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.
