CSAT 2020

Q. A digit n > 3 is divisible by 3 but not divisible by 6. Which one of the following is divisible by 4?

(a) 2n
(b) 3n
(c) 2n + 4
(d) 3n+ 1
Correct Answer: (d) 3n+ 1

Question from UPSC Prelims 2020 CSAT Paper

Explanation :Β 

Digit n > 3 is divisible by 3

Since n is divisible by 3 but not divisible by 6, it means that n must be an odd multiple of 3. Let’s take n=9 as an example.

(a) 2n = 2 * 9 = 18 which is not divisible by 4

(b) 3n = 3 * 9 = 27 which is not divisible by 4

(c) 2n +4 = (2 *9) +4 =22 which is not divisible by 4

(d) 3n +1 = (3 *9)+1=28 which is divisible by 4

So the correct answer is (d) 3n+1.

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