How do you convert $0.123\left( {123{\text{ repeating}}} \right)$ to a fraction?

Hint: Given a decimal number. We have to convert the number into fractional form. First, we will multiply the term by the number of zeros equivalent to the number of digits after the decimal. Then, subtract the given number from the new formed number. Then, find the prime factors of the numerator and denominator to simplify the expression. Then, by cancelling out the common terms, reduce the fraction in its lowest form.

Complete step-by-step answer:
We are given the repeating number. Write the number $0.123\left( {123{\text{ repeating}}} \right)$ in mathematical form.
Let the number be $n$
$n = 0.\overline {123} $ ……(1)
Here, the bar above the three digits shows that these digits are repeating.
Here, the number of digits after the decimal are three. Thus, we will multiply both sides of equation (1) by $1000$
$ \Rightarrow 1000n = 0.\overline {123} \times 1000$
On simplifying the equation, we get:
$ \Rightarrow 1000n = 123.\overline {123} $ ……(2)
Now, we will subtract the equation (1) from the equation (2).
$ \Rightarrow 1000n - n = 123.\overline {123} - 0.\overline {123} $
Now, we will simplify the equation to determine the value of $n$
$ \Rightarrow n\left( {1000 - 1} \right) = \left( {123 + 0.\overline {123} } \right) - 0.\overline {123} $
$ \Rightarrow n\left( {999} \right) = 123 +{{0.\overline {123} }} -{{0.\overline {123} }}$
Now, divide both sides of the equation by $999$
$ \Rightarrow \dfrac{{n\left( {999} \right)}}{{999}} = \dfrac{{123}}{{999}}$
$ \Rightarrow n = \dfrac{{123}}{{999}}$
Now, find the prime factors of the numerator and denominator of the RHS of the equation.
$ \Rightarrow n = \dfrac{{3 \times 41}}{{3 \times 333}}$
Now, cancel out the common terms, we get:
$ \Rightarrow n = \dfrac{{{3} \times 41}}{{{3} \times 333}}$
$ \Rightarrow n = \dfrac{{41}}{{333}}$

Final answer: Hence, the number in fraction form is $\dfrac{{41}}{{333}}$

Note:
 In such types of questions the students mainly don't get an approach on how to solve it. In such types of questions students mainly forget to make mistakes by determining the number by which the given number is multiplied. Then, students mainly get confused whether the equations containing a given number and new formed number are subtracted or added.