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# 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)** $\frac{23}{2^3\times5^2}$.

**(vii)** $\frac{129}{2^2\times5^7\times7^{17}}$.

**(viii)** $\frac{6}{15}$.

**(ix)** $\frac{35}{50}$.

**(x)** $\frac{77}{210}$.

To do:

Here, we have to check without actually performing the long division, whether the given rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion.

Solution:

If we have a rational number $\frac{p}{q}$, where $p$ and $q$ are co-primes and the prime factorization of $q$ is of the form $2^n.5^m$, where $n$ and $m$ are non-negative integers, then $\frac{p}{q}$ has a terminating expansion.

Now,

(i) In $\frac{13}{3125}$,

- $13$ and $3125$ are co-primes.
- $3125 = 5^5 \times 2^0$, which is in the form $2^n\ \times\ 5^m$.

So, $\frac{13}{3125}$ has a terminating decimal expansion.

(ii) In $\frac{17}{8}$,

- $17$ and $8$ are co-primes.
- $8= 2^3 \times 5^0$, which is in the form $2^n\ \times\ 5^m$.

So, $\frac{17}{8}$ has a terminating decimal expansion.

(iii) $\frac{64}{455}=\frac{64}{5\times7\times13}$

In $\frac{64}{455}$,

- $64$ and $455$ are co-primes.
- $455= 5 \times 7\times13$, which is not in the form $2^n\ \times\ 5^m$.

So, $\frac{64}{455}$ has a non-terminating repeating decimal expansion.

(iv) $\frac{15}{1600}=\frac{3\times5}{2^6\times5^2}$

$=\frac{3}{2^6\times5}$

In $\frac{3}{2^6\times5}$,

- $3$ and $2^6\times5$ are co-primes.
- $2^6\times5= 2^6 \times 5^1$, which is in the form $2^n\ \times\ 5^m$.

So, $\frac{15}{1600}$ has a terminating decimal expansion.

(v) $\frac{29}{343}=\frac{29}{7\times7\times7}$

In $\frac{29}{343}$,

- $29$ and $343$ are co-primes.
- $343= 7^3$, which is not in the form $2^n\ \times\ 5^m$.

So, $\frac{29}{343}$ has a non-terminating repeating decimal expansion.

(vi) In $\frac{23}{2^3\times5^2}$,

- $23$ and $2^3\times5^2$ are co-primes.
- $2^3 \times 5^2$ is in the form $2^n\ \times\ 5^m$.

So, $\frac{23}{2^3\times5^2}$ has a terminating decimal expansion.

(vii) In $\frac{129}{2^2\times5^7\times7^{17}}$:

- $129$ and $2^2\times5^7\times7^{17}$ are co-primes.
- $2^2 \times 5^7 \times 7^{17}$ is not in the form $2^n\ \times\ 5^m$.

So, $\frac{129}{2^2\times5^7\times7^{17}}$ has a non-terminating repeating decimal expansion.

(viii) $\frac{6}{15}=\frac{2\times3}{3\times5}$

$=\frac{2}{5}$

In $\frac{2}{5}$,

- $2$ and $5$ are co-primes.
- $5= 2^0 \times 5^1$, which is in the form $2^n\ \times\ 5^m$.

So, $\frac{6}{15}$ has a terminating decimal expansion.

(ix) $\frac{35}{50}=\frac{5\times7}{5\times10}=\frac{7}{10}$

In $\frac{7}{10}$:

- $7$ and $10$ are co-primes.
- $10 = 2^1 \times 5^1$, which is in the form $2^n\ \times\ 5^m$.

So, $\frac{35}{50}$ has a terminating decimal expansion.

(x) $\frac{77}{210}=\frac{7\times11}{7\times30}=\frac{11}{30}$

In $\frac{11}{30}$:

- $11$ and $30$ are co-primes.
- $30 = 2^1 \times 3^1 \times 5^1$, which is not in the form $2^n\ \times\ 5^m$.

So, $\frac{77}{210}$ has a non-terminating repeating decimal expansion.

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