kshitizbansal92 wrote:

If n > 0, is (4^n + 1) divisible by 5?

(1) n is the integral root of equation: \(n^2/3\) + n - 6 = 0

(2) n is an odd integer.

Statement (1)Multiply by 3 to get rid of the fraction. \(n^2\) +3n -18 = 0... Factors to (n+6)(n-3). Roots are 3 and -6. It is given that n>0, so n must = 3.

Plug in n=3...... \(4^3\) + 1 = 65. Is 65 divisible by 5? YES.

Statement (2)We know n has to be positive. so our possible values for n are 1,3,5,7,9 etc...

Make n=1 and we get \(4^1\) +1 =5. Divisible by 5? YES

Make n=3 and we get \(4^3\) + 1 = 65. Divisible by 5? YES

OR if you are sound in your number theory, you will know that 4 has a cyclicity of 2.

4^1=4

4^2=16

4^3=64

4^4=256

So for all

odd integers it will end in units digit 4.

For all

even integers it will end in units digit 6.

Answer: D

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