Binomial Theorem: Is 11n+2 + 122n+1 divisible by 113, 123, or 133 for n in N?

[Saitama]

Homework Statement

If n $\in$ N, then 11ⁿ⁺² + 12²ⁿ⁺¹ is divisible by:-

a)113
b)123
c)133

Homework Equations

The Attempt at a Solution

I did it by substituting different values of n and divided by each of the option. Answer came out to be 133.
But I want to do it step by step using binomial theorem.
When i tried to do it i did it the following way:-

$\Rightarrow$121(11ⁿ) + (12)²ⁿ(12)
$\Rightarrow$121(11ⁿ) + (1+11)²ⁿ(12)

Now i am not able to think what should i do next?

Thanks!

 

[HallsofIvy]

What I would have done is this- when n= 1 this is 11²+ 12= 121+ 12= 133 so if one of those given numbers divides 11ⁿ⁺²+ 12^{2n+ 1}, for all n, it must be 133.

To prove that is so, use induction on n. If for, say, n= k, 11^k+2+ 12^2k+1 is divisible by 133, we must have 11^k+2+ 12^2k+1= 133m for some integer m. Then 11^(k+1)+2+ 12^{2(k+1)+ 1}= 11(11^{k+ 2})+ 144(12^2k+1)= 11(11^k+2+ 12^2k+1)+ 133(12^2k+1= 11(133m)+ 133(12^2k+1)= 133(11m+ 12^2k+1) showing that if the expression for n=k is divisible by 133, so is the expression for n= k+1.

 

[Saitama]

But for proving how did you took 133?
I need to solve the question as if i don't know anything about the number 133..

 

[ehild]

Use the binomial theorem to expand (11+1)²ⁿ⁺¹.

ehild

 

[Saitama]

Use the binomial theorem to expand (11+1)²ⁿ⁺¹.

ehild

Till where should i expand it?

 

[ehild]

What is the last term in the expansion?

ehild

 

[Saitama]

What is the last term in the expansion?

ehild

The last term is (11)²ⁿ⁺¹ .
But what's the use of it?

 

[I like Serena]

The last term is (11)²ⁿ⁺¹ .
But what's the use of it?

Actually, that is the first term. ;)

What is the last, and let's say the one before the last?

Perhaps you'll see a pattern and perhaps you might be able to say something how to factorize the resulting expression...

 

[Saitama]

Actually, that is the first term. ;)

What is the last, and let's say the one before the last?

Perhaps you'll see a pattern and perhaps you might be able to say something how to factorize the resulting expression...

No, this is the last term only, if we solve it like this (1+11)²ⁿ⁺¹ we can use (1+x)ⁿ expansion...

 

[Ray Vickson]

But for proving how did you took 133?
I need to solve the question as if i don't know anything about the number 133..

He explained it very clearly: for n = 1 the expression 11^(n+2)+ 12^(2n+ 1) equals 133. You are allowed to substitute n = 1 into the expression, aren't you?

RGV

 

[ehild]

I meant the last term in (11+1)²ⁿ⁺¹. That one which is not divisible with 11.

ehild

 

[zonk]

No, this is the last term only, if we solve it like this (1+11)²ⁿ⁺¹ we can use (1+x)ⁿ expansion...

Depending on which summation formula you use it's either the first term or the last term. That is of course you're summing a positive integer power, which you are.

 

[Saitama]

I meant the last term in (11+1)²ⁿ⁺¹. That one which is not divisible with 11.

ehild

I will try to work on your suggestions, if i will get any doubt i will ask you...
Thanks for your help everyone!

 

[Saitama]

He explained it very clearly: for n = 1 the expression 11^(n+2)+ 12^(2n+ 1) equals 133. You are allowed to substitute n = 1 into the expression, aren't you?

RGV

When i substitute n=1 in 11ⁿ⁺¹ + 12²ⁿ⁺¹, i am not able to get 133...
How you get 133?

