Amber G. wrote:Okay - here is a simple problem (as easy as 1,2,3

) I posed in a group consisting of mainly Google Engineer's

Find three positive rational numbers such that their sum is 6 and the product is 6.One answer is 1,2,3 as 1+2+3 = 1*2*3 = 6

Any other answer?

(Only fractions - rational numbers which are positives - are allowed.

(Let us see if brfites visiting this forum can find the answer(s) before the other group... computers /books etc are allowed .. googling is discouraged but I can't stop it

)

(Request: Please For this problem (only) _ No invoking great mathematicians

and writing about fantastic theorems ... Just the answer(s) please.. if one answer appears find a different answer - Imagine your audience just know how to add and multiply ordinary fractions! I want to see how many answers we get here in brf ...)

Didn't google.

So this problem is:

abc=6;-------a+b+c=6

For a given a:

b+c=6-a;-------b(6-a-b)=6/a

Working with that last equation:

b^2-(6-a)*b+6/a=0

b=0.5*( (6-a) +/- sqrt( (6-a)^2 - 24/a ) )

For rational b, the discriminant has to be positive:

(6-a)^2 >= 24/a

(6-a)^2 is a parabola with a minimum at a=6; 24/a is a rectangular hyperbola with a discontinuity at a=0. We want all a where the hyperbola is below the parabola.

The hyperbola will be below the parabola for all a<0. There are also three points where the parabola and hyperbola have the same value (since this is a cubic equation). Too lazy to find the exact values of a, but roughly:

Root 1 (call it R1): a is between 0.9 and 1

R2: a is between 3 and 4

R3: a is between 7 and 8

For a<0, the hyperbola is of course below the parabola, but a can't be negative, so this case is out.

For 0<=a<R1, the hyperbola is above the parabola, hence this case is out.

For R1<=a<R2, the hyperbola is below the parabola (which is what we want).

For R2<=a<R3, the hyperbola is above the parabola, hence this case is out.

For R3<=a<Inf, the hyperbola is below the parabola, and a is also positive, but b and c will both turn out negative, hence this case is out.

So any value of a between R1 and R2 will work, except for the condition that a, b, and c all have to be rational. Not sure how to enforce this condition.

For a given a, it is easy to see that b will be one solution to the quadratic, and c will be the other solution - i.e.:

b= 0.5 * ( (6-a) - sqrt( (6-a)^2 - 24/a) )

c= 0.5 * ( (6-a) + sqrt( (6-a)^2 - 24/a) )

A further point - a=1, b=2, c=3 is a solution, but this is technically the same solution as a=2, b=1, c=3, or a=2, b=3, c=1, etc. So a further condition could be enforced, that a<=b<=c. This could modify the range of a (restrict it from the current R1<=a<=R2).