Jozrael

How do I prove that 2^n + 3^n - 5^n is always divisible by 6?

dlish

i couldnt remember this now. been far too long since my uni days. what are you trying to prove? is this homework?

sorry im no help, but theres a heap of highly intelligent people here that could help you.

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An injustice anywhere, is an injustice everywhere
"Romance is never having to apologize for getting it in her hair" - World's King

filtherton

What is n? Integer? Natural? How are you supposed to prove it? Induction? Contradiction?

Rekna

Ok here you go:

assume that 2^n + 3 ^n - 5^n is divisible by 6

base case: 2^2+3^2-5^2=-12

inductive step:

2^(n+1)+3^(n+1)-5^(n+1)
=2*2^n+ 3*3^n -5*5^n
=2^n+2^n+3^n+3^n+3^n-5^n-5^n-5^n-5^n-5^n
= (2^n+3^n+5^n)+(2^n+3^n+5^n) + (3^n -3*5^n)

The first 2 terms are both divisible by 6 (from our assumption) but is the 3rd?

We can try induction on that one to prove it

assume 3^n -3*5^n is divisible by 6

base case: 3^2 - 3*5^2 = -66

inductive step

3^(n+1) - 3*5^(n+1)
=3*3^n - 3*5*5^n
=3^n+3^n+3^n-3*(5^n+5^n+5^n+5^n+5^n)
=(3^n-3*5^n)+(3^n-3*5^n)+(3^n-3*5^n)-3*2(5^n)
=(3^n-3*5^n)+(3^n-3*5^n)+(3^n-3*5^n)-6(5^n)

The first 3 terms are divisible by 6 from the assumption and the 4th term is divisible by 6 because it is multiplied by 6. Thus 3^n -3*5^n is divisible by 6 and as a result 2^n + 3 ^n - 5^n is divisible by 6.

QED.

Please check my work.

dlish

does this last post deserve anyhing else but a smily?

well done rek...even if it is wrong. well done!

__________________
An injustice anywhere, is an injustice everywhere
"Romance is never having to apologize for getting it in her hair" - World's King