We can use the pattern of the unit digits of powers of 2 to find the unit digit of $2^{4n+3}$ ,

The unit digits of the powers of 2 follow a pattern:

$2^{1}$ has a unit digit of 2,

$2^{2}$ has a unit digit of 4,

$2^{3}$ has a unit digit of 8, and

$2^{4}$ has a unit digit of 6.

( $2^{5}$ =32) unit digit is 2

( $2^{6}$ =64) unit digit is 4

( $2^{7}$ =128) unit digit is 8

( $2^{8}$ =256) unit digit is 6

( $2^{9}$ =512) unit digit is2

( $2^{10}$ =1024) unit digit is 4

( $2^{11}$ =2048) unit digit is 8

( $2^{12}$ =4096) unit digit is 6

( $2^{13}$ =8192) unit digit is 2

( $2^{14}$ =16384) unit digit is 4

( $2^{15}$ =32768) unit digit is 8

...

As we can observe, the unit digits of powers of 2 repeat in a cycle: 2, 4, 8, 6. Every fourth power of 2 has the same unit digit.

Notice that the pattern repeats every four powers of 2.

So, we need to find the remainder when 4n+3 is divided by 4.

4n+3 = 4n + 4 – 1

4n+3=4n+1-1

Therefore, we can conclude that the unit digit of ( $2^{4n+3}$ ) will be the same as the unit digit of ( $2^{3}$ ), which is 8.

Thus, 4n+3 has a remainder of 3 when divided by 4.

Therefore, the unit digit of ( $2^{4n+3}$ ) is the same as the unit digit of ( $2^{3}$ ) , which is 8.

Therefore, the answer is D.

Arithmetic Data Sufficiency