Finding Last Digit

Question: The unit digit of $2^{100}$ is

$1.~2 \quad 2.~4 \quad3.~6 \quad4.~8$

Answer: This problem can be solved by basic idea of periodicity. Calculating few terms mod $10$,

$2^1$	$2^2$	$2^3$	$2^4$	$2^5$	$2^6$
$2$	$4$	$8$	$6$	$2$	$4$

We can see that the periodicity is $4$.
Hence $2^{k + 4n}= 2^k$ for $k =\{1,2,3,4\}$.
Here $2^{100} = 2^{4+ 24\cdot 4} = 2^4 = 6$.

Hence 3 is correct answer.

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Question: The last digit of $38^{2011}$ is

$1.~6 \quad 2.~2 \quad 3.~4 \quad 4.~8$

Answer: First note that $38 = 8$ (mod $10$).
Compute the periodicity of $8$ mod $10$.

$8^1$	$8^2$	$8^3$	$8^4$	$8^5$
$8$	$4$	$2$	$6$	$8$

Again the period is $4$.
Hence $38^{2011} = 8^{2011}= 8^{3 + \text{ multiple of }4} = 8^3 = 2$.

Hence 2 is the correct choice.