spoiler)
