Question

Find out the prime factorisation of the denominator of $27.\overline{142857}$.

Answer 1

Hint: We find the formula to convert non-terminating recurring decimals to fractions. After getting the fraction part we get the denominator. We use a prime factorisation method to find the solution of the problem.

Complete step-by-step answer:
We need prime factorisation of the denominator of $27.\overline{142857}$.
The given number is in decimal form. To find the denominator of the number, we need to convert it into fraction form. The lower part of the fraction needs to be factored.
The given decimal is also in non-terminating recurring form which means a certain part of the decimal is going to appear again and again till infinity. In this problem the whole part after the decimal is in repetition mode.
By breaking the form and expanding it we get \[27.\overline{142857}=27.142857142857142857...........\]
So, the 142857 comes back again and again one after another. There will be no change of form in between them. These are the always rational numbers.
Now we convert this type decimal form into fraction form by following a formula.
Let’s assume the form of the decimal is x. x is the part of the decimal form which is under the repetition and y is the part of the decimal form which is not under the repetition. The decimal is in between any two digits. We also need to note down two numbers. These are the number of digits after decimal being under the repetition and the number of digits after decimal not being under the repetition mode. Let’s take these numbers as ‘a’ and ‘b’.
So, the fraction form of $x.y$ is \[\dfrac{\left\{ \right\}-x}{{{a}_{9}}{{b}_{0}}}\].
Here the terms used, will be described.
\[\left\{ \right\}\] means writing down the whole number without using the decimal. x is the part of the decimal in which there is no repetition. Also, \[{{a}_{9}}\] defines the use of ‘a’ numbers of 9 and \[{{b}_{0}}\] defines use of ‘b’ numbers of 0. These will be written as a whole total number. Both of these \[{{a}_{9}}\] and \[{{b}_{0}}\] has to be only for the part after decimal.
Let’s take an example as $3.45\overline{421}$. Here only two numbers are in repetition which means $3.45\overline{421}=3.45421421421....$. Now we apply the formula.
The whole number is \[345421\]. The without repetition part is \[345\].
In the after decimal part there are three numbers which are in repetition. So, we are going to use three 9’s. Also, there are two numbers which are not in repetition. So, we are going to use two 0’s also.
So, a fraction part of $3.45\overline{421}$ will be $\dfrac{345421-345}{99900}$. Now we solve to get
$\dfrac{345421-345}{99900}=\dfrac{345076}{99900}=3\dfrac{11344}{24975}$.
For our given problem $27.\overline{142857}$ the fraction form will be $27.\overline{142857}=\dfrac{27142857-27}{999999}$.
Solving the number, we get $\dfrac{27142857-27}{999999}=\dfrac{27142830}{999999}=\dfrac{190}{7}=27\dfrac{1}{7}$.
Now we need to find the prime factorisation of the denominator which is 7.
7 is a prime number. So, the prime factorisation will be done using only 1.
So, the prime factorisation is $7=7\times 1$.

Note: The formula can also be derived in other ways. In that case the mixed decimal is not considered, only the part after the decimals. And at the end the integer part is added. There is no other divisor of a prime other than 1 and itself. So, the number of factorisations of prime numbers is 1.