Bunuel
If x and n are integers such that x = 1^1 + 2^2 + 3^3 + . . . + n^n, what is the remainder when x is divided by 5?
(1) n = 10
(2) 5 < n < 11
FROM STATEMENT - I ( SUFFICIENT )Remainder of a number divided by 5 will come only when the Units digit is not 0 or 5
1^1 = 1
2^2 = 4
3^3 = 7
4^4 = 6
5^5 = 5
6^6 = 6
7^7 = 3
8^8 = 6
9^9 = 9
10^10 = 0
Here the units digit will be 7 ( ie, 1 + 4 + 7 + 6 + 5 + 6 + 3 + 6 + 9 ), thus the remainder will be 2FROM STATEMENT - II ( INSUFFICIENT )Since, we need to know the units value of the series, it is essential to find the value of n for the series 1^1 + 2^2 + 3^3 + . . . +
n^nHere , 5 < n < 11 ; n can take multiple values -
n = { 6 , 7 , 8 , 9 , 10 }
Thus unique value of units digit can not be obtained, hence this statement alone is not sufficient to find the remainde of the series divided by 5
Hence, Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient to answer the question asked, answer will be (A)