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

Approach:
Given: n>0, is (4^n+1) divisible by 5 -> Means last digit should be 0 or 5
Cyclicity of 4^n = 2 (4,16,64,256.....) ..means n should be 1,3,5..... Odd number per se.
Stmnt 1:
on solving the expression values of n comes out as 3 or -6, Since n>0, n=3 satisfies the condition.
Sufficient

Stmnt 2:
n is odd integer, straightaway sufficient.

Hence, Answer should be Option D

know that n>0 and cyclicity of 4 is 4,6 odd and even power


#1
simplify given expression \(\frac{n^{2}}{3} + n - 6 = 0\)
we get n=0,-6 & 3
sufficient as n=3 ;
#2
n is odd sufficient
IMO D

n > 0, \(\frac{4^{n} + 1}{5}\) = Integer?

To rephrase the question, we are being asked whether the units digit of \(4^{n}\) + 1 is 0 or 5

St 1: \(\frac{n^{2}}{3}\) + n - 6 = 0 ==>\(n^{2}\) + 3n -18 = 0 ==> (n+6)(n-3) = 0 ==> n = -6 or 3 ==> n =3 since we are told n > 0
Therefore, \(4^{3}\) + 1 = 64 + 1 ==> 65 ==> Sufficient since the unit digit of 65 is 5

St 2: n is an odd integer

When 4 is raised to a positive integer, the unit digit repeats twice. For example, \(4^{1}\) = 4, \(4^{2}\) = 16, \(4^{3}\) = 64, \(4^{4}\) = 256 ==> when n = odd, unit digits is 4, when n = even, unit digit is 6 ==> the unit digit of \(4^{n}\) + 1 = 4 + 1 = 5 ==> sufficient

The answer is D

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