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If n and a are positive integers, what is the units digit of n^4a+2 – n^8a?
(1) n = 3
(2) a is odd.

Please assist with above problem.
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alanforde800Maximus
If n and a are positive integers, what is the units digit of n^4a+2 – n^8a?
(1) n = 3
(2) a is odd.

Please assist with above problem.

Merging topics. Please refer to the discussion above.
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This question takes a few seconds to solve with cyclicity.

What is the units digit of \(n^{4a+2} - n^{8a}\)? Lets look at the statements.

(1) n = 3
\(3^{4a+2} - 3^{8a}\)

Cyclicity of 3 = 3, 9, 7, 1

3^{4a+2} = units digit of 9
3^{8a} = units digit of 1

9 - 1 = units digit of 8. Sufficient.

(2) a is odd
We don't know what n is; insufficient.

Answer is A.
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Bunuel
If n and a are positive integers, what is the units digit of n^(4a+2) – n^(8a)?

(1) n = 3
(2) a is odd.


Kudos for a correct solution.

n^4a . n^2 - n^8a
= n^4a (n^2 - n^4a)

1) when a = 1, it becomes 3^4 (9 - 3^4) = 81 * -72 = .....2
when a = 2, it becomes 3^8 (9 - 3^8) =...1 (....2) = 2, since 3^8 will result in an integer ending with 1, it will give 2 as the unit digit when 9 is subtracted from it.
so, no matter what is the value of a is, the unit digit in this case will always be 2. Sufficient

2) the base value determines the unit digit. Without it, the power is not sufficient.

A is the answer.
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Hello from the GMAT Club BumpBot!

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