# Prove by Induction the 11^(n+1)+12^(2n-1) is dvisible by 133.

## Homework Statement

Prove by Induction the 11^(n+1)+12^(2n-1) is visible by 133.

For all n>1 or n=1

## The Attempt at a Solution

I have shown that the base case of n=1 holds.
Then I assumed that the original statement "11^(n+1)+12^(2n-1)is divisible by 133" holds.

Now I need to show the statement holds for n=n+1 ---> 11^(n+2)+12^(2n+1) must be shown to be divisible by 133. I see it equals

11^(n+1)*11+12^(2n-1)*12^2

I pull out 11*12^2 to create something times the original statement (which is divisible)

11*12^2*[11^(n+1)+12^(2n-1)]

multiplied out this equals 12^2*11^(n+2) + 11*12^(2n+1)

I then subtract from this the necessary multiples of 11^(n+2) and 12^(2n+1) to recreate(re-balance) the (n+1) form 6 lines above.

the two terms I subtract are -143*11^(n+2) - 10*12^(2n+1).

If I can show these are divisible by 133, I have completed the proof. I have checked via computer that their sum is divisible for n=1 through 15, but can't seem to find a way to show it.

SammyS
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## Homework Statement

Prove by Induction the 11^(n+1)+12^(2n-1) is visible by 133.

For all n>1 or n=1

## The Attempt at a Solution

I have shown that the base case of n=1 holds.
Then I assumed that the original statement "11^(n+1)+12^(2n-1)is divisible by 133" holds.

Now I need to show the statement holds for n=n+1 ---> 11^(n+2)+12^(2n+1) must be shown to be divisible by 133. I see it equals

11^(n+1)*11+12^(2n-1)*12^2

I pull out 11*12^2 to create something times the original statement (which is divisible)

11*12^2*[11^(n+1)+12^(2n-1)]
You can't do that! Those expressions are not equivalent. If they were equivalent, you would be done here.
multiplied out this equals 12^2*11^(n+2) + 11*12^(2n+1)

I then subtract from this the necessary multiples of 11^(n+2) and 12^(2n+1) to recreate(re-balance) the (n+1) form 6 lines above.

the two terms I subtract are -143*11^(n+2) - 10*12^(2n+1).

If I can show these are divisible by 133, I have completed the proof. I have checked via computer that their sum is divisible for n=1 through 15, but can't seem to find a way to show it.

That is why I then add

-143*11^(n+2) - 10*12^(2n+1)

to recreate it [11^(n+1)*11+12^(2n-1)*12^2]

though now with the intent of showing that all the terms are divisible by 133. It is an equivalent statement when the negative terms are attached. Apologies if I did not make my method clear.

SammyS
Staff Emeritus
Science Advisor
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That is why I then add

-143*11^(n+2) - 10*12^(2n+1)

to recreate it [11^(n+1)*11+12^(2n-1)*12^2]

though now with the intent of showing that all the terms are divisible by 133. It is an equivalent statement when the negative terms are attached. Apologies if I did not make my method clear.
... but then you cannot factor out 11*122 which is what the next line shows.

When I pulled the 11*12^2 out front, I did not legitimately factor it out. I just wrote it as that product times the original statement which I am assuming to be divisible. Because this was not a legitimate operation I then multiply out

11*12^2*[11^(n+1)+12^(2n-1)]

and see what I need to add or subtract back to make it equal the original (n+1) quantity

11^(n+1)*11+12^(2n-1)*12^2.

So I end up with this (the line above) is equal to

11*12^2*[11^(n+1)+12^(2n-1)] -143*11^(n+2) - 10*12^(2n+1)

where that in bold can be assumed by the original assumption to be divisible by 133, then if I can show the non-bold portion is also, I can say that the original (n+1) quantity is divisible by 133.

D H
Staff Emeritus
Science Advisor
You're trying too hard, I think, and you aren't taking advantage of the fact that 11^(n+1)+12^(2n-1) is divisible by 133.

Hint: For 11^(n+1)+12^(2n-1) to be divisible by 133 means there exists some integer kn such that 11^(n+1)+12^(2n-1) = 133*kn. This in turn means that 11^(n+1) = 133*kn - 12^(2n-1).

Thank you both. I substituted for 12^(n-1)*12^2 and was able to write both terms as multiples of 133.