Quick problem to do with powers

1. Apr 22, 2004

hi, can someone try and represent:

3(3^(2n+4) - 2^(2n))

as a multiple of five. Been workin on it but I'm not very good at this sorta thing.

Cheers

Pat

2. Apr 22, 2004

Njorl

It is easy to prove it is so, but I haven't seen any quick method for expressing it as a factor of 5, ie 5*f(x). The last digit of the "3" exponential term alternates between 9 and 1. The last digit of the "2" exponential term alternates between 4 and 6. The subtraction always yields a 5 for the last digit, x9-y4 or x1-y6.

Njorl

3. Apr 22, 2004

uart

Sorry, no obvious factors of five there.

How about 5 ( 48.6 * 3^(2n) - 0.6 * 2^(2n) )

:p

4. Apr 22, 2004

Muzza

That's cheating, uart ;)

5. Apr 22, 2004

uart

Hehe that's why I put the ":p" there, it means I was poking my tongue out.

Anyway it's actually pretty easy. You can readily reduce the problem to proving that (9^n - 4^n) is a multiple of 5. Just express the 9^n as (5+4)^n and use the binomial expansion formula. The 4^n terms cancel out you're left with only terms that are multiples of 5

Last edited: Apr 22, 2004
6. Apr 22, 2004

cheers for your help, altho I need to use the 3 because the original proof was by induction to prove f(x) = 3^(2n+4) - 2^(2n) is always a multiple of five, and I've been told to do it by:

f(k+1) - f(k) = f(k+1) - 4f(k) + 3f(k)

Iv proved f(k+1) - 4f(k) is a multiple of 5 by rearranging the powers so need to do it with the 3f(k) now. There should be a way.

Pat

7. Apr 22, 2004

uart

Yes there's a way, you've already done it. The whole basis of proof by induction is that you can proof f(k+1) by assumming f(k). Of course you must verifiy f(1) yourself, but that is very easy.

So you actually just assume that f(k) is a multiple of 5 and then show that this implys that f(k+1) is also a multiple of 5. You've done all the required work already. :)

8. Apr 23, 2004

uart

Just to clarify to steps.

1. Prove that g(k) = f(k+1) - 4*f(k) is a multiple of 5 (without making ANY assumptions that f(k) is a multiple of 5).

2. Note that f(k+1) = g(k) + 4*f(k)

3. Using 1. and 2. show that f(k) a mult of 5 implies that f(k+1) is a mult of 5.

4. Verify f(1) is a mult of 5 and you're done.