 

[I like Serena]

When i substitute n=1 in 11ⁿ⁺¹ + 12²ⁿ⁺¹, i am not able to get 133...
How you get 133?

You have the wrong expression. ;)

It is: 11ⁿ⁺² + 12²ⁿ⁺¹

And try n=0...

 

[Saitama]

You have the wrong expression. ;)

It is: 11ⁿ⁺² + 12²ⁿ⁺¹

And try n=0...

Sorry, i will be careful next time but if i put n=1, i am not able to get 133.
And i cannot substitute n=0 because n$\in$N

 

[I like Serena]

Sorry, i will be careful next time but if i put n=1, i am not able to get 133.
And i cannot substitute n=0 because n$\in$N

Ah well, I have been taught to avoid the set ℕ.
That is because different books say different things about whether 0 is included or not, and you did not specify! So I guess you have been taught that ℕ = ℤ⁺?

Still, with n=0 you'll get 133, and with n=1 you'll get a multiple of 133!

 

[Saitama]

Yes we have been taught N=1,2,3,...
But then too, can we substitute n=0?

 

[I like Serena]

Yes we have been taught N=1,2,3,...
But then too, can we substitute n=0?

Look at it this way: if you can proof it for {0,1,2,3,...} which you can, it will surely be true for N too, wouldn't it?

 

[Saitama]

Look at it this way: if you can proof it for {0,1,2,3,...} which you can, it will surely be true for N too, wouldn't it?

I think i can't substitute n=0 because Natural numbers is a set of {1,2,3,4...} and it is mentioned in the question n$\in$N

 

[ehild]

It is possible that the expression is not divisible by any of those numbers. Calculate the expression for n=1 and see. If the result is divisible by 133 or by any of the other ones prove that it is so for any n by induction. But you can cancel the other ones at once because of the binomial theorem.

ehild

 

[HallsofIvy]

But for proving how did you took 133?
I need to solve the question as if i don't know anything about the number 133..

That was the first thing I answered:
"What I would have done is this- when n= 1 this is 11²+ 12= 121+ 12= 133 so if one of those given numbers divides 11ⁿ⁺²+ 12^{2n+ 1}, for all n, it must be 133. "

 

[ehild]

That was the first thing I answered:
"What I would have done is this- when n= 1 this is 11²+ 12= 121+ 12= 133 so if one of those given numbers divides 11ⁿ⁺²+ 12^{2n+ 1}, for all n, it must be 133. "

Sorry, HallsofIvy, I intended to refer to you, but I just forgot somehow. My apologies.

But the expression is 133 for n=0, and not for n=1.


ehild

 

[HallsofIvy]

Ah- and 0 is not a positive integer. Okay, if n= 1, $11^3+ 12^3= 3059$ which is divisible by 133 but not by 113 or 123.

 

[Saitama]

Thanks for your help everyone!

 

[Samo84]

IF n is a natural number you can substitute it by o , n= 0 ; of course you get 133 in your expression.
then you use the demonstration by recurrence as hallsofIvy did show, there is another method to do this kind of problems but i can't remember its name in English. But the recurrence method works fine so no need for it, just focus and you will understand it.

 

[Ray Vickson]

He should have said n = 0. That gives 133.

RGV

 

[Ray Vickson]

Of course you can substitute n = 0 into f(n) (to get 133). You could substitute n = 3*pi/2 or n = log(147) or anything you want. The question just says that something happens for n in N; it does NOT say you are forbidden from looking at n not in N. Anyway,putting n = 0 give you a number, 133. You can then say: "Oh, let's look at the number 133 to see what happens". You will see that 133 divides f(1), and that is a hint that you might have something that will work for other n. You can try the other two suggestions to see that they fail, even for n = 1.

Anyway, when you said before that you are not allowed to "know" about the number 133, that is false: it is given to you as one of the possibilities.

RGV