For what integers n is3^(2n+1) + 2^(n+2)divisible by7?

Can anybody help me?(Crying)

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- August 21st 2010, 06:51 AMyuudDivisible by 7
For what integers n is

**3^(2n+1) + 2^(n+2)**divisible by**7**?

Can anybody help me?(Crying) - August 21st 2010, 07:06 AMyeKciM
- August 21st 2010, 07:16 AMundefined
Another approach is to reduce the exponents mod 6 making use of Euler's theorem. (Actually you can just use Fermat's little theorem which is a special case of Euler's theorem.) There are 6 congruence classes of n to list out.

- August 21st 2010, 07:31 AMmelese
- August 21st 2010, 07:54 AMyuud
Yeah I tried n=0...

Then n=k

And then n=k+1

So that's it??? So n>=0??

I still haven't learnt modulo mathematics or Euler's Theorem.... - August 21st 2010, 08:01 AMyeKciM
hm...

depends, how did you show that ? if not problem post it then we will see is that it :D

P.S. that's not true !!!

and induction is based (actually that induction comes from definition of natural numbers) on natural numbers - August 21st 2010, 08:22 AMyuud
But it's mentioned integers in the question...So 0 also must be included??

My Proof...

first the base case.

n = 0

3^1 +2^2 = 3+4 = 7

now the more general case.

let k = n. assume 3^(2*k+1) +2^(k+2) is divible by 7.

prove 3^(2*(k+1)+1) +2^((k+1)+2) is divisable by 7.

3^(2*k +3) +2^(k+3)

3^2 *3^(2*k +1) +2* 2^(k+2)

9* 3^(2*k +1) +2* 2^(k+2).

so, based on our assumption, we know 3^(2*k+1) +2^(k+2) is divisable by 7.

we also know based on our formula, that the power for 3 will always increase by 2, and the power for 2 will always increase by 1, for subequent k's. therefore 9* 3^(2*k+1) +2* 2^(k+1) will also be divisble by 7 - August 21st 2010, 08:35 AMyeKciM
if one is to use mathematical induction, one must know that :

set of natural numbers is subset of set with "features" :

1° (1) is natural number,

2° any natural number (n) have his follower n'=n+1 which is also natural number

3° (1) is not follower of any naturan number

4° if m'=n' than m=n, meaning every natural number is follower of one natural number (and onley one)

5° if and if in M apply properties 1° and 2° than M=N

axiom 5 is known as the principle of induction :D

and so on...

Edit : sorry for this... i was going somewhere else with this.

anyway can't do you harm to know this :D ( because this is sub forum "university math" so i thought that you would, and should know this :D ) - August 21st 2010, 08:38 AMyuud
That's too advanced for me... hehe... Is my proof good?

- August 21st 2010, 08:50 AMundefined
Yes you should include 0 because the question asks about integers. This poses no problem with respect to using induction.

I don't see how you justified your conclusion at the end. I would rewrite

9* 3^(2*k +1) +2* 2^(k+2)

as

7 * 3^(2k + 1) + 2 * [3^(2k + 1) + 2^(k + 2)]

See? - August 21st 2010, 09:41 AMyuud
- August 21st 2010, 09:54 AMyeKciM
that is it :D:D:D

that's it ... because

is divided by 7 ..

and

is also divided by 7 because you assumed that n=k is true :D - August 21st 2010, 03:14 PMSoroban
Hello, yuud!

Here is a sneaky method . . . if you know modulo arithmetic.

Quote:

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It is divisible by 7 forpositive intergers .*all*

- August 21st 2010, 03:31 PMundefined
- August 21st 2010, 03:51 PMmelese