Divisible by 7

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• Aug 21st 2010, 05:51 AM
yuud
Divisible by 7
For what integers n is 3^(2n+1) + 2^(n+2) divisible by 7?

Can anybody help me?(Crying)
• Aug 21st 2010, 06:06 AM
yeKciM
Quote:

Originally Posted by yuud
For what integers n is 3^(2n+1) + 2^(n+2) divisible by 7?

Can anybody help me?(Crying)

try using induction :D

$\displaystyle n=1$

$\displaystyle 3^{2n+1}+2^{n+2}= 3^3 + 2^3 = 27+8 = 35 = 7\cdot 5$ so it's divisible by 7

$\displaystyle n=2$

$\displaystyle 3^{2n+1}+2^{n+2}= 3^5 + 2^4 = 243+16 = 259 = 7 \cdot 37$

and so on...
then assume that is true for $\displaystyle n=k$

and if you assume that's true then show that is true and for $\displaystyle n=k+1$ and if you show that... it will mean that it's true for any natural number :D
• Aug 21st 2010, 06:16 AM
undefined
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.
• Aug 21st 2010, 06:31 AM
melese
Quote:

Originally Posted by yuud
For what integers n is 3^(2n+1) + 2^(n+2) divisible by 7?

Can anybody help me?(Crying)

It may be simple to reduce $\displaystyle 3^{2n+1}$ and $\displaystyle 2^{n+2}$ modulo 7. Then try to find the remainder of $\displaystyle 3^{2n+1}+2^{n+2}$ when divided by 7. Good Luck!
• Aug 21st 2010, 06:54 AM
yuud
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....
• Aug 21st 2010, 07:01 AM
yeKciM
hm...

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

P.S. $\displaystyle 0 \in \mathbb{N}$ that's not true !!!
and induction is based (actually that induction comes from definition of natural numbers) on natural numbers
• Aug 21st 2010, 07:22 AM
yuud
Quote:

Originally Posted by yeKciM
hm...

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

P.S. $\displaystyle 0 \in \mathbb{N}$ that's not true !!!
and induction is based (actually that induction comes from definition of natural numbers) on natural numbers

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
• Aug 21st 2010, 07:35 AM
yeKciM
Quote:

Originally Posted by yuud
But it's mentioned integers in the question...So 0 also must be included??

if one is to use mathematical induction, one must know that :

set of natural numbers $\displaystyle \mathbb{N}$ is subset of set $\displaystyle \mathbb{R}$ with "features" :

1° (1) is natural number, $\displaystyle 1\in \mathbb{N}$

2° any natural number (n) have his follower n'=n+1 which is also natural number $\displaystyle n\in \mathbb{N} \Rightarrow (n+1)\in \mathbb{N}$

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 $\displaystyle M\subset N$ 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 )
• Aug 21st 2010, 07:38 AM
yuud
That's too advanced for me... hehe... Is my proof good?
• Aug 21st 2010, 07:50 AM
undefined
Quote:

Originally Posted by yuud
That's too advanced for me... hehe... Is my proof good?

Quote:

Originally Posted by yuud
But it's mentioned integers in the question...So 0 also must be included??

Yes you should include 0 because the question asks about integers. This poses no problem with respect to using induction.

Quote:

Originally Posted by yuud
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

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?
• Aug 21st 2010, 08:41 AM
yuud
Quote:

Originally Posted by undefined
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?

So who can give me a correct answer,,, I'm not able to prove it...(Worried)
• Aug 21st 2010, 08:54 AM
yeKciM
that is it :D:D:D

$\displaystyle A = 3^{2k+1} + 2^{k+2}$

$\displaystyle 7\cdot 3^{2k+1} + 2\cdot A$

that's it ... because

$\displaystyle 7\cdot 3^{2k+1}$ is divided by 7 ..

and

$\displaystyle 2\cdot A$ is also divided by 7 because you assumed that n=k is true :D
• Aug 21st 2010, 02:14 PM
Soroban
Hello, yuud!

