Bunuel wrote:
For how many integer values of n from 1 to 100, inclusive, is the value of \( 1^{n}+2^{n}+3^{n}\) divisible by 4?
(A) 16
(B) 32
(C) 49
(D) 56
(E) 64
Are You Up For the Challenge: 700 Level QuestionsWe want \(1^n + 2^n + 3^n\) to be divisible by 4
\(1^n\) and \(3^n\) are both odd, while \(2^n\) is even
For any positive odd value of n, \(1^n + 3^n\) will always be divisible by (1 + 3), i.e. 4
Also, for any integer value of n > 1, \(2^n\) is divisible by 4
Thus, for all odd numbers except 1, \(1^n + 2^n + 3^n\) is divisible by 4
=> Number of odd numbers till 100 (except 1) = 100/2 - 1 = 49
Considering the even numbers:
For any positive even value of n, \(1^n + 3^n\) will only divisible by 2
Also, for any integer value of n > 1, \(2^n\) is divisible by 4
Thus, for any even value, \(1^n + 2^n + 3^n\) is NOT divisible by 4
Thus, there are 49 such numbers.
Answer C _________________
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