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) $\frac{13}{3125}$ (ii) $\frac{17}{8}$ (iii) $\frac{64}{455}$ (iv) $\frac{15}{1600}$ (v) $\frac{29}{343}$ (vi) $\frac{23}{2^3 5^2}$ (vii) $\frac{129}{2^{2 }{5}^{7 }{7}^5}$ (viii) $\frac{6}{15}$ (ix) $\frac{35}{50}$ (x) $\frac{77}{210}$

(i) Writing the factors for the denominator,

3125 = 5 × 5 × 5 × 5 × 5 = 5⁵

Since the denominator is in the form of 2^m × 5ⁿ, thus $\frac{13}{3125}$ has a terminating decimal expansion.

(ii) Writing the factors for the denominator,

8 = 2×2×2 = 2³

Since the denominator is in the form of 2^m × 5ⁿ, thus $\frac{17}{8}$ has a terminating decimal expansion.

(iii) Writing the factors for the denominator,

455 = 5×7×13

Since the denominator is not in the form of 2^m × 5ⁿ, thus $\frac{64}{455}$ has a non-terminating decimal expansion.

(iv) Writing the factors for the denominator,

1600 = 2⁶×5²

Since the denominator is in the form of 2^m × 5ⁿ, thus $\frac{15}{1600}$ has a terminating decimal expansion.

(v) Writing the factors for the denominator,

343 = 7×7×7 = 7³

Since the denominator is not in the form of 2^m × 5ⁿ, thus $\frac{29}{343}$ has a non-terminating decimal expansion.

(vi) Here the factors for denominator are already given.

Since the denominator is in the form of 2^m × 5ⁿ, thus $\frac{23}{2^3 5^2}$ has a terminating decimal expansion.

(vii) Here the factors for denominator are already given.

Since the denominator is not in the form of 2^m × 5ⁿ, thus $\frac{129}{2^{2 }{5}^{7 }{7}^5}$ has a non-terminating decimal expansion.

(viii) Writing the factors for the denominator,

$\frac{6}{15}=\frac{2}{5}$

Since the denominator is in the form of 2^m × 5ⁿ, thus $\frac{6}{15}$ has a terminating decimal expansion.

(ix) Writing the factors for the denominator,

$\frac{35}{50}=\frac{7}{10}=\frac{7}{2\times 5}$

Since the denominator is in the form of 2^m × 5ⁿ, thus $\frac{35}{50}$ has a terminating decimal expansion.

(x) Writing the factors for the denominator,

$\frac{77}{210}=\frac{7\times 11}{7\times 30}=\frac{11}{30}=\frac{11}{2\times 3\times 5}$

Since the denominator is not in the form of 2^m × 5ⁿ, thus $\frac{77}{210}$ has a non-terminating decimal expansion.