Exponential problem

hey guy help me to prove this question:


3^(2n+1) + 2^(n+2) is dividable by 7.
 

Zurtex

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What have you tried so far? And if nothing, may I suggest you try it with induction.
 
Can u prove it ?
 
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3^(2n+1)+2^(n+2)=3*9^n+4*2^n. To be divisible by 7, you must prove
(9^n mod 7) =(2^n mod 7).
 

Zurtex

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quentinchin said:
Can u prove it ?
Yes (assuming it's true), but this forum isn't about doing your work for you. It's just about helping.
 
sorry? can you clarify (9^n mod 7) =(2^n mod 7) and finish it?
 

Zurtex

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LittleWolf said:
3^(2n+1)+2^(n+2)=3*9^n+4*2^n. To be divisible by 7, you must prove
(9^n mod 7) =(2^n mod 7).
I'm a bit dubious about the logical jump there.
 

shmoe

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Zurtex said:
I'm a bit dubious about the logical jump there.
It's fine, you could then collect the common 2^n terms working mod 7, etc., but if you're already familiar with modular aritmetic, there shouldn't be much to prove.


It appears the quentinchin isn't familiar with modular arithmetic yet, so I offer the following hint: 9^n=(7+2)^n
 

Zurtex

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shmoe said:
It's fine, you could then collect the common 2^n terms working mod 7, etc., but if you're already familiar with modular aritmetic, there shouldn't be much to prove.


It appears the quentinchin isn't familiar with modular arithmetic yet, so I offer the following hint: 9^n=(7+2)^n
Well whether or not it's fine, as you mention it would appear that quentinchin doesn't appear too familiar with modular arithmetic. Which is why I suggested just a simple proof by induction at the start of the thread.
 

shmoe

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Zurtex said:
Which is why I suggested just a simple proof by induction at the start of the thread.
Which he seemed to ignore (sigh) so I offered another suggestion. Here's hoping he actually gives either way a try and posts his efforts.
 
hehe i found it but a bit complex.
3^(2n+1) + 2^(n+2)
= 3(9^n) + 4(2^n)
= 3(9^n) - 3(2^n) + 7(2^n)
= 3(9-2)[9^(n-1) + 9^(n-2)*2 + ... + 9*2^(n-2) + 2^(n-1)] + 7(2^n)
= 7[3(9^(n-1) + 9^(n-2)*2 + ... + 9*2^(n-2) + 2^(n-1) + 2^n]
hence the equation 3^(2n+1) + 2^(n+2) is dividable by 7.
 
shame to tell that i still do not understand what is modular arithmetic and induction.
Can give me an example?
 

Zurtex

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quentinchin said:
= 3(9^n) - 3(2^n) + 7(2^n)
= 3(9-2)[9^(n-1) + 9^(n-2)*2 + ... + 9*2^(n-2) + 2^(n-1)] + 7(2^n)
hence the equation 3^(2n+1) + 2^(n+2) is dividable by 7.
It may be because I'm very tired, but I don't see how you made that step.

Anyway, proof by induction. First of all you prove it for n = 1, then assuming it's true for n = k you prove it for n = k + 1. That proves it for all natural numbers.

Here is a nice page on it: http://en.wikipedia.org/wiki/Mathematical_induction

Really good for this exact sort of problem.
 
Zurtex said:
It may be because I'm very tired, but I don't see how you made that step.

Anyway, proof by induction. First of all you prove it for n = 1, then assuming it's true for n = k you prove it for n = k + 1. That proves it for all natural numbers.

Here is a nice page on it: http://en.wikipedia.org/wiki/Mathematical_induction

Really good for this exact sort of problem.
I really appreciate it, but do you guys overlook the question? The question is to <PROVE> that 3^(2n+1) + 2^(n+2) is divisible by 7 but not using induction method.
 

Zurtex

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quentinchin said:
I really appreciate it, but do you guys overlook the question? The question is to <PROVE> that 3^(2n+1) + 2^(n+2) is divisible by 7 but not using induction method.
No, proof by induction here is clearly the easiest way to prove it. Even if you use some other method to reduce the problem, induction is probably the easiest way to prove what you have.

