No need to raise 13 and 12 to any digits. Since we are interested in finding the unit digits only we can work with 3 and 2.

We should also keep in mind that the unit digits of any number will reveal a pattern when raised to certain numbers.

For eg.

\(3^1 =3\)

\(3^2 = 9\)

\(3^3 = 27\)

\(3^4 = 81\)

\(3^5 = 243\)

\(3^6 = 729\)

You can notice that the units digit of \(3^5\) is same as \(3^1\) and units digit of \(3^2\) is same as \(3^6\)

Hence the units digits repeat in an interval of 4

Similar is the case for the powers of 2

\(2^1 =2\)

\(2^2 = 4\)

\(2^3 = 8\)

\(2^4 = 16\)

From here on the units digit of powers of 2 will repeat.

Now we just need to add the respective units digit of 2 and 3 for a given digit of power

For

x = 1; 3+2 = 5

for

x = 2; 9+4 = 13 or, 3

for

x = 3; 7+8 = 15 or, 5

for

x = 4; 1+6 = 7

since 5 is

repeated we count it as 1.

Therefore there are in total 3 distinct unit digits

option B
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This is my response to the question and may be incorrect. Feel free to rectify any mistakes