Serious Daily math problem #0003

[Divergent_Integral]

@WorldYamataizer @Diocel @grondilu @mental_out @SelfCrucified @your personality

I've decided to make a new thread for every daily math problem, so as to have each problem on the thread's first page. I hope it's okay by the mods.

Problem #0003​

Prove that, for all natural numbers n,
3^(2n + 1) + 2^(n - 1) is divisible by 7 (with remainder 0).

Hint: you may want to use mathematical induction. First, you prove that the statement is true for n = 1. Next, you prove that, assuming the statement is true for some natural number N, it is also true for N + 1.

 

[Mountainbikecel]

n=1: 3^3 + 1 = 28

 

[grondilu]

3^(2(n+1)+1)+2^(n+1 - 1) =
3^(2n+1+2)+2^n =
9*3^(2n+1) + 2^n=
9*(7k - 2^(n-1)) + 2^n=
9*7k - 9*2^(n-1) +2^n=
9*7k + 2^(n-1)(-9 + 2)=
9*7k - 7*2^(n-1)=
7*(9k - 2^(n-1))
which is manifestly a multiple of 7

PS. It would arguably be more elegant and short to do it all in Z/7Z but that may be too obscure for most people.

 

[Deleted member 6657]

3^(2(n+1)+1)+2^(n+1 - 1) =
3^(2n+1+2)+2^n =
9*3^(2n+1) + 2^n=
9*(7k - 2^(n-1)) + 2^n=
9*7k - 9*2^(n-1) +2^n=
9*7k + 2^(n-1)(-9 + 2)=
9*7k - 7*2^(n-1)=
7*(9k - 2^(n-1))
which is manifestly a multiple of 7

I did plan on taking an approach like this one where I would show that the sequence can be written as 7*y with y being the sequence in question some factor in terms of n. I only just got the chance to sit down but you already have written the solution I had in mind. What you wrote is a good demonstration of the induction method where you evaluate for n = n + 1. tbh I havent solved such a problem since 9 years ago early in high school.

Thanks for the notification, btw @Divergent_Integral. This problem only took about 35 minutes for grondilu to come up with a solution. I imagine that is a shorter time than the first and second problems posted, as we were expecting a problem and received a forum notification.

 

[grondilu]

I did plan on taking an approach like this one where I would show that the sequence can be written as 7*y with y being the sequence in question some factor in terms of n. I only just got the chance to sit down but you already have written the solution I had in mind. What you wrote is a good demonstration of the induction method where you evaluate for n = n + 1. tbh I havent solved such a problem since 9 years ago early in high school

I started to do it in my mind but I quickly lost track. Still I was fairly confident this was the way to go so I started writing it and the result showed up quite straightforwardly.

I happens sometimes in math. There is not much to think about, just start making the calculations and everything unfolds automatically.

 

[Deleted member 19971]

Everyone in this thread mogs me

 

[Deleted member 27204]

3^(2(n+1)+1)+2^(n+1 - 1) =
3^(2n+1+2)+2^n =
9*3^(2n+1) + 2^n=
9*(7k - 2^(n-1)) + 2^n=
9*7k - 9*2^(n-1) +2^n=
9*7k + 2^(n-1)(-9 + 2)=
9*7k - 7*2^(n-1)=
7*(9k - 2^(n-1))
which is manifestly a multiple of 7

Where did you get the idea to multiply the 3^(2n+1+2) term by 9 but not the 2^n term? And how is that even possible that you could only multiple a single term of the equation by a number and not all other terms?

 

[grondilu]

Where did you get the idea to multiply the 3^(2n+1+2) term by 9 but not the 2^n term? And how is that even possible that you could only multiple a single term of the equation by a number and not all other terms?

The 9 comes from the 2 in the exponent of 3. It's not a term that was multiplied everywhere, rather it was factored out of the first term.

3^(2n+1+2)=3^(2n+1)*3^2 = 9*3^(2n+1)

 

[Deleted member 27204]

The 9 comes from the 2 in the exponent of 3. It's not a term that was multiplied everywhere, rather it was factored out of the first term.