If you prove it for n = 1, i.e:

33 + 23 = 35

Then assume it's true for n = k, i.e:

32k + 1 + 2k + 2 = 7a (for some integer a)

With the above prove it for n = k + 1, i.e:

32(k + 1) + 1 + 2(k + 1) + 2 = 7b (for some integer b)

If you can do it for that, that proves it [itex]\forall n \in \mathbb{N}[/itex].
 

Galileo

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To prove [itex]3^{2k+3}+2^{k+3}[/itex] is divisible by 7 you have to use the induction hypothesis that [itex]3^{2k+1}+2^{k+2}[/itex] is divisible by 7.
If you want to use that equation, just try to massage the first one so you can get the expression of the induction hypothesis in there. After that, the rest is trivial.
 

shmoe

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quentinchin said:
I really appreciate it, but do you guys overlook the question? The question is to <PROVE> that 3^(2n+1) + 2^(n+2) is divisible by 7 but not using induction method.
Your orignial question put no restrictions on what techniques could be used. Besides, you've solved it already so what's the problem?

(By the way, my hint had nothing to do with induction, but the binomial theorem.)
 

mathwonk

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in my opinion, it is impossible to prove any statement true for all integers without somehow using induction. even defining addition for all integers needs induction.
 
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I agree with Shmoe-BINOMIAL THEOREM. (7+2)^n=...?
 
Maybe mathematic induction is the best way to prove this sort of question but if we too often using this kind of method, our mathematic level wouldn't go any further.

shmoe said:
Your orignial question put no restrictions on what techniques could be used. Besides, you've solved it already so what's the problem?

(By the way, my hint had nothing to do with induction, but the binomial theorem.)
I really do not know how to solve this question and I just get the answer from my friend after I asked this question.

Is that
3*9^n + 4*2^n
=3(7+2)^n + 4*2^n
=3[7^n + n*7^(n-1)*2 + ... + n*7*2^(n-1) + 2^n] + 4*2^n
=3[7^n + n*7^(n-1)*2 + ... + n*7*2^(n-1)] + 7*2^n
=7{3[7^(n-1) + n*7^(n-2) + ... + n*2^(n-1)] + 2^n}

this is also another good method.
 

Zurtex

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quentinchin said:
Maybe mathematic induction is the best way to prove this sort of question but if we too often using this kind of method, our mathematic level wouldn't go any further.



I really do not know how to solve this question and I just get the answer from my friend after I asked this question.

Is that
3*9^n + 4*2^n
=3(7+2)^n + 4*2^n
=3[7^n + n*7^(n-1)*2 + ... + n*7*2^(n-1) + 2^n] + 4*2^n
=3[7^n + n*7^(n-1)*2 + ... + n*7*2^(n-1)] + 7*2^n
=7{3[7^(n-1) + n*7^(n-2) + ... + n*2^(n-1)] + 2^n}

this is also another good method.
Seems reasonable but I certainly would not want to do that for every time I got asked to prove:

[tex]a x^{f(n)} + b y^{g(n)} \quad \text{is a multiple of k} \, \, \forall n \in \mathbb{N}[/tex]

I'm curious, what level of maths do you do?
 
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hello all

after reading through this thread 2 things came to mind, i would like to say firstly, there are always opportunities to learn something new, but what i find amazing is that some people choose to close there doors on such opportunities even though knowledge is humans most valuable asset, secondly not all proofs have to be Rigorous to be a proof, if you can prove any expression for all natural numbers through mathematical induction what more could you want, if you want to build your mathematical mind you should try to challenge your self to prove something in as many ways as possible weither simple or hard, coming from a person who wants to be a mathematician would i be wrong to say such things?

Steven
 
Zurtex said:
I'm curious, what level of maths do you do?
I know that my level just pre-U, but in my point of view about math is, it's doesn't matter how much knowledge or formula I know but what the most important thing is I must try to apply what I gained in surrounding of me. Just like mathematic induction I will also using it to solve other question.

ps: ( I really appreciate what you taught me and comment on me.)

steven187 said:
if you want to build your mathematical mind you should try to challenge your self to prove something in as many ways as possible weither simple or hard, coming from a person who wants to be a mathematician would i be wrong to say such things?
i totally agree with you.
 

mathwonk

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i continue to claim no proof is possible without induction. i.e. the binomial theorem requires induction, etc, etc...

the only logical possibilty would be to assume several significant results about integers, all requiring induction themselves, and then deduce the result from those.
 

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