+0

# Number Theory

0
173
2

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

Oct 3, 2021

#2
+2401
+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
+13571
+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.

!

Oct 3, 2021
#2
+2401
+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.

=^._.^=

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