Question

Without actually performing the long division, state whether $\dfrac{13}{3125}$ has terminating decimal expansion or not.
If true enter 1 and if false then enter 0
A. True
B. False

Answer 1

Hint: We need to find whether $\dfrac{13}{3125}$ has terminated decimal expansion or not. Firstly, we start to solve the given question by finding the prime factorization of the denominator. If the prime factorization of the denominator is of the form, ${{2}^{m}}\times {{5}^{n}}$ then it has a terminating decimal expansion.

Complete step-by-step answer:
We are given a fraction and are asked to find whether it has a terminating decimal expansion or not.
Let us consider the value of the fraction $x=\dfrac{13}{3125}$
We will be solving the given question by finding the prime factorization of the denominator.
Fractions, in mathematics, are used to represent the portion or the part of the entire or whole thing. They are generally represented as follows,
$= \dfrac{a}{b}$
Here,
$a$ is the numerator of the fraction
$b$ is the denominator of the fraction
For Example:
$= \dfrac{1}{2},\dfrac{2}{5}$
According to the question,
Numerator = 13
Denominator = 3125
We need to find if the fraction $\dfrac{13}{3125}$ has terminating decimal expansion or not. We will have to find the prime factorization of the denominator of the fraction to state whether it has terminating decimal expansion or not.
Prime factors of a number can be found by prime factorization. It is a method of finding all the prime factors which are multiplied to get the original number.
The prime factors of 3125 are given as follows,
$= 3125=5\times 5\times 5\times 5\times 5$
Writing the right-hand side of the above expression to the power of 5, we get,
$= 3125={{5}^{5}}$
The above equation can be also written as follows,
$\therefore 3125={{2}^{0}}\times {{5}^{5}}$
From the above,
The prime factorization of the denominator is of the form ${{2}^{m}}\times {{5}^{n}}$ (m=0, n=5). So, it has terminating decimal expansion.
$\therefore$ The fraction $\dfrac{13}{3125}$ has a terminating decimal expansion.

Note: We must remember that any number raised to the power of zero is 1. It is given as follows,
$= {{n}^{0}}=1$
Here, n is any number except 0.
The prime factors of a given number are only correct if their product results in a given number.
In the given question,
$= 3125={{2}^{0}}\times {{5}^{5}}$
LHS:
$= 3125$
RHS:
$= {{2}^{0}}\times {{5}^{5}}$
Simplifying the above expression, we get,
$= 1\times 3125$
$= 3125$
LHS=RHS. The prime factorization of 3125 is correct.