[Guest]

Determine all possible triplet(s) (a, b, c) of positive integers, with a <= b <= c, that satisfy this equation:

a*b*c – (a+b+c) = 2

[Guest]

1*2*6-(2+3+5)=2

[Guest]

1*2*6-(2+3+5)=2

Nice try.

But the assignment of the triplets should be uniform throughout the entire equation.

For example, if (a, b, c) = (1, 2, 6), then: a*b*c – (a+b+c) = 1*2*6 – (1+2+6) = 3, which does not satisfy the given equation.

Also, if (a, b, c) = (2, 3, 5), then: a*b*c – (a+b+c) = 2*3*5 – (2+3+5) = 20, which does not satisfy the given equation.

This problem has more than one answer.

Keep trying, and I am confident that you will be able to derive them.

(a,b,c) = (1,2,6) is very very close to one of the triplets.

[Guest]

(1, 2, 5), (1, 3, 3) & (2, 2, 2)

Can't think of any others... and Merry Christmas to everyone celebrating it now!

[Guest]

=W=,

You are right. By method of elimination these are the only possible solutions. Merry Chrismas.

[Guest]

Who needs eliminations? The problem is solved very simply from elementary analysis.

Let a, b, c be positive integers as described in the OP.

Then a (bc 1) = b + c +2, so that 1 <= (b + c + 2) / (bc 1).

From this, c <= (b+3) / (b-1).

b cannot be 4 or greater since it would then follow that b would be larger than c. Also b cannot be 1 since then one would find that 0 = 4.

Hence either b is 2 or 3.

If b = 2, then c lies between 2 and 5, and knowing that b <= (c+3) / (c-1) also,

(a, b, c) = (1, 2, 5) or (2, 2, 2)

while b = 3 yields (1, 3, 3).

These must be the only solutions.

Edited December 30, 2009 by jerbil