Question

prove that for all n in natural numbers, 6^n+7^(2n+3) is divisible by 43

Answer 1

Have you used exponent rules to simplify it at all?

Answer 2

What have you tried so far?

Answer 3

i tried the mathematical induction...but i'm still a bit confused :S

Answer 4

i think thats what i need to use to solve but i get a weird result

Answer 5

Okay, so for induction, what was your base case?

Answer 6

if n=1, and its so far divisible to 43
i think its proving if n=k+1 thats confusing me the most

Answer 7

You have to prove that if k is true, then k+1 is true.

Answer 8

So try plugging in k+1, and getting it back to the original statement in terms of k.

Answer 9

Hopefully you'd get something like 43 times the original statement in terms of k.

Answer 10

so if i have:
6^1+k + 7^2(k+1)+5, how do i break that up and find out if its true?

Answer 11

You use exponent rules like:\[\Large x^{n+m}=x^n\cdot x^m\]\[\Large (x^n)^m=x^{n\cdot m}\]

Answer 12

so...it would be 6(6^k) + 7^2k+3 + 7^2k+4?? is that close?

Answer 13

Hold on let me look...

Answer 14

i don't i was even taught the exponent rule before...lol

Answer 15

\[\Large 6^{(k+1)}+7^{2(k+1)+3}\]\[\Large 6^{(k+1)}+7^{2k+2+3}\]\[\Large 6^k\cdot 6^1+7^{2k+3}\cdot 7^2\]

Answer 16

So we've got to isolate stuff a bit.

Answer 17

oh ok so then what happens to the 6 and the 7^2?

Answer 18

I'm not completely sure how to deal with this part, or if we're going in the right direction with the induction.

Answer 19

do i just say taht its still true because 6^k +7^2k+3 is divisble by 43?

Answer 20

No, we can't say that, because we don't know that it is true.

Answer 21

I tried another question like this and i would get extra expressions so i didn't know what to do if its the same for this question.

Answer 22

Can write 49 as (6+43)...

Answer 23

Oh, estudier just did the hard part for us.

Answer 24

so overall it becomes (6^k+7^2k+3)(6) + (6+43)...does this prove that its true?

Answer 25

\[\Large 6^k\cdot 6+7^{2k+3}\cdot 6+7^{2k+3}\cdot 43\]

Answer 26

No.... do you know how to prove that something is divisible? Did you learn the definition of divisible?

Answer 27

not really...my prof just gave us a question and keeps telling me to read the textbook :P i'm using bartles

Answer 28

Suppose we have two numbers, \(k\) and \(n\). To say \(n\) is divisible by \(k\) is to say there exists a natural number \(m\), such that \(n = m \cdot k\)

Answer 29

We know this is true for \[ 6^n+7^{2n+3}\]

Answer 30

Well, by inductive hypothesis

Answer 31

Now we need to isolate that expression.

Answer 32

ok sorta getting it

Answer 33

We need to isolate
\[\Large 6^k+7^{2k+3}\] From
\[\Large 6^k\cdot 6+7^{2k+3}\cdot 6+7^{2k+3}\cdot 43\]

Answer 34

another question ...how did you end up getting 7^2k+3 twice? sorry for asking for so much, i just didn't learn this :P

Answer 35

\[\Large 7^{2k+3} \cdot 7^2 \Rightarrow 7^{2k+3} \cdot 49 \Rightarrow 7^{2k+3}\cdot (6+43) \]\[\Large \Rightarrow 7^{2k+3}\cdot 6+7^{2k+3}\cdot 43\]

Answer 36

So now we have:
\[\Large 6(6^k+7^{2k+3})+7^{2k+3}\cdot 43\]Are you following?

Answer 37

yup! :) i'm getting it so far

Answer 38

By our inductive hypothesis, we know that \(6^k+7^{2k+3}\) is divisible by 43. That means there is some \(m\) such that \(6^k+7^{2k+3} = 43 m\). We substitute this into what we have. This step is the reason for using induction.

Answer 39

So we have\[\Large 6(43m)+7^{2k+3}\cdot 43\]

Answer 40

Can you factor out the 43?

Answer 41

yes. then you're left with 43(6m +7^2k+3)

Answer 42

Okay, so why do You think I asked you to do that?

Answer 43

because the whole thing is now divisible by 43? therefore its true?

Answer 44

cause if i divided the whole thing b 43 i would be left with 6m+7^2k+3. if i know that m is divisble by 43 so is 7^2k+3...does that make sense?

Answer 45

Actually, we don't know that \(m\) is divisible by 43 and the same goes for \(7^{2k+3}\)

Answer 46

What's important is that \(6m+7^{2k+3}\) is a natural number.

Answer 47

So we have found the \(m_2 = 6m+7^{2k+3}\), such that \(6^{(k+1)}+7^{2(k+1)+3} = 43m_2\).

Answer 48

This is what proves that the \(P(k)\Rightarrow P(k+1)\).

Answer 49

So, with this and your base case combined, we are done.

Answer 50

Sorry for not replying back earlier but thank you very much! that really helped me! :)