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

Quote:

$\displaystyle \text{For what integers }n\text{ is }\,3^{2n+1} + 2^{n+2}\,\text{ divisible by 7?}$

$\displaystyle 3^{2n+1} + 2^{n+2} \;\;=\;\left(3^3\right)^\frac{2n+1}{3}} + \left(2^3\right)^{\frac{n+2}{3}}$

. . . . . . . . . . .$\displaystyle =\;(27)^{\frac{2n+1}{3}} + (8)^{\frac{n+2}{2}}$

. . . . . . . . . . .$\displaystyle \equiv\; (\text{-}1)^{\frac{2n+1}{3}} + (1)^{\frac{n+2}{2}}\text{ (mod 7)}$

. . . . . . . . . . .$\displaystyle \equiv\;\left(\sqrt[3]{\text{-}1}\right)^{2n+1} + \left(\sqrt[3]{1}\right)^{n+2}\text{ (mod 7)}$

. . . . . . . . . . .$\displaystyle \equiv\; (\text{-}1)^{2n+1} + (1)^{n+2}\text{ (mod 7)}$

. . . . . . . . . . .$\displaystyle \equiv\; -1 + 1\text{ (mod 7)}$

$\displaystyle 3^{2n+1} + 2^{2n+1} \;\equiv\;0 \text{ (mod 7)}$

It is divisible by 7 for all positive intergers $\displaystyle n$.

• Aug 21st 2010, 02:31 PM
undefined
Quote:

Originally Posted by Soroban
Hello, yuud!

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

$\displaystyle 3^{2n+1} + 2^{n+2} \;\;=\;\left(3^3\right)^\frac{2n+1}{3}} + \left(2^3\right)^{\frac{n+2}{3}}$

. . . . . . . . . . .$\displaystyle =\;(27)^{\frac{2n+1}{3}} + (8)^{\frac{n+2}{2}}$

. . . . . . . . . . .$\displaystyle \equiv\; (\text{-}1)^{\frac{2n+1}{3}} + (1)^{\frac{n+2}{2}}\text{ (mod 7)}$

. . . . . . . . . . .$\displaystyle \equiv\;\left(\sqrt[3]{\text{-}1}\right)^{2n+1} + \left(\sqrt[3]{1}\right)^{n+2}\text{ (mod 7)}$

. . . . . . . . . . .$\displaystyle \equiv\; (\text{-}1)^{2n+1} + (1)^{n+2}\text{ (mod 7)}$

. . . . . . . . . . .$\displaystyle \equiv\; -1 + 1\text{ (mod 7)}$

$\displaystyle 3^{2n+1} + 2^{2n+1} \;\equiv\;0 \text{ (mod 7)}$

It is divisible by 7 for all positive intergers $\displaystyle n$.

That's a cool trick; the only problem is that for n < 0, we would not interpret the original expression

$\displaystyle 3^{2n+1} + 2^{n+2}$

mod 7 but rather in such a way as to yield non-integers, thus the only integers that satisfy are the non-negative ones.
• Aug 21st 2010, 02:51 PM
melese
Quote:

Originally Posted by Soroban
Hello, yuud!

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

$\displaystyle 3^{2n+1} + 2^{n+2} \;\;=\;\left(3^3\right)^\frac{2n+1}{3}} + \left(2^3\right)^{\frac{n+2}{3}}$

. . . . . . . . . . .$\displaystyle =\;(27)^{\frac{2n+1}{3}} + (8)^{\frac{n+2}{2}}$

. . . . . . . . . . .$\displaystyle \equiv\; (\text{-}1)^{\frac{2n+1}{3}} + (1)^{\frac{n+2}{2}}\text{ (mod 7)}$

. . . . . . . . . . .$\displaystyle \equiv\;\left(\sqrt[3]{\text{-}1}\right)^{2n+1} + \left(\sqrt[3]{1}\right)^{n+2}\text{ (mod 7)}$

. . . . . . . . . . .$\displaystyle \equiv\; (\text{-}1)^{2n+1} + (1)^{n+2}\text{ (mod 7)}$

. . . . . . . . . . .$\displaystyle \equiv\; -1 + 1\text{ (mod 7)}$

$\displaystyle 3^{2n+1} + 2^{2n+1} \;\equiv\;0 \text{ (mod 7)}$

It is divisible by 7 for all positive intergers $\displaystyle n$.

I'm not quite familar with the extenstion of operations for modular arithmetic. Maybe a simpler way is: $\displaystyle 3^{2n+1}+2^{n+2}=3\cdot 9^n+4\cdot 2^n$; since $\displaystyle 9\equiv 2 \bmod 7$ and $\displaystyle 4\equiv -3 \bmod 7$ it follows that $\displaystyle 3^{2n+1}+2^{n+2}\equiv 3\cdot 2^n-3\cdot 2^n=0\bmod 7$. Hence disibility by 7 holds for all nonnegative integers $\displaystyle n$.
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