3^(2n+1+2)=3^(2n+1)*3^2 = 9*3^(2n+1)

Thanks tbh that wasn't intuitive at first but I can see your logic in that now.

 

[Deleted member 28148]

haven't learned this yet

 

[Deleted member 31630]

I wish I could contribute a meaningful solution to this problem, but unfortunately, I didn’t really learn much number theory to feel comfortable tackling this problem.

I knew how to solve it the way @grondilu did, but I don’t want to parrot his method. (maybe we could use modular arithmetic?) But I am a stranger to number theory so I wouldnt really know.

 

[Divergent_Integral]

I wish I could contribute a meaningful solution to this problem, but unfortunately, I didn’t really learn much number theory to feel comfortable tackling this problem.

I knew how to solve it the way @grondilu did, but I don’t want to parrot his method. (maybe we could use modular arithmetic?) But I am a stranger to number theory so I wouldnt really know.

First of all: well done, @grondilu. Actually @WorldYamataizer touches on an interesting point. I've seen a few other problems of the same kind, all variations on the theme of a linear combinations of prime powers being divisible by a prime (usually 7). While these can all be proved by induction, one wonders whether they arise out of a more general theorem or principle.

 

[grondilu]

I have to say that although I appreciate the opportunity to mathmog other members of the forum , tbh there are websites dedicated to that kind of stuff, like project euler. I'm on it btw :

 

[Michinomiya Hirohit]

n=1: 3^3 + 1 = 28

 

[Deleted member 23712]

@WorldYamataizer @Diocel @grondilu @mental_out @SelfCrucified @your personality

I've decided to make a new thread for every daily math problem, so as to have each problem on the thread's first page. I hope it's okay by the mods.

Problem #0003​

Prove that, for all natural numbers n,
3^(2n + 1) + 2^(n - 1) is divisible by 7 (with remainder 0).

Hint: you may want to use mathematical induction. First, you prove that the statement is true for n = 1. Next, you prove that, assuming the statement is true for some natural number N, it is also true for N + 1.

GrAY ass nigger

 

[Divergent_Integral]

I have to say that although I appreciate the opportunity to mathmog other members of the forum, tbh there are websites dedicated to that kind of stuff, like project euler. I'm on it btw :

View attachment 386626

Of course there are dozens of dedicated sites for this kind of stuff; I know that. But I see incels.co as a little internet town full of likeminded people, and every socially healthy town has clubs devoted to all manner of subjects. A chess club, a book club, etc. Let this be our math club then.

 

[grondilu]

Of course there are dozens of dedicated sites for this kind of stuff; I know that. But I see incels.co as a little internet town full of likeminded people, and every socially healthy town has clubs devoted to all manner of subjects. A chess club, a book club, etc. Let this be our math club then.

This forum is the incel club. I'm not sure it's a good idea to mix clubs. You don't bring your chessboard to your book club.

 

[Divergent_Integral]

This forum is the incel club. I'm not sure it's a good idea to mix clubs. You don't bring your chessboard to your book club.

If that were so, there wouldn't be much sense in separate subforums for incel discussion proper and mere "lounging". People discuss all sorts of shit in the Lounge (that's what it's for), so why not math?

 

[grondilu]

If that were so, there wouldn't be much sense in separate subforums for incel discussion proper and mere "lounging". People discuss all sorts of shit in the Lounge (that's what it's for), so why not math?

Fair point. For a moment I had forgotten this is indeed the Lounge.

 

[Deleted member 31630]

@Divergent_Integral

Mind if i handle problem #0004 tomorrow?
I have a cheeky geometry problem in mind.

 

[Divergent_Integral]

Fair point. For a moment I had forgotten this is indeed the Lounge.

Thanks. It's nice to have you on board. You seem like a really clever fellow, for whom it's fun to set problems.

@Divergent_Integral

Mind if i handle problem #0004 tomorrow?
I have a cheeky geometry problem in mind.

No, not at all. I'm curious what you have in mind for us.

 

[happiless]

No math for your face