Question

Let N be the integer next above \((\sqrt{3} + 1)^{2014}\). The greatest integer p such that 16^p divides N is _____.

Answer 1

Okay I am pretty sure that I am struck again. I cannot figure it out that why does this always happen with me ?? Its really frustrating...

Answer 2

I am going to do something about this N first

Answer 3

First of all for finding N, I can write N as:

N = [\(\sqrt{3} + 1)^{2014}\)] + 1

Where, [x] means greatest integer function

Answer 4

Now I have to find out [\(\sqrt{3} + 1)^{2014}\)].

Okay, so, uhm,

[(\(\sqrt{3} + 1)^{2014}\)] = (\(\sqrt{3} + 1)^{2014}\) - {(\(\sqrt{3} + 1)^{2014}\)}

Where, {x} represents fractional part of x.

Answer 5

Now, I know this,

\(\sqrt{3} = 1.732\)

So, I can write it as 1 + some pure fraction (let f). So, I have,

\(\sqrt{3} = 1+f\)

So,

\((\sqrt{3} + 1)^{2014} = (f+2)^{2014}\)

Answer 6

Okay so I stuck again. And I am going to wait with my hands folded and cursing maths in my mind until something else appears in my mind.

Answer 7

Are you guys doing the same too ? lol

Answer 8

yes :))

Answer 9

Okay, lets do it together.

Answer 10

Okay I think there is something better than that:

\(\large{(\sqrt{3} + 1)^{2014} = {(\sqrt{3})^{2014}}_i + {(\sqrt{3})^{2013}}_f + {(\sqrt{3})^{2012}}_i + ... + {(\sqrt{3})^{0}}_i}\)

Answer 11

I have put \(i\) to represent that the term is integer and \(f\) to represent that the term is fractional.

Answer 12

Now, I know for sure, that integer will come from both the terms. So, I am going to simplify only the odd terms.

Answer 13

*odd powered terms

Answer 14

Now, I know that,

\(\sqrt{3} = 1 + f\) where, f<1
So, for some odd power of (\(\sqrt{3})\), say, 2n - 1, I am going to write its expansion.

Answer 15

\(\large{(\sqrt{3})^{2n-1} = (1 + f)^{2n-1}}\)

Answer 16

I think its safe to assume that:

\(N = (\sqrt{3} + 1)^{2014} + (\sqrt{3} - 1)^{2014}\)

As, I know that \((\sqrt{3} + 1)^{2014}\) is going to sum of terms with "i" which would be integer and sum of terms with "f" which would be fractional. Now, for N is an integer, I must add some positive number to it to make it an integer and that number to be added must be the smallest possible number which in my view is \((\sqrt{3} - 1)^{2014}\), as it removes all the fractional terms (i.e. terms marked as "f").

Answer 17

Now,

\[\large{(\sqrt{3} + 1)^{2014} + (\sqrt{3} - 1)^{2014}}\]

\[\large{= 2^{1007}[(2+\sqrt{3})^{1007} + (2-\sqrt{3})^{1007}]}\]

\[\large{= 2^{1009}[2^{1006} + \left(\begin{matrix}1007 \\ 2\end{matrix}\right)*2^{1004}*3 + ... + \left(\begin{matrix}1007 \\ 1006\end{matrix}\right)*3^{503}]}\]

Now, the part in [ ] brackets is not divisible by . So,

\[\large{2^{1009} = 2*2^{1008} = 2* 16^{252}}\]

So, required value of p = 252.

Answer 18

*sigh* Its done !!!!!!!!

Answer 19

o.O you wrote LaTeX ?

Answer 20

Hey I resent that statement ?

Answer 21

Well, uhm, sort of.

Answer 22

Now, the only thing remains is getting this solution checked. Suggestions for who to ask for help ?

Answer 23

@nipunmalhotra93 @dan815 may help you!

Answer 24

I am going to add some tags from my side as well.

@BSwan , @ganeshie8 , @mukushla , @mathmale .

Answer 25

I just want to know if my answer is correct or not.

I only have question and my approach to the problem. I don't have the answer.

Answer 26

Well I have to go to have dinner. Sorry friends.

Bye

Answer 27

quite right :-) you nailed it

Answer 28

Thanks @mukushla

Answer 29

no problem

Answer 30

wow ! the mighty mukushla :)

Answer 31

Hail mukushla lol

Answer 32

lol, gane vish solved it not me :-)

Answer 33

;)

Answer 34

Hey its not "pick on vishu" day.. :P

Answer 35

:p

Answer 36

good night guys. Its time to go to bed. :)

Answer 37

Enjoy the life 'coz its worth enjoying !!