Decimal expansion of rational numbers

(0.\overline{045}=\frac{45}{999}=\frac{5}{111}), so the denominator is (111). An initial zero inside the repeating block is also counted as a digit.

(0.\overline{045}=\frac{45}{999}), and reducing by (9) gives (\frac{5}{111}). First form the denominator with (9)'s according to the repeating digits.

Both (3^0) and (17^0) equal (1), so the effective denominator is (2^4\cdot 5^3). The larger exponent is (4), so the decimal terminates after (4) places.

The factor (31) makes the decimal recurring, and the larger exponent of (2) and (5) is (6), giving the non-repeating start. In mixed denominators, the larger exponent gives the delay.

(0.01875=\frac{1875}{100000}=\frac{3}{160}), and (160=2^5\cdot 5). The correct prime factorisation is (2^5\cdot 5), so complete the calculation before choosing.

(0.01875=\frac{1875}{100000}), and dividing by (625) gives (\frac{3}{160}). Convert the decimal to a fraction and reduce fully.

A rational number has a terminating decimal only when, after cancellation, its denominator has no prime factors other than 2 and 5. If another prime factor remains, the division cannot end; because remainders repeat, the decimal becomes non-terminating recurring. This rule helps classify the decimal without carrying out a long division.

Here, the numerator is 200 = \(2^3\times5^2\). Cancelling common factors with \(2^3\times5^3\times7\) leaves the denominator \(5\times7\), since one factor 5 and the factor 7 remain. The factor 7 is not allowed in a terminating denominator, so the decimal is non-terminating recurring. Therefore option B follows.

Since (58=2\cdot 29), the factor (29) cancels and the reduced denominator is (2^2\cdot 5^2). If an extra prime appears, check cancellation first.

(0.\overline{27}=\frac{27}{99}) and (0.\overline{72}=\frac{72}{99}), so their sum is (1). The sum of two recurring decimals can be terminating.

Since (11) remains, the decimal is non-terminating recurring. The larger exponent in (2^3\cdot 5^3) gives (3) initial non-repeating digits.

For a terminating decimal, the reduced denominator (q) can contain only (2) and (5). In (q^3), powers increase but no new prime factor appears.

A rational number has a terminating decimal only when, after reducing the fraction to lowest terms, its denominator has no prime factors other than 2 and 5. If any other prime remains in the denominator, the decimal division cannot end; because the remainders eventually repeat, the decimal is non-terminating recurring.

Here, cancel the common factors in the numerator and denominator: \(2^4\) cancels part of \(2^7\), and one factor 13 cancels part of \(13^2\). The reduced denominator is \(2^3\cdot 5^3\cdot 13\). Since the prime factor 13 remains, the decimal expansion is non-terminating recurring. Therefore, option B is correct; it is not terminating and cannot be non-recurring because the number is rational.

(0.03125=\frac{3125}{100000}=\frac{1}{32}). Convert a terminating decimal to a fraction and reduce the denominator.

(0.0003125=\frac{3125}{10000000}), and reducing by (3125) gives (\frac{1}{3200}). Do not forget to cancel common factors in large denominators.

(224=2^5\cdot 7), so (5) non-repeating digits appear before the recurring part. For comparison, check the larger power of (2) and (5).

Reason: \(10^6=2^6\cdot 5^6\). The given denominator has \(2^6\) but only \(5^4\), so it is short by \(5^2\). Multiply numerator and denominator by \(5^2=25\) to get denominator \(10^6\). Thus \(N=7\times25=175\). Note on distractors: 35 comes from multiplying by \(5^1\) (a common slip), 700 from multiplying by 100, and 875 from multiplying by \(5^3=125\). Exam tip: compare prime-power factors of the denominator with those of \(10^n\) to find the exact factor to multiply quickly.

After cancellation, the denominator is (2^3\cdot 3\cdot 5^4\cdot 11), which contains (3) and (11). If primes other than (2) and (5) remain in the reduced denominator, the decimal is non-terminating recurring.

Since (10^5=2^5\cdot 5^5), the denominator lacks (5^3). Therefore (N=19\cdot 125=2375); multiply by the missing factor when making a power of (10).

One non-repeating zero and three repeating digits give (\frac{125}{9990}), which reduces to (\frac{25}{1998}). In mixed recurring decimals, do not treat the first denominator as the final one.

Since (q) divides (10^6), it has only (2) and (5), but not dividing (10^5) means the larger exponent is (6). Therefore the decimal terminates exactly after (6) places.

The denominator has only (2) and (5), so the decimal terminates with the larger exponent (9). In exams, use the larger exponent instead of adding exponents.

Since (484=2^2\cdot 11^2), the reduced denominator is (2^2\cdot 5^3). The larger exponent is (3), so reduce first and then count decimal places.

For termination, (3^2) and (17^2) must cancel completely, so (n=2601). For the least value, cancel only the factors other than (2) and (5).

(0.4272727\ldots=\frac{423}{990}=\frac{47}{110}), so the denominator is (110). In mixed recurring decimals, the final fraction must be reduced.

(0.\overline{108}=\frac{108}{999}=\frac{4}{37}). First form the denominator with (9)'s according to the repeating digits and then reduce.