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Find the largest integer n for which 12^n evenly divides 20.

 Oct 3, 2021

Best Answer 

 #2
avatar+2246 
+2

To add a bit, the reason 12^n can't be wholly divisable by 20 is because 20 = 2^2 * 5 and 12 = 2^2 * 3. 

That means that 12^n will only have factors of 2 and 3 (assuming n is an integer). 

Therefore, it will never be able to divide out the 5 in 20. 

 

=^._.^=

 Oct 3, 2021
edited by catmg  Oct 3, 2021
 #1
avatar+12396 
+2

Find the largest integer n for which 12^n evenly divides 20.

 

Hello Guest!

 

\( n\in \mathbb Z\\ z=\dfrac{12^n}{20}\\\)

\(\large z\notin \mathbb Z\)  

\(12 ^ n\) is in no case wholly divisible by 20.

 

laugh  !

 Oct 3, 2021
 #2
avatar+2246 
+2
Best Answer

To add a bit, the reason 12^n can't be wholly divisable by 20 is because 20 = 2^2 * 5 and 12 = 2^2 * 3. 

That means that 12^n will only have factors of 2 and 3 (assuming n is an integer). 

Therefore, it will never be able to divide out the 5 in 20. 

 

=^._.^=

catmg  Oct 3, 2021
edited by catmg  Oct 3, 2021

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