A challenging problem about finding common denominators

[kac3pro]

Alice, Bob, and Carol each think of an expression that is a fraction with 1 as a numerator and a constant integer times some power of x as the denominator. The simplest common denominator of Alice's and Bob's expressions is 4x². The simplest common denominator Bob's and Carol's expressions is 12x³.. The simplest common denominator of Alice's and Carol's expressions is 6x³. Find all possible expressions that could be Carol's expression. The way I did it is just brute forcing through all the possible constant terms and then the powers of x that the Carol's expression could be. This way I arrived at the solutions [MATH]C = 1/(3x^3)[/MATH] and [MATH]C = 1/(6x^3)[/MATH]The downside of this method is that you can easily miss some combination especially as this numbers become bigger, so I wonder if there's some more elegant solution to this problem that assures you always find all of the answers. This is the problem from a 6-9 grades algebra textbook, although it is presented as an extra challenging one.

 

[tkhunny]

How would you do it without the 'x's?

 

[kac3pro]

Now I realized that in order for Carol's number's denominator (or C as I'm gonna call it from now on) has to be a divisor of both 6 and 12. That leaves me with 4 options 1,2,3 or 6. Following this process for both A and B I find that they have to be either 1, 2 or 4. Then I try to substitute each of the values of C into the equations and what I find out is that for the values of C=1 and C=2 there's no way for the math to work out given the 3 possible values of A and B. So this way i arrive at the same conclusion that the constant term of C has to be either 3 or 6. I don't know if that is what I was to conclude, but it definitely narrows down the search compared to just trying out random values and then tweaking them so that the math works out. Thanks for your help and if you have any more advice for me I'll be waiting on them here.

 

[tkhunny]

More systematic. Think Prime Factorization

Alice and Bob 4x2. -- 2 2 x x
Bob and Carol 12x3 -- 2 2 3 x x x
Alice and Carol 6x3 -- 2 3 x x x

Assuming no one's denominator is 1:

The ONLY possibilities for Alice and Bob are
(2)(2 x x)
(x)(2 2 x)
(2 2)(x x)
(2 x)(2 x)
(x x)(2 2)
(2 x x)(2)
(2 2 x)(x)


They ALL have denominators that can be expressed as [math]2^{n}3^{m}x^{p}[/math], where n, m, and p are integers.

 

[kac3pro]

Looking at that clearly shows that:
1. C has to include x³, because A cannot.

2. The greatest common factor of 4x² and 6x³ and all of its factors are are all the possible values of A (and for the B and see you can go through the same process). That means that the values of A are factors of 2x² so:
2 x 2x x² 2x²
Plugging this 5 values into the equations gives me yet another time the answer that C has to be either 3x³ or 6x³.

What I still can't figure out is if there's a way to do it without having to try out any of the cases and what the formula you gave at the end [MATH]2^n3^mx^p[/MATH] is useful for (other than that it's indeed well pointed out).

 

[tkhunny]

Perhaps those are not the only possibilities? Were we expecting all natural numbers for exponents?

What if Bob's denominator is [math]2^{1/2}3^{0}x^{1}[/math]?

Were you expecting a 1/2 in there?

 

[kac3pro]

I'll admit I didn't think of that perhaps because rational exponents have not yet been covered in the textbook this problem comes from. Sadly I don't have access to the solutions manual to check if we should include those.

 

[tkhunny]

I'll admit I didn't think of that perhaps because rational exponents have not yet been covered in the textbook this problem comes from. Sadly I don't have access to the solutions manual to check if we should include those.

You should,