Without actually performing the long division, state whether the following rational number will have terminating decimal expansion or a non-terminating repeating decimal expansion. Also find the number of places of decimal after which the decimal expansion terminates.
$
\dfrac{{13}}{{3125}} \\
{\text{a}}{\text{. 3}} \\
{\text{b}}{\text{. 4}} \\
{\text{c}}{\text{. 5}} \\
{\text{d}}{\text{. 6}} \\
$
Answer
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Hint: -Factorize the denominator and convert it into prime factorization form. [I.e.$\left( {\dfrac{1}{{{2^m} \times {5^n}}}} \right)$]
Convert the fraction into prime factorization form. [I.e.$\left( {\dfrac{1}{{{2^m} \times {5^n}}}} \right)...........\left( 1 \right)$]
Calculate the factors of the numbers and convert them into multiplication of prime numbers.
A number is called a prime number if the factor of the number is either 1 or itself.
Factorize the denominator
$\therefore $The factors of 3125 is $\left[ {5 \times 5 \times 5 \times 5 \times 5} \right] = \left( {{2^0} \times {5^5}} \right)$(where 2 and 5 is a prime number)
$ \Rightarrow \dfrac{{13}}{{3125}} = \left( {\dfrac{{13}}{{{2^0} \times {5^5}}}} \right)$
Compare above equation from equation (1)
$ \Rightarrow m = 0,{\text{ }}n = 5$
Now as we know $\dfrac{1}{{{5^1}}} = 0.2$(i.e. complete division)
$\therefore \dfrac{1}{{{5^1}}}$ is terminating decimal expansion and it terminates after one place from the decimal.
$\therefore \dfrac{{13}}{{3125}} = \dfrac{{13}}{{{5^5}}}$ is also a terminating decimal expansion.
And terminating decimal place ${\text{ = }}\max \left( {m,n} \right){\text{ = }}\max \left( {0,5} \right) = n = 5$
In general we can say if a denominator has ${5^n}$ then it terminates after n places from the decimal.
So, it terminates after 5 decimal places.
Hence option (c) is the correct answer.
Note: -In such types of questions the key concept we have to remember is that first convert the fraction into prime factorization form then check whether the prime factors of the denominator has complete division or not if yes then the fraction is a terminating decimal place otherwise not, and the value of the terminating decimal place is the power of greatest prime factor.
Convert the fraction into prime factorization form. [I.e.$\left( {\dfrac{1}{{{2^m} \times {5^n}}}} \right)...........\left( 1 \right)$]
Calculate the factors of the numbers and convert them into multiplication of prime numbers.
A number is called a prime number if the factor of the number is either 1 or itself.
Factorize the denominator
$\therefore $The factors of 3125 is $\left[ {5 \times 5 \times 5 \times 5 \times 5} \right] = \left( {{2^0} \times {5^5}} \right)$(where 2 and 5 is a prime number)
$ \Rightarrow \dfrac{{13}}{{3125}} = \left( {\dfrac{{13}}{{{2^0} \times {5^5}}}} \right)$
Compare above equation from equation (1)
$ \Rightarrow m = 0,{\text{ }}n = 5$
Now as we know $\dfrac{1}{{{5^1}}} = 0.2$(i.e. complete division)
$\therefore \dfrac{1}{{{5^1}}}$ is terminating decimal expansion and it terminates after one place from the decimal.
$\therefore \dfrac{{13}}{{3125}} = \dfrac{{13}}{{{5^5}}}$ is also a terminating decimal expansion.
And terminating decimal place ${\text{ = }}\max \left( {m,n} \right){\text{ = }}\max \left( {0,5} \right) = n = 5$
In general we can say if a denominator has ${5^n}$ then it terminates after n places from the decimal.
So, it terminates after 5 decimal places.
Hence option (c) is the correct answer.
Note: -In such types of questions the key concept we have to remember is that first convert the fraction into prime factorization form then check whether the prime factors of the denominator has complete division or not if yes then the fraction is a terminating decimal place otherwise not, and the value of the terminating decimal place is the power of greatest prime factor.
Last updated date: 22nd Sep 2023
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