AkashdeepDeb
  • AkashdeepDeb
HELP NEEDED: If X = {8^n - 7n - 1|n ∈ N} and Y = {49(n-1)|n ∈ N}, then prove that X ⊆ Y. Do we just have to Plug-in values for n or is there any alternative method?
Mathematics
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SOLVED
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jamiebookeater
  • jamiebookeater
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AkashdeepDeb
  • AkashdeepDeb
@satellite73 @experimentX
AkashdeepDeb
  • AkashdeepDeb
Anyone? :|
anonymous
  • anonymous
lets see what we can do somehow it must be true that if \(z=8^n - 7n - 1\) then \(z\) can be written as \(49(n-1)\)

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More answers

AkashdeepDeb
  • AkashdeepDeb
Yeah
AkashdeepDeb
  • AkashdeepDeb
Can we do this By combination?
anonymous
  • anonymous
i don't know i am not that fast my first thought was to factor \(8^n-1\) as the difference of two cubes
anonymous
  • anonymous
but now i think @experimentX has a better idea
experimentX
  • experimentX
lol ... no i don't I just came back :) after walk
AkashdeepDeb
  • AkashdeepDeb
heheh
AkashdeepDeb
  • AkashdeepDeb
Thanks @satellite73 ! I'll go through it once more and see!
anonymous
  • anonymous
maybe it would be easier by induction?
AkashdeepDeb
  • AkashdeepDeb
Principal Of Mathematical Induction? I'll try THAT out! Thanks!! :D
anonymous
  • anonymous
i will bet induction and some algebra will work certainly true for \(n=1\)
experimentX
  • experimentX
show that \[ 1 + 8 + ... +8^{n-1} = n \mod 7 \]
AkashdeepDeb
  • AkashdeepDeb
It DOES work with 1 and 2
experimentX
  • experimentX
for all n , 8^n = 1 mod 7 add it n times from 0 to n-1
anonymous
  • anonymous
Certainly works by induction: you can prove that \(8^n - 7n - 1\) is divisible by 49 for any \(n\). But experimentX's method looks neater...
anonymous
  • anonymous
k but how to prove that \(8^k\equiv 1 (7)\)?
anonymous
  • anonymous
and you are not allowed to say "by induction!"
experimentX
  • experimentX
ab mod c = ((a mod c)(b mod c)) mod c
anonymous
  • anonymous
yeah, but what about the \(k\)? ok i am being silly
experimentX
  • experimentX
hmm ... (abcd ... k) mod c = (a mod c)(b mod c) ... (k mod c) mod c since all terms are 8 and the remainder is 1, the remainder is also 1. I think this can be done better via euler totient function. I am little out of touch on this.
anonymous
  • anonymous
@satellite73 I guess your point is that 8^n = 1 mod 7 implicitly requires induction (e.g. ab mod c = ((a mod c)(b mod c)) mod c applied "repeatedly") to be proved formally? ;-)
AkashdeepDeb
  • AkashdeepDeb
Thanks for your help. I got it by using Principal of Mathematical Induction.

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