Without actually performing the long division, state whether the following rational numbers expansion.

Solution:

(i) $\frac{13}{3125}=\frac{13}{5^{5}}$

Hence, $\mathrm{q}=5^{5}$, which is of the form $2^{\mathrm{n}} 5^{\mathrm{m}}(\mathrm{n}=0, \mathrm{~m}=5) .$ So, the rational number $\frac{\mathbf{1 3}}{\mathbf{3 1 2 5}}$ has a terminating decimal expansion.

(ii) $\frac{17}{8}=\frac{17}{2^{3}}$

Hence, $\mathrm{q}=2^{3}$, which is of the form $2^{\mathrm{n}} 5^{\mathrm{m}}(\mathrm{n}=3, \mathrm{~m}=0) .$ So, the rational number $\frac{\mathbf{1 7}}{\mathbf{8}}$ has a terminating decimal expansion.

(iii) $\frac{64}{455}=\frac{64}{5 \times 7 \times 13}$

Hence, $\mathrm{q}=5 \times 7 \times 13$, which is not of the form $2^{\mathrm{n}} 5^{\mathrm{m}}$. So, the rational number $\frac{\mathbf{6 4}}{\mathbf{4 5 5}}$ has a nonterminating repeating decimal expansion.

(iv) $\frac{15}{1600}=\frac{15}{2^{6} \times 5^{2}}=\frac{3 \times 5}{2^{6} \times 5^{2}}$

Hence, $\mathrm{q}=2^{6} \times 5$, which is of the form $2^{\mathrm{n}} \times 5^{\mathrm{m}}(\mathrm{n}=6, \mathrm{~m}=1) .$ So, the rational number $\frac{15}{1600}$ has a terminating decimal expansion.

(v) $\frac{29}{343}=\frac{29}{7^{3}}$

Hence, $\mathrm{q}=7^{3}$, which is not of the form $2^{\mathrm{m}} \times 5^{\mathrm{n}}$. So, rational number $\frac{29}{\mathbf{7}^{3}}$ has a non-terminating repeating decimal expansion.

(vi) $\frac{23}{2^{3} \times 5^{2}}$

Hence, $\mathrm{q}=2^{3} \times 5^{2}$, which is of the form $2^{n} \times 5^{m}(n=3, m=2)$. So, the rational number has a terminating decimal expansion.

(vii) $\frac{129}{\mathbf{z}^{\mathbf{2}} \times \mathbf{5}^{\mathbf{7}} \times \mathbf{7}^{\mathbf{5}}}$

Hence, $q=2^{2} \times 5^{7} \times 7^{5}$, which is not of the form $2^{\mathrm{m}} \times 5^{\mathrm{n}} .$ So, rational number $\frac{129}{\mathbf{2}^{2} \times \mathbf{5}^{7} \times \mathbf{7}^{5}}$ has a non-terminating repeating decimal expansion.

(viii) $\frac{6}{15}=\frac{2 \times 3}{3 \times 5}=\frac{2}{5}$

Hence, $\mathrm{q}=5$, which is of the form $2^{\mathrm{m}} \times 5^{\mathrm{n}}(\mathrm{m}=0, \mathrm{n}=1) .$ So, the rational no $\frac{\mathbf{2}}{\mathbf{5}}$ has a terminating decimal expansion.

(ix) $\frac{35}{50}=\frac{5 \times 7}{2 \times 5^{2}}=\frac{7}{2 \times 5}$

Hence, $\mathrm{q}=2 \times 5^{1}$, which is of the form $2^{\mathrm{m}} \times 5^{\mathrm{n}}(\mathrm{m}=1, \mathrm{n}=1)$. So, the rational number $\frac{35}{50}$ has a terminating decimal expansion.

(x) $\frac{77}{210}=\frac{7 \times 11}{2 \times 3 \times 5 \times 7}=\frac{11}{2 \times 3 \times 5}$

Hence, $\mathrm{q}=2 \times 3 \times 5$, which is not of the form $2^{\mathrm{m}} \times 5^{\mathrm{n}} .$ So, the rational number $\frac{\mathbf{7 7}}{\mathbf{2 1 0}}$ has a non-terminating repeating decimal expansion.