<s>[INST] 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?
 [/INST] The last digit of $2^{2006}$ is $2^{2006} (\text{mod }10)$. So let's look at the behavior of powers of $2$ mod $10$.
$2^2 = 4$,
$2^3 = 8$ (nothing interesting so far),
$2^4 = 16 = 6$ (remember we are working mod $10$, we only keep the last digit).
$2^5 = 32 = 2$, Now THIS is interesting. This suggests we divide $2006$ by $5$.
So let's do that: $2006 = (401)(5) + 1$
going back to working mod $10$, we have:
$2^{2006} = 2^{(5)(401)+1} = (2^5)^{401}(2)= 2^{402}$ so we have knocked down the size of our exponent a great deal. Repeating this procedure again leads to:
$2^{402} = 2^{(5)(80) + 2} = (2^5)^{80}(2^2) = 2^{82}$ (still mod $10$).
Another go:
$2^{82} = 2^{(5)(16)+2} = (2^5)^{16}(2^2) = 2^{18}$ (mod $10$),
One last time:
$2^{18} = 2^{(5)(3)+3} = (2^5)^3(2^3) = 2^6$ We can stop now, it's clear that the last digit we are looking for is $4$ (the last digit of $64 = 2^6$). But if we wished (and we were for some reason unwilling to physically compute $2^6$) we could continue for one last step:
$2^6 = (2^5)(2^1) = (2)(2) = 4$ (mod $10$).
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