FP1 Proof by divisibilitiy

Original post by Damien_Dalgaard

I get 3(4^k) + 6 I suspect that I have gone wrong? :/

thats right man. you can finish it easily from there. just use modular arithmetic: consider the remainder when you divide 4^k by 18, and observe that it cycles through a fixed set of integers. Then prove it cycles through thsi fixed set of integers, and show that they are always congruent to 12mod18 when multiplied by 3, and then you are done

Reply 5

Original post by Damien_Dalgaard

I get 3(4^k) + 6 I suspect that I have gone wrong? :/

Give me like 5 minutes to screw about with Latex?

Reply 6

Damien_Dalgaard

OP

16

Original post by Jai Sandhu

Give me like 5 minutes to screw about with Latex?

You legend :biggrin:

Reply 7

Original post by CancerousProblem

thats right man. you can finish it easily from there. just use modular arithmetic: consider the remainder when you divide 4^k by 18, and observe that it cycles through a fixed set of integers. Then prove it cycles through thsi fixed set of integers, and show that they are always congruent to 12mod18 when multiplied by 3, and then you are done

That is a bit much for FP1 no? This is not STEP :tongue:

Reply 8

Original post by Jai Sandhu

That is a bit much for FP1 no? This is not STEP :tongue:

i haven't done fp1, but I tried the problem anyway and that's exactly what I did. I don't see an easier way out if you have to use induction.

Alternately you can just use induction again on 3(4^k)+6 to show it's always divisible by 18,
3(4^(k+1)+6
=3(4^k)+6 + 3(3)(4^k)
=3(4^k)+6 + 9(2^k)(2^k) which is obviously divisible by 18

(edited 8 years ago)

Reply 9

Original post by CancerousProblem

i haven't done fp1, but I tried the problem anyway and that's exactly what I did. I don't see an easier way out if you have to use induction

There is, writing it up atm.

Reply 10

Original post by Jai Sandhu

There is, writing it up atm.

Induction twice is not easier than modular arithmetic lol
modular arithmetic is really straightforward to use here, can't see a reason not to

Reply 11

Original post by CancerousProblem

Induction twice is not easier than modular arithmetic lol
modular arithmetic is really straightforward to use here, can't see a reason not to

He wont get the marks :/

Btw sorry for delay, latex is killing me T_T

Reply 12

Original post by Jai Sandhu

He wont get the marks :/

Btw sorry for delay, latex is killing me T_T

seriously? modular arithmetic banned in FP1? Are you kidding lol...

I hope step doesn't ban AM-GM since their questions use it so much, it would be annoying to have to prove AM-GM everytime i want to use it

edit: yes the question said to use induction, but you can induct once and once you get the 3(4^k)+6 component you can use modular arithmetic to sort it out. That would still be induction, right?

(edited 8 years ago)

Reply 13

Original post by Damien_Dalgaard

https://7cba9babeb0db0ff9468853e0b2d0a80708ec59c.googledrive.com/host/0B1ZiqBksUHNYZGxseFBIQkphV0k/June%202014%20(IAL)%20QP%20-%20F1%20Edexcel.pdf

9ii

f(k+1)-f(k)= 4^k+1 + 6(k+1) + 8 - (4^k +6k +8)

I don't know where to go from here, anyone help please :smile:

4n
+ 6n + 8

[br][br]f(k) = 4^k + 6k + 8[br][br]f(k+1) + f(k) = 4^{k+1} + 6(k+1) + 8 + 4^(k) + 6k + 8[br][br]f(k+1) + f(k) = 4^k +4 \times 4^k + 6(k+1) +6k + 8 + 8[br][br]f(k+1) + f(k) = 5 \times 4^k + 12k + 24[br][br]f(k+1) = 5 \times 4^k + 12k + 22 - f(k) [br][br]f(k+1) = 5 \times 4^k + 12k + 22 - (4^k + 6k + 8)[br][br]f(k+1) = 4 \times 4^k + 6k + 14[br][br]f(k+1) = 4 \times (4^k + 6k + 8) - 18k - 18[br][br]f(k+1) = 4 \times f(k) - 18(k - 1)[br][br]

I think you can do the rest?

(edited 8 years ago)

Reply 14

Patrick2810

16

Didn't you already get A in further maths?

Reply 15

Original post by Jai Sandhu

4n
+ 6n + 8

removed latex quote
I think you can do the rest?

meh honestly tat looks more complicated to me but that's just my opinion

(edited 8 years ago)

Reply 16

OK I fixed the latex!!!! Goddamnit.

Reply 17

Original post by CancerousProblem

meh honestly tat looks more complicated to me but that's just my opinion

Can you requote that with the edited version, I published a latex form which is a bit broken... thanks. Yeah it is more complicated but hey, its what the examiners want.

Reply 18

lukejoshjames

7

Original post by CancerousProblem

meh honestly tat looks more complicated to me but that's just my opinion

It is more complicated, as you have to find exactly what THEY want, and not the way you want to do it. Sometimes they have alternative methods but they're mostly as complicated.

Reply 19