NeilWatson wrote:If n and m are positive integers, what is the remainder when 3^(4n+2) + m is divided by 10?

1) n=2
2) m=1

When a positive integer is divided by 10, the remainder is the UNITS digit:
13/10 = 1 R3.
197/10 = 19 R7.
5264/10 = 526 R4.
In the problem above, we need to determine the units digit of 3^(4n+2) + m.

If x, y and z are positive integers, then the units digit of their product (xyz) is equal to the units digit of the following product:
(units digit of x)(units digit of y)(units digit of z).

3^(4n + 2) = 3^[2 * (2n+1)] = 9^(2n+1) = 9^(odd).
When 9 is raised to an odd power, the units digit is 9:
9¹ = 9 --> units digit of 9.
9³ = 9 * 9² = 9 * 81 --> (units digit of 9)(units digit of 1) --> units digit of 9.
9� = 9² * 9³ = 81 * 729 --> (units digit of 1)(units digit of 9) --> units digit of 9.

Thus:
3^(4n + 2) + m = (integer with a units digit of 9) + m.
In order to determine the units digit of 3^(4n + 2) + m, we need to know the value of m.

Question stem, rephrased: What is the value of m?

Statement 1: n=2
INSUFFICIENT.

Statement 1: m=1
SUFFICIENT.

The correct answer is B.

=(3^4n x 3^2) + 1
=3^4n x (3^2+1)

The portion in red isn't quite correct.
Your reasoning assumes that the following relationship is valid:
(xy) + 1 = (x)(y+1).
To determine whether the equation above works, test x=2 and y=3:
(2*3) + 1 = (2)(3+1)
7 = 8.
Doesn't work.

The issue here is PEMDAS:
The operation inside the parentheses -- (3^4n x 3^2) -- must be performed BEFORE any operations are performed outside the parentheses (such as adding 1).
(3^4n x 3^2) + 1 is not the same as 3^4n x (3^2 + 1).