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

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Homework Help Overview

The discussion revolves around number theory and algebraic expansions, specifically addressing the primality of a large number composed of repeated digits and a summation problem involving powers of two.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to determine the primality of a number formed by repeating the digit '1' 91 times and seeks methods for factorization. They express uncertainty about how to apply the binomial theorem in this context. For the second problem, they identify the left-hand side as an arithmetic-geometric progression and explore manipulation of terms to find a solution.

Discussion Status

Participants have provided hints regarding the composite nature of 91, suggesting that this influences the divisibility of the large number. Some participants are exploring the implications of this factorization, while others are attempting to clarify the reasoning behind it. There is ongoing exploration of the summation problem, with suggestions to consider partial sums and established series formulas.

Contextual Notes

Participants are grappling with the implications of the number 91 being composite and its effect on the primality of the repeated digit number. There is also a mention of needing to perform calculations independently to arrive at conclusions for the second problem.

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Homework Statement


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

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




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 [tex]\ n\times2^n[/tex] to RHS and then dividing by 2^2...but that doesn't seem to help ?
 
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any suggestions ?
 
Here are some hints...
1.
a) 91 is composite
b) Perhaps you can start by taking partial sums and seeing what happens...
 
NateTG said:
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
 
f(x) said:
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
[tex]\underbrace{\underbrace{1111111}_{7\; times}\underbrace{1111111}_{7\; times}\ldots\underbrace{1111111}_{7\; times}}_{13\; times}[/tex]

and if 1111111 is not a prime then you are done. Otherwise more work needed.
 
[tex]11 + 1100 + 11 0000 + 11 000000 ...[/tex]

O god damn it, 91's an odd number nvm me didn't notice
 
Last edited:
111,111,111,111,111=111*1,001,001,001,001
 
For problem (b) you might want to consider the sum to n terms of
[tex]1+2x+3x^2+4x^3+...+nx^{n-1}[/tex]
 
If i am not mistaken (b)'s answer is [tex]2^9+1[/tex]
but you need to do the workings yourself..
 
  • #10
NateTG said:
111,111,111,111,111=111*1,001,001,001,001

How did you get that ?


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

Yeah 513 was surely an option...
 
  • #11
mjsd said:
one reason that it helps is because 91=7x13 means that you can chop the number into segments like
[tex]\underbrace{\underbrace{1111111}_{7\; times}\underbrace{1111111}_{7\; times}\ldots\underbrace{1111111}_{7\; times}}_{13\; times}[/tex]

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

1111111 = 239 x 4649
 

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