The last digit of a positive integer is the same as the last digit of n2 only when the last digit is 0, 1, 5, or 6. For example, The last digit of 10 is 0 and the last digit of 102 (= 100) is also 0. The last digit of 11 is 1 and the last digit of 112 (= 121) is also 1. The last digit of 15 is 5, and the last digit of 152 (= 225) is also 5. The last digit of 16 is 6, and the last digit of 162 (= 256) is also 6. Since we have multiple cases with Statement (1) alone, it is not sufficient.
Now, the last digit of an even number is 0, 2, 4, 6, or 8. Hence, Statement (2) alone is not sufficient.
The common solutions from the two statements are 0 and 6 (which means n ends with either 0 or 6). Since
we do not have a single solution, the statements together are not sufficient. The answer is (E).