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.

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