BTGmoderatorDC wrote:If x is a positive integer, is the remainder 0 when \(3^x + 1\) is divided by 10?

(1) x = 4n + 2, where n is a positive integer.
(2) x > 4

\(3^x + 1\) will yield a remainder of 0 when divided by 10 if it has a units digit of 0.

3¹ --> units digit of 3.
3² --> units digit of 9. (Since the product of the preceding units digit and 3 = 3*3 = 9.)
3³ --> units digit of 7. (Since the product of the preceding units digit and 3 = 9*3 = 27.)
3� --> units digit of 1. (Since the product of the preceding units digit and 3 = 7*3 = 21.)

From here, the units digits will repeat in the same pattern:
3, 9, 7, 1...3, 9, 7, 1...3, 9, 7, 1....
The units digit repeat in a CYCLE OF 4.

Implication:
When an integer with a units digit of 3 is raised to a power that is a multiple of 4, the units digit will be the END of the cycle: 1.
When an integer with a units digit of 3 is raised to a power that is TWO MORE than a multiple of 4, the units digit will be the SECOND VALUE in the cycle: 9.

Thus, \(3^x + 1\) will have a units digit of 0 only if x is two more than a multiple of 4. with the result that \(3^x + 1\) = (integer with a units digit of 9) + 1 = integer with a units digit of 0.
Question stem, rephrased:
Does x = 4n + 2, where n is a nonnegative integer?

Statement 1:
Since x = 4n + 2, the answer to the rephrased question stem is YES.
SUFFICIENT.

Statement 2:
No way to determine whether x = 4n + 2.
INSUFFICIENT.

The correct answer is A.