Gmat_mission wrote:If x and y are positive integers and n = 5^x + 7^(y + 3), what is the units digit of n?

(1) y = 2x - 16
(2) y is divisible by 4.

5^x will have a units digit of 5, regardless of the value of x:
5¹ = 5
5² = 25
5³ = 125
In every case, the units digit is 5.

To determine the units digit of n, we need to know the units digit of 7^(y+3).
Question stem, rephrased:
What is the units digit of 7^(y+3)?

When 7 is raised to consecutive powers, the unit digit repeat in a CYCLE OF 4:
7¹ --> units digit of 7.
7² --> units digit of 9. (Since the product of the preceding units digit and 7 = 7*7 = 49.)
7³ --> units digit of 3. (Since the product of the preceding units digit and 7 = 9*7 = 63.)
7� --> units digit of 1. (Since the product of the preceding units digit and 7 = 3*7 = 21.)
From here, the units digits will repeat in the same pattern: 7, 9, 3, 1.
Thus, the units digit repeat in a CYCLE OF 4.
Implication:
When 7 is raised to a power that is a multiple of 4, the units digit will be 1.
When 7 is raised to a power that is 1 more than a multiple of 4, the units digit will be 7.
When 7 is raised to a power that is 2 more than a multiple of 4, the units digit will be 9.
When 7 is raised to a power that is 3 more than a multiple of 4, the units digit will be 3.
The result will be the cycle shown above -- 7, 9, 3, 1 -- where every exponent that is a multiple of 4 yields a units digit of 1.

Statement 1:
Case 1: x=9, with the result that y = (2*9) - 16 = 2
In this case, 7^(y+3) = 7�.
Since the exponent is 1 more than a multiple of 4, the units digit will be 7.
Case 2: x=10, with the result that y = (2*10) - 16 = 4
In this case, 7^(y+3) = 7�.
Since the exponent is 3 more than a multiple of 4, the units digit will be 3.
Since the units digit for can be different values, INSUFFICIENT.

Statement 2:
Since the exponent for 7^(y+3) is 3 more than a multiple of 4, the units digit will be 3.
SUFFICIENT.

The correct answer is B.