Find the last digit of (2^222 + 2^333 + 2^444)/2

[Jameson]

What is the last digit of $$\frac{2^{222}+2^{333}+2^{444}}{2}$$?

(This problem is aimed at 9th and 10th graders. If you find it easy, I suggest giving a shot at http://mathhelpboards.com/potw-secondary-school-high-school-students-35/problem-week-70-july-29th-2013-a-5791.html using modular arithmetic)
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[Jameson]

Congratulations to the following members for their correct solutions:

1) MarkFL
2) eddybob123

Solution (from MarkFL):

Spoiler

If we look at the first few powers of 2:

$$2^1=2$$

$$2^2=4$$

$$2^3=8$$

$$2^4=16$$

We see this cycle of last digits, with a period of 4, will repeat. If we observe that:

$$111=4\cdot28-1$$

Then we might hypothesize that:

$$2^{4n-1}=10m_n+8$$ where $$m_n\in\mathbb{N}$$

Multiplying this induction hypothesis by:

$$2^4=16$$

we obtain:

$$2^{4(n+1)-1}=160m_n+120+8$$

If we define:

$$m_{n+1}=16m_{n}+12$$ where $$m_1=0$$

then we may write:

$$2^{4(n+1)-1}=10m_{n+1}+8$$

thereby completing the proof by induction.

Thus, we may write:

$$2^{111k}=\left(2^{111} \right)^k=\left(10m_{28}+8 \right)^k$$ where $$k\in\mathbb{N}$$

By the binomial theorem (and the fact that $m_n$ is always even), we find that:

$$\left(10m_{28}+8 \right)^k=\sum_{j=0}^k\left({k \choose j}\left(10m_{28} \right)^{k-j}2^{3j} \right)=20n_k+2^{3k}$$

where $$n_k\in\mathbb{N}$$

Hence:

$$2^{222}=20n_2+2^6=20\left(n_2+3 \right)+4$$

$$2^{333}=20n_3+2^9=20\left(n_3+25 \right)+12$$

$$2^{444}=20n_4+2^{12}=20\left(n_4+204 \right)+16$$

Adding, we then find:

$$2^{222}+2^{333}+2^{444}=20(n_2+n_3+n_4+232)+32=20(n_2+n_3+n_4+233)+12$$

And so, dividing by 2, we find:[/I]

$$\frac{2^{222}+2^{333}+2^{444}}{2}=10(n_2+n_3+n_4+233)+6$$

Thus, the last digit is 6.

Alternate solution (from eddybob123):

Spoiler

We notice a pattern in the last digit of $$2^n$$:

$$\text{If}~n\equiv 1~\text{(mod}~4)~\text{then}~2^n\equiv 2~\text{(mod}~10)$$
$$\text{If}~n\equiv 2~\text{(mod}~4)~\text{then}~2^n\equiv 4~\text{(mod}~10)$$
$$\text{If}~n\equiv 3~\text{(mod}~4)~\text{then}~2^n\equiv 8~\text{(mod}~10)$$
$$\text{If}~n\equiv 0~\text{(mod}~4)~\text{then}~2^n\equiv 6~\text{(mod}~10)$$

Since $221\equiv 1~\text{(mod}~4)$, $332\equiv 0~\text{(mod}~4)$, and $443\equiv 3~\text{(mod}~4)$, then $2^{221}+2^{332}+2^{443}~\text{(mod}~10)\equiv 2+6+8~\text{(mod}~10)\equiv 6 ~\text{(mod}~10)$