# Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion :

Question : Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion :

(i) $$13\over 3125$$

(ii)  $$17\over 8$$

(iii)  $$64\over 455$$

(iv)  $$15\over 1600$$

(v)  $$29\over 343$$

(vi)  $$23\over {2^3 5^2}$$

(vii)  $$129\over {2^2 5^7 7^5}$$

(viii)  $$6\over 15$$

(ix)  $$35\over 50$$

(x)  $$77\over 210$$

Solution :

(i)  Since the factors of the denominator 3125 are $$2^0 \times 5^5$$. Therefore $$13\over 3125$$ is a terminating decimal.

(ii)  Since the factors of the denominator 8 are $$2^3 \times 5^0$$. So, $$17\over 8$$ is a terminating decimal.

(iii)  Since the factors of the denominator 455 is not in the form $$2^n \times 5^m$$. Therefore $$64\over 55$$ is a non-terminating repeating decimal.

(iv)  Since the factors of the denominator 1600 are $$2^6 \times 5^2$$. Therefore $$15\over 1600$$ is a terminating decimal.

(v)  Since the factors of the denominator 343 is not of the form $$2^n \times 5^m$$. Therefore it is a non-terminating repeating decimal.

(vi)  Since the factors of the denominator is of form $$2^3 \times 5^2$$. Therefore it is a terminating decimal.

(vii)  Since the factors of the denominator $$2^2 \times 5^7 \times 7^5$$ is not of the form $$2^n \times 5^m$$. Therefore it is a non-terminating repeating decimal.

(viii)  $$16\over 5$$ = $$2\over 5$$ here the factors of the denominator 5 are $$2^0 \times 5^1$$. Therefore it is a terminating decimal.

(ix)  Since the factors of the denominator 50 are $$2^1 \times 5^2$$. Therefore $$35\over 30$$ is a terminating decimal.

(x)  Since the factors of the denominator 210 is not of the form $$2^n \times 5^m$$. Therefore $$77\over 210$$ is a non-terminating rrepeating decimal.