The last digit of $2^{2006}$ My $13$ year old son was asked this question in a maths challenge. He correctly guessed $4$ on the assumption that the answer was likely to be the last digit of $2^6$. However is there a better explanation I can give him?

Well, looking at successive powers of two, starting at $2^1$ we have $$2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, \dots$$
Now, looking at only the final digit we have $$2, 4, 8, 6, 2, 4, 8, 6, 2, 4, \dots$$
You may be able to guess that the pattern repeats the segment $2, 4, 8, 6$ forever. If you're correct, then you can work out the last digit by figuring where $2^{2006}$ sits in this pattern. Note that the first term in the sequence is $2$, and every four terms along it is also $2$. That is the $(4k+1)^{\textrm{th}}$ term is always $2$. Likewise, the $(4k + 2)^{\textrm{nd}}$ term is always $4$, the $(4k+3)^{\textrm{rd}}$ term is always $8$, and the $(4k+4)^{\textrm{th}}$ term is always $6$. Now you just need to figure out whether $2006$ is of the form $4k + 1, 4k+2, 4k + 3$, or $4k + 4$. As $2006$ is even, it is either of the form $4k+2$ or $4k+4$. As $4k+2 = 2006$ has an integer solution $(k = 501)$ and $4k + 4 = 2006$ doesn't, $2006$ is of the form $4k+2$ and therefore $2^{2006}$ ends in a $4$.
A guess is all well and good, but it ain't no proof. How do we know the pattern of digits continues to repeat forever (or at least up to $2006$)? Well, you can use modular arithmetic to prove the pattern continues, but if your son knew modular arithmetic, he could have used it to solve the problem in the first place. 
Another way to see the pattern continues is to think about what happens when you multiply numbers. Consider multiplying the number $43$ by $2$. How do you do it? First you multiply $3$ and $2$ and put the result in the ones column, then you move onto the $4$ and multiply it by $2$ and put it in the tens column. Had we chosen $46$ instead of $43$ then it is slightly more complicated because $6\times 2 = 12$. In this situation we put the $2$ in the ones column and we carry the $1$ to the tens column. This all sounds a little bit boring but the point is this: 

To find out the last digit of $a\times 2$, where $a$ could have lots of digits, we just need to know what (the last digit of $a$)$\, \times\, 2$ is. In particular, the last digit of $($(the last digit of $a$)$\, \times\, 2)$ is the same as the last digit of $(a\times 2)$. 

This is true because of the way we multiply as explained above. When multiplying $43$ by $2$, we only put something in the ones column when we did $3\times 2$. In the case of $46$ multiplied by $2$, we get $6\times 2 = 12$ and put the $2$ in the ones column. So irrespective of the size of the number, we only put something in the ones column when we multiply the rightmost (i.e. last) digit by $2$, and in the case that the result has more than one digit, we only take the last digit of the result and put it in the ones column.
With this in mind, we have the final digits of the first four powers of $2$ are $2, 4, 8, 6$. As $6 \times 2  = 12$, we know that the final digit of $2^5 = 2^4\times 2$ is $2$. Then as $2\times 2$ we know that the final digit of $2^6 = 2^5\times 2$ is $4$. Then as $4\times 2= 8$ we know the final digit of $2^7 = 2^6\times 2$ is $8$. Then as $8\times 2 = 6$, we know that the final digit of $2^8=2^7\times 2$ is $6$. Now we're back to $6$ and can see that we will continue the pattern forever. So the initial guess was correct and hence the argument that $2^{2006}$ ends in a $4$ is valid.

Note, similar considerations to those made above can be used to show that the last digit of $a\times b$ is the same as the last digit of (last digit of $a$)$\, \times\, $(last digit of $b$). As before, you can also prove this using modular arithmetic.
