Challenge VII: A bit of number theory solved by Boorglar

[micromass]

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.

 

[Boorglar]

Spoiler

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.

 

[micromass]

Well, that was fast.

 

[Boorglar]

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.

 

[CellCoree]

This is a very interesting and challenging problem in number theory. It involves the concept of rational numbers and their relationship to natural numbers. In order to solve this problem, we must first understand the properties of rational numbers.

A rational number is any number that can be expressed as a ratio of two integers, such as 3/4 or 5/6. In order to prove or find a counterexample for this problem, we must consider the properties of rational numbers and how they interact with natural numbers.

After analyzing the problem, I believe that there exists a natural number L for any positive rational number q. This is because any rational number can be written as a decimal with a finite number of digits after the decimal point. For example, 3/4 can be written as 0.75 and 5/6 can be written as 0.8333... with the 3 repeating infinitely.

Therefore, if we multiply any rational number q by a large enough natural number L, we can create a decimal with a repeating pattern of nines and zeros. For example, if we take q=3/4 and L=100, we get 75 which is equivalent to 0.75 with two trailing zeros. Similarly, if we take q=5/6 and L=1000, we get 833 which is equivalent to 0.833 with three trailing zeros.

In order to prove this more rigorously, we can use the concept of continued fractions. Every rational number has a unique continued fraction representation, which is a finite or infinite sequence of integers. By manipulating this continued fraction, we can find a natural number L that satisfies the given equation.

In conclusion, I believe that there exists a natural number L for any positive rational number q such that Lq has a repeating pattern of nines and zeros. However, further analysis and proof may be required to confirm this solution.