Deducing information based on common denominators between unknown fractions.

Question

I'm trying to answer a 3-part question from the "Art of Problem Sovling: Introduction to Algebra" book.

The problem is outlined as:

Fiona, George and Henry each think of a different fraction. The simplest common denominator between Fiona's and George's fraction is $10ab^{2}$. The simplest common denominator between George's and Henry's fraction is $20a^{3}b^{2}$. The simplest common denominator between Fiona's and Henry's fraction is $4a^{3}b$

The questions are:

(a)Whose fraction has the highest power of $b$? What is the power?

(b)Whose fraction has the largest constant? (Assuming all constants are positive)

(c)What is the simplest common denominator between all 3 fractions.

Here's how I attempted to answer this.

(a)

It's George because all the common denominators between George and the other two result in more $b$'s being factored out. I assume the power is $b^{2}$ but I'm not too sure, because Henry's and Fiona's $b$ power could be preventing from further factoring George's fraction.

(b) My guess is either George or Henry, but which one of the two I'm not sure.

(c) $LCM(10,4,20) = 20 \therefore 20ab$

I'd appreaciate if someone could clear this up in most explanative manner.

Answer 1

a) I believe you are right and it is George

b) It would be George again, as Fiona and Henry are able to make a common denominator with a constant of 4, so one (Henry) has a constant of 4, and the other (Fiona) has a constant of 1. George's 2 common denominators have constants of 10 and 20. His fraction has a constant of 10, because lcm(10,1)=10, and lcm(10,4)=20.

c) Your constant is correct, but the a's and b's are incorrect. lcm(a, a³) = a³, and lcm(b, b²) = b², so it would have a common demominator of 20a³b²

Answer 2

We may suppose that Fiona, George and Henry each think of $$\frac{1}{n_1a^{a_1}b^{b_1}},\quad \frac{1}{n_2a^{a_2}b^{b_2}},\quad \frac{1}{n_3a^{a_3}b^{b_3}}$$ respectively where $n_1,n_2,n_3$ are constants.

Then, we have $$10ab^2=\text{lcm}(n_1,n_2)a^{\max(a_1,a_2)}b^{\max(b_1,b_2)}$$ $$20a^3b^2=\text{lcm}(n_2,n_3)a^{\max(a_2,a_3)}b^{\max(b_2,b_3)}$$ $$4a^3b=\text{lcm}(n_3,n_1)a^{\max(a_3,a_1)}b^{\max(b_3,b_1)}$$ i.e. $$\text{lcm}(n_1,n_2)=10,\qquad \text{lcm}(n_2,n_3)=20,\qquad \text{lcm}(n_3,n_1)=4$$ $$\max(a_1,a_2)=1,\qquad \max(a_2,a_3)=3,\qquad \max(a_3,a_1)=3$$ $$\max(b_1,b_2)=2,\qquad \max(b_2,b_3)=2,\qquad \max(b_3,b_1)=1$$

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(a)

From $\max(b_3,b_1)=1$, we see that $b_3\le 1$ and $b_1\le 1$. From $\max(b_1,b_2)=2$, we get $b_2=2$.

So, George's fraction has the highest power of $b$, and the power is $b^2$.

(b)

From $\text{lcm}(n_3,n_1)=4$, we get $n_3\le 4$ and $n_1\le 4$. It follows that $n_3$ is not divisible by $5$. From $\text{lcm}(n_2,n_3)=20=2^2\cdot 5$, we see that $n_2$ has to be divisible by $5$, so we get $n_2\ge 5$.

So, George's fraction has the largest constant.

(c)

Using

$$\max(A,B,C)=\max(\max(A,B),\max(B,C))\tag1$$

$$\text{lcm}(n_1,n_2,n_3)=\text{lcm}(\text{lcm}(n_1,n_2),\text{lcm}(n_2,n_3))\tag2$$

(The proof is written at the end of this answer)

we get $$\begin{align}&\text{lcm}(n_1,n_2,n_3)a^{\max(a_1,a_2,a_3)}b^{\max(b_1,b_2,b_3)} \\\\&=\text{lcm}(\text{lcm}(n_1,n_2),\text{lcm}(n_2,n_3))a^{\max(\max(a_1,a_2),\max(a_2,a_3))}b^{\max(\max(a_1,a_2),\max(a_2,a_3))} \\\\&=\text{lcm}(10,20)a^{\max(1,3)}b^{\max(2,2)} \\\\&=20a^3b^2\end{align}$$

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Proof for $(1)$ :

We may suppose that $A\le C$.

If $A\le C\le B$, then we have $\max(A,B,C)=B$ and $$\max(\max(A,B),\max(B,C))=\max(B,B)=B$$

If $A\le B\le C$, then we have $\max(A,B,C)=C$ and $$\max(\max(A,B),\max(B,C))=\max(B,C)=C$$

If $B\le A\le C$, then we have $\max(A,B,C)=C$ and $$\max(\max(A,B),\max(B,C))=\max(A,C)=C$$

Proof for $(2)$ :

One can consider the prime factorization of $n_1,n_2,n_3$, and use $(1)$ for the exponent of each prime factor.