Challenge VII: A bit of number theory solved by Boorglar

Click For Summary

Discussion Overview

The discussion revolves around a number theory challenge concerning the existence of a natural number \( L \in \mathbb{N} \) such that \( Lq = 99\ldots9900\ldots00 \) for an arbitrary positive rational number \( q \in \mathbb{Q} \). Participants are invited to prove the statement or provide a counterexample.

Discussion Character

  • Exploratory, Debate/contested, Mathematical reasoning

Main Points Raised

  • One participant notes the challenge's nature and its focus on number theory.
  • Another participant expresses confidence in solving such problems quickly, indicating familiarity with the topic.
  • A participant mentions an oversight regarding a multiplication factor in the formulation but suggests it does not significantly affect the problem.

Areas of Agreement / Disagreement

The discussion does not appear to have reached a consensus, as participants have not yet provided proofs or counterexamples, and the challenge remains open for exploration.

Contextual Notes

There may be missing assumptions regarding the specific forms of \( L \) and \( q \), and the implications of the number of nines and zeros in the expression are not fully explored.

micromass
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Messages
22,170
Reaction score
3,333
This new challenge was suggested by jostpuur. It is rather number theoretic.

Assume that [itex]q\in \mathbb{Q}[/itex] is an arbitrary positive rational number. Does there exist a natural number [itex]L\in \mathbb{N}[/itex] such that

[tex]Lq=99…9900…00[/tex]

with some amounts of nines and zeros? Prove or find a counterexample.
 
Last edited by a moderator:
Mathematics news on Phys.org
So we want to get a number of the form [itex]L \frac{a}{b} = 10^m(10^n-1)[/itex].
Write [itex]a = 2^x5^yd[/itex], where d is relatively prime with 10.

Then [itex]L = \frac{10^m}{2^x5^y} \frac{10^n-1}{d}[/itex].
The left fraction is obviously an integer if we choose m larger than max( x, y ).
The right fraction can be made an integer since d is relatively prime with 10, and therefore 10 is in the multiplicative group modulo d. Let n be the order of 10 in U(d), then [itex]10^n-1[/itex] is divisible by d so L is an integer.
 
Well, that was fast. :-p
 
Yeah I tend to be good at those types of problems haha.
Ah by the way I forgot the b multiplying the fractions but it doesn't really matter.
 

Similar threads

  • · Replies 125 ·
5
Replies
125
Views
20K
  • · Replies 86 ·
3
Replies
86
Views
14K
  • · Replies 67 ·
3
Replies
67
Views
12K
  • · Replies 61 ·
3
Replies
61
Views
13K
  • · Replies 52 ·
2
Replies
52
Views
13K
  • · Replies 43 ·
2
Replies
43
Views
13K
  • · Replies 83 ·
3
Replies
83
Views
22K
  • · Replies 86 ·
3
Replies
86
Views
24K
  • · Replies 121 ·
5
Replies
121
Views
25K
  • · Replies 105 ·
4
Replies
105
Views
15K