find a counterexample to the statement 4^n+1 is divisible by 5.?

Question

show how u got ur answ.plz

Answer 1

This statement is only true for few n. Those, in fact, where 4^n results in a number that ends in ...9 or ...4 (because only numbers ending in ...0 or ...5 are divisible by 5). But multiples of 4 will always be even, so there's none ending in ...9. And you get powers of 4 ending in ...4 only for uneven n.

Examples:
n=1: 4^1+1 = 5
n=3: 4^3+1 = 65
n=5: 4^5+1 = 1'025
...
n=11: 4^11+1 = 4'194'305

Counter-examples:
n=0: 4^0+1 = 2, which is not divisible by 5
n=2: 4^2+1 = 17, which is not divisible by 5
n=4: 4^4+1 = 257, and so on.

What do I get for doing your home work? :-)

Answer 2

I assume the statement is ( 4^n ) + 1 as 4^(n+1) is never divisible by 5 (as all the factors of any number produced is 2)

this statement is only true if 4^n is equal to a number with a digit ending in 4 or 9. As 9 is odd, only a number ending in 4 can possibly be correct for this statement.

Looking at the pattern of 4^n, it alternates between final digits of 4 and 6. As a result, the digit 4 only occurs when n is odd and so the number is divisible by 5.

So any even number is a counterexample, any odd number is an example and any negative number is a counterexample.

Answer 3

for n=0, the statement 4^n+1 is equal to 2. Since 2 is not divisible by 5, this is a counterexample.

If you mean 4^(n+1), n=0 is also a counterexample

Answer 4

4² + 1 = 17, 17 is not divisible by 5.