Least Common Multiples of A. 23^31,23^17 B. 3^7*5^3*7^3,2^11*3^5*5^9 C. 3^13*5^17,2^12*7^21 Please I am very confused now

Least Common Multiples(LCM)  ?

A. 23^31,23^17

$$\small{ \begin{array}{l|rclclclcl} \text{Number 1:} & 23^{31} &=& 23^{\textcolor[rgb]{1,0,0}{31}} &&&&&& \\ \text{Number 2:} & 23^{17} &=& 23^{17} &&&&&&\\ \hline \text{The greatest Exponent} \\ \text{ of each prime number} &\text{LCM}&=& 23^{\textcolor[rgb]{1,0,0}{31}}&&&&&& \end{array} }}$$

B. 3^7*5^3*7^3,2^11*3^5*5^9

$$\small{ \begin{array}{l|rclclclcl} \text{Number 1:} & 3^{7}*5^3*7^3 &=& 2^0 &*& 3^{\textcolor[rgb]{1,0,0}{7}} &*& 5^3 &*& 7^{\textcolor[rgb]{1,0,0}{3}} \\ \text{Number 2:} & 2^{11}*3^5*5^9 &=& 2^{\textcolor[rgb]{1,0,0}{11}}&*& 3^5 &*& 5^{\textcolor[rgb]{1,0,0}{9}} &*& 7^0\\ \hline \text{The greatest Exponent} \\ \text{ of each prime number} &\text{LCM}&=& 2^{\textcolor[rgb]{1,0,0}{11}}&*&3^{\textcolor[rgb]{1,0,0}{7}} &*&5^{\textcolor[rgb]{1,0,0}{9}} &*& 7^{\textcolor[rgb]{1,0,0}{3}} \end{array} }}$$

C. 3^13*5^17,2^12*7^21

   $$\small{ \begin{array}{l|rclclclcl} \text{Number 1:} & 3^{13}*5^{17} &=& 2^0 &*& 3^{\textcolor[rgb]{1,0,0}{13}} &*& 5^{\textcolor[rgb]{1,0,0}{17}} &*& 7^0 \\ \text{Number 2:} & 2^{12}*7^{21} &=& 2^{\textcolor[rgb]{1,0,0}{12}}&*& 3^0 &*& 5^0 &*& 7^{\textcolor[rgb]{1,0,0}{21}}\\ \hline \text{The greatest Exponent} \\ \text{ of each prime number} &\text{LCM}&=& 2^{\textcolor[rgb]{1,0,0}{12}}&*&3^{\textcolor[rgb]{1,0,0}{13}} &*&5^{\textcolor[rgb]{1,0,0}{17}} &*& 7^{\textcolor[rgb]{1,0,0}{21}} \end{array} }}$$  

A)

. $$\\23^{31}\quad and \quad 23^{17}\\\\ 23^{31}=23^{17}*23^{14}\\ 23^{17}=23^{17}*1\\\\ so\;\; 23^{17} $ is the lowest common denominator$$ $

miss melody he is looking for LCM not LCD :)

B.

$$\\3^7*5^3*7^3\quad and \quad 2^{11}*3^5*5^9\\\\ 3^7*5^3*7^3=3^5*3^2*5^3*7^3\\ 3^7*5^3*7^3=\textcolor[rgb]{0,1,0}{3^5*5^3}*7^3*3^2\\\\ 2^{11}*3^5*5^9=\textcolor[rgb]{0,1,0}{3^5*5^3}*5^6*2^{11}\\\\ $So the lowest common denominator is $\textcolor[rgb]{0,1,0}{3^5*5^3}$$

Thanks anon.    I messed up   LOL

the first one 

LCM=  23^31 because they both go into that. 

2)

$$\\3^7*5^3*7^3\quad and \quad 2^{11}*3^5*5^9\\\\ 3^7*5^3*7^3=3^5*3^2*5^3*7^3\\ 3^7*5^3*7^3=\textcolor[rgb]{0,1,0}{3^5*5^3}*7^3*3^2\\\\ 2^{11}*3^5*5^9=\textcolor[rgb]{0,1,0}{3^5*5^3}*5^6*2^{11}\\\\ $So the lowest common multiple will be $\\ \textcolor[rgb]{0,1,0}{3^5*5^3}*7^3*3^2*5^6*2^{11}$$

You can try the last one for yourself anon unless one of the others on the forum would like to have a go.

And I am not meaning our other full blown mathematicians either.

Why don't one of you amateurs have a go at it?

Don't forget to show your logic though  

Here's B

3^7*5^3*7^3, 2^11*3^5*5^9

We take each different base along with the highest power associated with that base.....so we have....

2^11 x 3^7 x 5*9 x 7^3 = LCM = 3,000,564,000,000,000