enigma123 wrote:

If a and b are positive integers, what is the remainder when \(4^{2a+1+b}\) is divided by 10?

(1) a = 1

(2) b = 2

We need to determine the remainder when 4^(2a+1+b) is divided by 10. Since the remainder when an integer is divided by 10 is always equal to the units digit of that integer, we need to determine the units digit of 4^(2a+1+b).

We can also simplify 4^(2a+1+b):

4^(2a+1+b) = (4^2a)(4^1)(4^b)

Let’s now evaluate the pattern of the units digits of 4^n for positive integer values of n. That is, let’s look at the pattern of the units digits of powers of 4. When writing out the pattern, notice that we are ONLY concerned with the units digit of 4 raised to each power.

4^1 = 4

4^2 = 6

4^3 = 4

4^4 = 6

The pattern of the units digit of powers of 4 repeats every 2 exponents. The pattern is 4–6. In this pattern, all exponents that are odd will produce a 4 as its units digit and all exponents that are even will produce a 6 as its units digit. Thus:

4^1 has a units digit of 4 and 4^2a has a units digit of 6.

Thus, to determine the units digit of (4^2a)(4^1)(4^b), we only need the value of b.

Statement One Alone:a = 1

Since we do not have the value of b, statement one alone is not sufficient to answer the question.

Statement Two Alone: b = 2

Since we have the value of b, statement two is sufficient to determine the remainder when (4^2a)(4^1)(4^b) is divided by 10.

Answer: B

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