How do you prove that the number in 28 is divisible by 6? The same goes for the one in 29, but by 19. I've been searching all around the internet how to go about it, but I'm really put off by it.
Also, should this go in the stupid questions thread? If yes, I will make sure next time I will post there.
they're not. make a stupid questions for a stupid questions thread
Well, according to my book, they are, and I have to prove it.
And if that is the case, I shall do so. Thank you for replying, user!
These are both pretty trivial when you consider the sums modulo the relevant number (6 or 19).
use induction
I know they are trivial, they are in the exercises chapter for 8th grade for fuck's sake, but I can't remember these things as I haven't used them in a while and I'm a brainlet.
Also...my English is extremely poor when it comes to maths, so what you said sounds really...exotic.
If
[eqn]a^n = r + 6d [/eqn]
then
[eqn](a-6)^n = \sum_{k=0}^n {n \choose k} a^{n-k} 6^k \\
= r + 6d + 6 \sum_{k=1}^n {n \choose k} a^{n-k} 6^{k-1}[/eqn]
So if you divide 13^n by 6 then you get the same remainder as when you divide (13 - 6)^n = 7^n by 6 which also has the same remainder as (7 - 6)^n = 1^n.
It's not hard to see if you think about it for a bit.
13^n + 7^n -2 = (6*2 + 1)^n + (6+1)^n -2 and by using Binomial Theorom you show that this is just the sum of a bunch of terms that are divisible by 6 so the sum is divisible by 6.
Which is just a long-winded way of saying:
[eqn]\begin{align*}13^n+7^n-2&\equiv1^n+1^n-2 &\mod 6 \\ &\equiv 1+1-2 &\mod 6 \\ &\equiv 0 &\mod 6\end{align*}[/eqn]
and
[eqn]\begin{align*}7*25^n+6^{n+1}&\equiv7*6^n+6*6^n &\mod 19 \\ &\equiv19*6^n &\mod 19 \\ &\equiv 0 &\mod 19\end{align*}[/eqn]
^ [math]2*6^{n+1}[/math]
Whatever bro I just think it's more elegant using only Binomial Theorom and not modular arithmetic since if OP is asking these sort of questions then he probably doesn't know what that is yet.
[math]6|(7^n+13^n-2) \\ P(1):7^1+13^1-2=18=3*6 \\ P(k)=7^k+13^k-2=6k \\
P(k+1)=7^{k+1}+13^{k+1}-2=7(6k)+6(7^k)+12=6(7k+7^k+2) [/math]
That is true. I'm pretty lost...should I look up modular arithmetic or just go with newton's binome? I mean I have plenty of time, but it looks really alien to me...
Although judging from the previous questions, the proof by induction method () is probably the one being required.
On your last line it should be 7(6k) + 6(13^k) + 12 by the way.
Start with Binomial Theorom that is a lot easier to understand.
I shall do so then. Thank you very much! Also, in hindsight, I also think it is supposed to be done with induction.
Yeah you're right. Do you know how to do simple induction? outlined the solution nicely except in the last line it should be 6(13^k) not 6(7^k).
n is a subset of what?
Never mind, I just woke up and my attention is bad.
Sorry for retarded question.
I ended up solving it with binomial. I know the steps for induction, I just lost my way at p(k+1) as I don't really know how to solve induction for divisiblity. I know I'm pretty dumb. I don't really wanna bother you any longer as you've helped me a great deal, so I'm not gonna enquire any further and maybe just gonna ask my teacher on monday.
However, if you don't mind it, can you explain in kids' steps why is it 6*(13^k) in the last line and where did that 12 come from? Yeah, this is how bad I am at the moment.
It's fine, this is a retarded question thread (sort of) anyway.
Since we already have P(k) = 7^k + 13^k - 2 = 6k and we want to show that P(k+1) is divisible by 6 then we need to write P(k+1) in the form 6*n where n is some integer. We want to get 7^(k+1) for a start so if we multiply P(k) by 7 we get 7^(k+1) + 7*13^k -14 = 7*6k. Next we want to get to 13^(k+1) so add on 6*13^k to each side to get 7^(k+1) + (7+6)*13^k - 14 = 7*6k + 6*13^k and so 7^(k+1) + 13^(k+1) - 14 = 7*6k + 6*13^k. We are almost there, we just need the -2 on the end so add 12 to each side to get 7^(k+1) + 13^(k+1) - 2 = 7*6k + 6*13^k + 12 = 6*(7k + 13^k + 2) which is in the form 6n so P(k+1) is divisible by 6. I don't know Latex so this doesn't look pretty but I hope you can follow, we have basically used P(k) to recreate P(k+1) and at every step the r.h.s is still divisible by 6. Stuff like this becomes easier with a lot of practice.
OP, there are many ways you may go.
Either the ones already listed or you could prove that it is divisible by 2 and 3.
It's really easy to see that it is indeed divisible by 2.
Now you just need to prove it's divisible by 3, which is just a bit harder, really.
Divisible by 2 and 3 => divisible by 6. Q.E.D.
Would write the exact formulation, but too lazy.
I can't believe you actually did this.
Really, thanks a lot. Now I actually get what I am supposed to do. I'm sorry for wasting your time, but you helped me immeasurably. I'll try solving the second one the same way and see what I can make of it.
I'm really tired and I'm clearly not at my prime, maybe had this thread been more during the day I wouldn't have required such babysitting, but I really appreciate your help. Hope you have a good day user! Stay safe.
OK man I'm glad my ramblings made sense haha. Good luck with the next one.
29. 2.6^n+1 + 7*25^n = (19 - 7)*6^n + 7*25^n
= 19*6^n + 7*(25^n - 6^n)
= 19*6^n + 7*((19 +6)^n - 6^n)