Question

prove for every 'n' {3^3n-2+3^6n-4+1} divide by 13

Answer 1

lord this is going to be a raft of algebra because you probably have to do it by induction right?

Answer 2

yea..

Answer 3

i can't possibly write it all out here, and the algebra is going to suck, but the idea is this
first show it is true if \(n=1\) which it is because you get \(13\)

Answer 4

then replace \(n\) by \(k\)
that is what you get to assume is true
\[\large 3^{3k-2}+3^{6k-4}-1\] is divisible by 13

Answer 5

yea and after that i did for k+1 and i stuck

Answer 6

next replace \(n\) by \(k+1\) and get
\[\large 3^{3k+1}+3^{6k+2}-1\]

Answer 7

true.. but i hve problem to make it like the basis of the induction

Answer 8

then see if you can rewrite this as \(3^{3k-1}+3^{6k-2}-1+something\)

Answer 9

there i need your help man

Answer 10

typo there, but you get the idea
should be
\[\large 3^{3k-1}+3^{6k-4}-1+something\] the something will obviously be divisible by 13, and the first part is by induction

Answer 11

ok let me try with pencil and paper, i can't do it on the fly

Answer 12

great! take your time.. its a lot of games between the numbers..

Answer 13

almost there

Answer 14

got it
you still there?

Answer 15

yeaa man..

Answer 16

ok we are at this step
\[\large 3^{3k+1}+3^{6k+2}-1\] right

Answer 17

why -1?

Answer 18

oh it should be \[\large 3^{3k+1}+3^{6k+2}+1\]

Answer 19

it is a \(+1\) out at the end yes?

Answer 20

yea

Answer 21

ok now we want the previous exponents \(3k-2\) and \(6k-4\) so step one is to write each as
\[\large 3^3\times 3^{4k-2}+3^6\times 3^{6k-4}+1\]

Answer 22

so far so good?

Answer 23

yea

Answer 24

ok now factor out the \(3^3\) and insert a \(1\) in the parentheses to make it look like the previous case \[\large 3^3\times 3^{4k-2}+3^6\times 3^{6k-4}+1\]
\[=3^3\left(3^{4k-2}+3^{6k-4}+1\right)-3^3+(3^6-3^3)\times 3^{6k-4}+1\]

Answer 25

the \(3^6-3^3\) is probably the confusing part, or maybe not
but when you factor out the \(3^3\) you have inside the parentheses \(3^3\times 3^{6k-4}\) but you need \(3^6\times 3^{6k-4}\) so you have to add back \((3^6-3^3)\times 3^{6k-4}\) at the end

Answer 26

once we have
\[\large =3^3\left(3^{4k-2}+3^{6k-4}+1\right)-3^3+(3^6-3^3)\times 3^{6k-4}+1\]we are pretty much done
the first part is divisible by 13 by induction, as it is \(27\) times something we know is divisible by 13

it turns out luckily that \(3^6-3^3=702\) is also divisible by 13

Answer 27

and also \(1-3^3=-26\) is also divisible by 13

Answer 28

guess it would be better to write
\[\large =3^3\left(3^{4k-2}+3^{6k-4}+1\right)-3^3+(3^6-3^3)\times 3^{6k-4}+1\]
\[=\large =3^3\left(3^{4k-2}+3^{6k-4}+1\right)+702\times 3^{6k-4}-26\]

Answer 29

or even \[\large =3^3\left(3^{4k-2}+3^{6k-4}+1\right)+13\times(54\times 3^{6k-4}-2)\]

Answer 30

told you the algebra was going to suck, and i didn't lie

Answer 31

lol yea! thanks a lot for your time . you really helped me! thx again!

Answer 32

yw