Binomial theorem question

[Pranav-Arora]

1. The problem statement, all variables and given/known data

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

a)113
b)123
c)133

2. Relevant equations



3. 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:-

\Rightarrow121(11n) + (12)2n(12)
\Rightarrow121(11n) + (1+11)2n(12)

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

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+ 122n+ 1, for all n, it must be 133.

To prove that is so, use induction on n. If for, say, n= k, 11k+2+ 122k+1 is divisible by 133, we must have 11k+2+ 122k+1= 133m for some integer m. Then 11(k+1)+2+ 122(k+1)+ 1= 11(11k+ 2)+ 144(122k+1)= 11(11k+2+ 122k+1)+ 133(122k+1= 11(133m)+ 133(122k+1)= 133(11m+ 122k+1) showing that if the expression for n=k is divisible by 133, so is the expression for n= k+1.

[Pranav-Arora]

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..:confused:

[ehild]

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

ehild

[Pranav-Arora]

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

[Pranav-Arora]

What is the last term in the expansion?

ehild

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

[I like Serena]

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

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...

[Pranav-Arora]

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..:confused:

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:smile:.

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.

[Pranav-Arora]

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

ehild

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

[Pranav-Arora]

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... :smile:

[Pranav-Arora]

You have the wrong expression. ;)

It is: 11n+2 + 122n+1

And try n=0... :smile:

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\inN :smile:

[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\inN :smile:

Ah well, I have been taught to avoid the set ℕ. :smile:
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! :wink:

[Pranav-Arora]

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

[I like Serena]

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

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? :wink:

[Pranav-Arora]

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? :wink:

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\inN :confused:

[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..:confused:
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+ 122n+ 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+ 122n+ 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.

[Pranav-Arora]

Thanks for your help everyone!! :smile:

[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 cant 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