Question

Which of the following rational numbers is expressible as a now terminating repeating decimal?

(a) $\frac{1351}{1250}$
(b) $\frac{2017}{250}$
(c) $\frac{3219}{1800}$
(d) $\frac{1723}{625}$

Answer 1

(c) $\frac{3219}{1800}$

$\frac{1351}{1250}=\frac{1351}{5^4\times 2}$
We know 2 and 5 are not the factors of 1351.
So, the given rational is in its simplest form.
And it is of the form $(2^m\times 5^n)$ for some integers $m,n$.
So, the given number is a terminating decimal.
∴ $\frac{1351}{5^4\times 2}=\frac{1351\times 2}{5^4\times 2^4}=\frac{10808}{10000}=1.0808$

$\frac{2017}{250}$ = $\frac{2017}{5^3\times 2}$
We know 2 and 5 are not the factors of 2017.
So, the given rational number is in its simplest form.
And it is of the form $(2^m\times 5^n)$ for some integers $m,n$.
So, the given rational number is a terminating decimal.
∴ $\frac{2017}{5^3\times 2}=\frac{2017\times 2^2}{5^3\times 2^3}=\frac{8068}{1000}=8.068$

$\frac{3219}{1800}=\frac{3219}{2^3\times 5^2\times 3^2}$
We know 2, 3 and 5 are not the factors of 3219.
So, the given rational number is in its simplest form.
∴ $(2^3\times 5^2\times 3^2)$ ≠ $(2^m\times 5^n)$
Hence, $\frac{3219}{1800}$ is not a terminating decimal.
$\frac{3219}{1800}=1.78833333.....$
Thus, it is a repeating decimal.

$\frac{1723}{625}=\frac{1723}{5^4}$
We know 5 is not a factor of 1723.
So, the given rational number is in its simplest form.
And it is not of the form $(2^m\times 5^n)$.
Hence, $\frac{1723}{650}$ is not a terminating decimal.