# Math Help - Divisible by 7

1. ## Divisible by 7

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

Can anybody help me?

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

Can anybody help me?
try using induction

$n=1$

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

$n=2$

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

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

and if you assume that's true then show that is true and for $n=k+1$ and if you show that... it will mean that it's true for any natural number

3. 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.

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

Can anybody help me?
It may be simple to reduce $3^{2n+1}$ and $2^{n+2}$ modulo 7. Then try to find the remainder of $3^{2n+1}+2^{n+2}$ when divided by 7. Good Luck!

5. 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....

6. hm...

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

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

7. Originally Posted by yeKciM
hm...

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

P.S. $0 \in \mathbb{N}$ 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

8. 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 $\mathbb{N}$ is subset of set $\mathbb{R}$ with "features" :

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

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

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 $M\subset N$ and if in M apply properties 1° and 2° than M=N

axiom 5 is known as the principle of induction

and so on...

Edit : sorry for this... i was going somewhere else with this.
anyway can't do you harm to know this ( because this is sub forum "university math" so i thought that you would, and should know this )

9. That's too advanced for me... hehe... Is my proof good?

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

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?

11. 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...

12. that is it

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

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

that's it ... because

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

and

$2\cdot A$ is also divided by 7 because you assumed that n=k is true

13. Hello, yuud!

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

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

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

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

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

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

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

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

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

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

14. Originally Posted by Soroban
Hello, yuud!

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

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

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

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

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

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

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

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

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

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

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

mod 7 but rather in such a way as to yield non-integers, thus the only integers that satisfy are the non-negative ones.

15. Originally Posted by Soroban
Hello, yuud!

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

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

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

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

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

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

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

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

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

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

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