An induction question for u all (1 Viewer)

kooltrainer

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prove by induction 7^k + 3K(7^k) -1 is divisible by 9
shuldnt be too hard ? gimi sum suggestions
 

lolokay

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do it the normal way, but I think at some point you need to do an induction proof within the induction proof
 

lolokay

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@kurt: do you know induction (for divisibility)?
(you let the expression for k=n equal, say 9m, then show that the expression is also true for k=n+1 if it is true for k=n)

so yeah, you get the k+1 expression in the form of 9M, but in order to do this you will need to do a separate induction proof to show a part of the expression is also divisible by 9
 
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kurt.physics

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@kurt: do you know induction (for divisibility)?
(you let the expression for k=n equal, say 9m, then show that the expression is also true for k=n+1 if it is true for k=n)

so yeah, you get the k+1 expression in the form of 9M, but in order to do this you will need to do a separate induction proof to show a part of the expression is also divisible by 9
I believe this is a kind of divisibility/number theory/modular arithmetic/ induction problem. You would find these kind of problems at national mathematical olympiads.
 

addikaye03

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I believe this is a kind of divisibility/number theory/modular arithmetic/ induction problem. You would find these kind of problems at national mathematical olympiads.
bahaha umm.. also quite common in NSW HSC. National Olympiad pfft

P.S: 500th post, glad i dedicated it to a maths based forum
 
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VenomP

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lolokay, could you please explain to me how you got from here:

7^(n+1) + 3(n+1)7^(n+1) - 1 is divisble by 9

to here?:

7 x 7^n + 7 x 3n x 7^n - 7 + 3 x 7^(n+1) + 6

In particular, i can't quite see where you got the bolded parts from... I know it has to eventually equal 1, as in the previous line, but how do you get 6 and 7?
 

lolokay

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yeah i just did that so i could get a multiple of 9M. I just broke the -1 up, using -1 = -7 + 6

also i screwed up my post in the second part. i think it's fixed now
 

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