Question: How can I solve these Number Theory/Expansion homework problems?

[f(x)]

Homework Statement

a)True or False :
\underbrace{111111111...1} is a prime number .
91 times

b) Find n such that -:
2\times 2^2+ 3\times2^3+4\times2^4+ \cdots + n\times2^n=2^{n+10}

Homework Equations

The Attempt at a Solution

I have no idea about a). The number is not divisible by 3,7,11...but i can't go on dividing all the way like this. How do i resolve this into prime factors(if possible) ? Do i use binomial theorem and how ?

About b), LHS is an AGP . I tried taking the \ n\times2^n to RHS and then dividing by 2^2...but that doesn't seem to help ?

 

[f(x)]

any suggestions ?

 

[NateTG]

Here are some hints...
1.
a) 91 is composite
b) Perhaps you can start by taking partial sums and seeing what happens...

 

[f(x)]

Here are some hints...
1.
a) 91 is composite

Umm i don't understand...how does 91 being composite influence divisibility ?
Could you please explain in abit more detail
Thx

 

[mjsd]

Umm i don't understand...how does 91 being composite influence divisibility ?
Could you please explain in abit more detail
Thx

one reason that it helps is because 91=7x13 means that you can chop the number into segments like
\underbrace{\underbrace{1111111}_{7\; times}\underbrace{1111111}_{7\; times}\ldots\underbrace{1111111}_{7\; times}}_{13\; times}

and if 1111111 is not a prime then you are done. Otherwise more work needed.

 

[Gib Z]

11 + 1100 + 11 0000 + 11 000000 ...

O god damn it, 91's an odd number nvm me didn't notice

 

[NateTG]

111,111,111,111,111=111*1,001,001,001,001

 

[truewt]

For problem (b) you might want to consider the sum to n terms of
1+2x+3x^2+4x^3+...+nx^{n-1}

 

[truewt]

If i am not mistaken (b)'s answer is 2^9+1
but you need to do the workings yourself..

 

[f(x)]

111,111,111,111,111=111*1,001,001,001,001

How did you get that ?

If i am not mistaken (b)'s answer is 2^9+1

Yeah 513 was surely an option...

 

[Gib Z]

one reason that it helps is because 91=7x13 means that you can chop the number into segments like
\underbrace{\underbrace{1111111}_{7\; times}\underbrace{1111111}_{7\; times}\ldots\underbrace{1111111}_{7\; times}}_{13\; times}

and if 1111111 is not a prime then you are done. Otherwise more work needed.

1111111 = 239 x 4649