Deducing information based on common denominators between unknown fractions. 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.
 A: 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
A: 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.
$$\text{lcm}(n_1,n_2)=10,\qquad \text{lcm}(n_2,n_3)=20,\qquad \text{lcm}(n_3,n_1)=4$$
$$\max(a_1,a_2)=1,\qquad \max(a_2,a_3)=3,\qquad \max(a_3,a_1)=3$$
$$\max(b_1,b_2)=2,\qquad \max(b_2,b_3)=2,\qquad \max(b_3,b_1)=1$$

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

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.
