/sci/ - Science & Math

[Anonymous No.5100679]

hey /sci/
need some help with this prrof by induction divisibility problem please
prove for every +ve integer n, 3^(3n-2)+2^(3n+1) is a multiple of 19

>done base step
>did f(n+1)-f(n)
>simplified to f(n+1)=3^(3n-2)+8f(n)
>stuck

please help
pic unrelated

[Anonymous No.5100892]

not gonna do any work to check this but I was doing some hs algebra heavy inducting yesterday and noticed for a few problems it was a lot easier to do (n-1)->n rather than n->(n+1)

[Anonymous No.5100899]

TIP: 27 - 8 = 19.

[Anonymous No.5101340]

not seeing it....
could you give me the next line of working?

[Anonymous No.5102498]

>>5100899
okay so thats
3^3-2^3 = 19....

[Anonymous No.5102596]

not sure if this helps, but the final stage of your proof should be 19^n

3^(3n+2) + 2^(3n+1)
=(3^(3n))(3^2) + 2^(3n)*2^1
=9(27^n) +2(8^n)

[Anonymous No.5103090]

bumping with gifs
its annoying me that i cant do a divisivilty induction

[Anonymous No.5104529]

[Anonymous No.5104594]

base case is trivial

induction: 3^(3k-2) + 2^(3k+1) is multiple of 19
multiply the whole thing by 27, so we have 27*3^(3k-2) + 27*2^(3k+1)
27*3^(3k-2) + 27*2^(3k+1) - 19*2^(3k+1) is still a multiple of 19 (we're just subtracting a multiple of 19 from a multiple of 19)
27*3^(3k-2) + 27*2^(3k+1) - 19*2^(3k+1) = 27*3^(3k-2) + 8*2^(3k+1) = 3^(3k+1) + 2^(3k+4) = 3^(3(k+1)-2) + 2^(3(k+1)+1)
and we're done

[Anonymous No.5104614]

>>5104594
thank you very much!

[Anonymous No.5104633]

Ok. So you did the base step. Now assume for some integer k, <div class="math"> 3^{3n-2}+2^{3n+1} = 19 k </div> Then multiply both sides by 19, keeping in mind that 19 = 27 - 8 = 3^3-2^3. <div class="math"> (3^{3n-2}+2^{3n+1}) * (27 - 8) = 19^2 k </div> <div class="math"> 27 (3^{3n-2}+2^{3n+1}) = 19^2 k + 8 (3^{3n-2}+2^{3n+1}) </div> Use the original equation to get rid of n on the right side, and multiply through on the right side. <div class="math"> 27 \times 3^{3n-2} + 27 \times 2^{3n+1} = 19^2 k + 8 \times 19 k</div>
<div class="math"> <div class="math"> 3^{3 \left( n+1 \right) - 2} + 27 \times 2^{3n + 1} = 19 \times 27 k </div> Now subtract <span class="math"> 19 \times 2^{3n+1} [/spoiler] from both sides. <div class="math"> 3^{3 \left( n+1 \right) - 2} + 8 \times 2^{3n + 1} = 19 (27 k - 2^{3n+1})</div> <div class="math"> 3^{3 \left( n+1 \right) - 2} + 2^{3 \left( n + 1 \right) + 1} = 19 (27 k - 2^{3n+1})</div> and then you're done.</div>

[Anonymous No.5104641]

Ok. So you did the base step. Now assume for some integer k, <div class="math"> 3^{3n-2}+2^{3n+1} = 19 k </div> Then multiply both sides by 19, keeping in mind that 19 = 27 - 8 = 3^3-2^3. <div class="math"> (3^{3n-2}+2^{3n+1}) * (27 - 8) = 19^2 k </div> <div class="math"> 27 (3^{3n-2}+2^{3n+1}) = 19^2 k + 8 (3^{3n-2}+2^{3n+1}) </div> Use the original equation to get rid of n on the right side, and multiply through on the right side. <div class="math"> 27 \times 3^{3n-2} + 27 \times 2^{3n+1} = 19^2 k + 8 \times 19 k</div>
<div class="math"> 3^{3 \left( n+1 \right) - 2} + 27 \times 2^{3n + 1} = 19 \times 27 k </div> Now subtract <span class="math"> 19 \times 2^{3n+1} [/spoiler] from both sides. <div class="math"> 3^{3 \left( n+1 \right) - 2} + 8 \times 2^{3n + 1} = 19 (27 k - 2^{3n+1})</div> <div class="math"> 3^{3 \left( n+1 \right) - 2} + 2^{3 \left( n + 1 \right) + 1} = 19 (27 k - 2^{3n+1})</div> and then you're done.

[Anonymous No.5104653]

>+ve integer
>+ve
fuck you

[Anonymous No.5105915]

>>5104641
thanks a lot :)

>>5104653
so?