If x is an even number, then $x^{4n}$, where n is a positive
integer, will always have

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Accepted Answer

Let $$ x = 2y $$, where $$ y $$ is any positive integer. Then,

$x^{4n}\ =\ \left(2y\right)^{4n}\ =\ \left(2y\right)^{4^{^n}}\ =\ \left(16y^4\right)^n$

Now, observe the unit digit of $$ y^4 $$ for different values of $$ y $$:  
- For $$ y = 1, 2, 3, 4, 6, 7, 8, 9 $$, the unit digit of $$ y^4 $$ is always **6**.  
- For $$ y = 5 $$, the unit digit of $$ y^4 $$ is **0**.

Thus, the unit digit of $$ (16y^4)^n $$ depends on the unit digit of $$ y^4 $$:  
- If the unit digit of $$ y^4 $$ is **6**, then the unit digit of $$ x^{4n} $$ will be **6**.  
- If the unit digit of $$ y^4 $$ is **0**, then the unit digit of $$ x^{4n} $$ will be **0**.

Hence, the unit digit of $$ x^{4n} $$ will always be either **0 or 6**, depending on the value of $$ y $$.

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