YES, But it is the better one for clarifying the unit digit conception.

\(8\) has a repeating pattern of \(4\)

\(8^1 = 8\)
\(8^2 = 64\)
\(8^3 = 512\)
\(8^4 = 4,096\)
\(8^5 = 32,768\)

i.e. 8, 4, 2, and 6 repeat at the units place every 4 powers

Hence, C, E, G, and I

Fixed typo in the question. it was NOT \(98^x\) but

\(9^{8^x}\)

Thank you

Does this change to the answer to the question? Shouldn't it only be 1 as the only possible units digit since nine will always be raised to an even exponent?

Not really, because we do have just a double step to perform but essentially the way to solve is the same

8^x if you to assess the pattern of 8

Turns out, you have to asses the pattern of nine based on the pattern of 8

Regards

I tried to solve the question as it is considered 9^8^x
1) 9^0=1
9^1=9
9^2=81 so in every two steps units digit repeats as "1"

2)8^x is the exponent of the question so
For the x>1, 8^x/2 such as 64/2, 512/2, 4096/2.... remainder would be "0"
Note: I diveded numbers every time to 2 because in the first explanation it repeats in every 2 steps.

3)For every circumustance remainder is "0" so 9^0=1 means for 9^8^x units digit is 1

İs it correct?

Thank You

Essentially yes

Could you change the answer on the problem? It's still showing up as the answer to the question (98)^x

The answer here should be B

Why ?