# Quick problem to do with powers

1. Apr 22, 2004

### padraig

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

### padraig

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.

Thanks for your help

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.

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