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

Question

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

(1)  \frac{n^{2}}{3} + n - 6 = 0 
(2) n is an odd integer.

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

(1) n is the integral root of equation:  \frac{n^{2}}{3} + n - 6 = 0 
(2) n is an odd integer.

satya2029 · Answer

Given n>0
4^n
when power is odd number- last digit is 4
when power is even number(not 0)-last digit is 6
1. n^2+3n-18=0
this gives n=3 or n=-6
since n>0, n=3= odd
means 4^odd= last digit is 4
4^n+1= last digit is 5 means divisible by 5
sufficient
2. n is odd
sufficient
D:)

chetan2u · Answer

If n > 0, is  (4^{n}  + 1) divisible by 5?
  For a number to be divisible by 5, the UNIT's digit should be 0 or 5..  
 Now, power of 4 have a cyclicity 4, 6, 4, 6... that is  4^1  will have 4 as units digit ,  4^2  will have 6 and so on. 
So, 
  (a) If n is ODD,  4^n+1  will have 4+1 or 5 as units digit and thus will be divisible by 5
(b) If n is EVEN,  4^n+1  will have 6+1 or 7 as units digit and thus will NOT be divisible by 5

(1)  \frac{n^{2}}{3} + n - 6 = 0 
 \frac{n^{2}}{3} + n - 6 = 0......n^2+3n-18=0.....n^2+6n-3n-18=0.......(n+6)(n-3)=0 , so n=3 or -6
Since n>0, n=3.....Since n is ODD, Answer is YES from (a) above..

(2) n is an odd integer.
Sufficient as answer is YES from point (a) above

D

fauji · Answer

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

Archit3110 · Answer

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

bebs · Answer

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}  = 1 6 ,  4^{3}  = 6 4 ,  4^{4}  = 25 6  ==> 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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If n > 0, is (4^n + 1) divisible by 5?

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