Decimal expansion of rational numbers

(0.\overline{027}=\frac{27}{999}=\frac{1}{37}). An initial zero inside the repeating block is counted as a digit.

Both (3^0) and (11^0) equal (1), so the effective denominator is (2^3\cdot 5^2). The larger exponent is (3), so the decimal terminates after (3) places.

The factor (37) makes the decimal recurring, and the larger exponent of (2) and (5) is (4), giving the non-repeating start. In mixed denominators, the larger exponent gives the delay.

The governing concept is converting a terminating decimal to lowest fractional form and then prime-factorising the reduced denominator. Since 0.0375 has four decimal places, write it as 375/10000. Divide numerator and denominator by their greatest common divisor, 125: 375 ÷ 125 = 3 and 10000 ÷ 125 = 80. Thus 0.0375 = 3/80. Now factor the denominator: 80 = 8 × 10 = 2³ × (2 × 5) = 2⁴ × 5. Therefore option B is correct. Option A misses one factor of 2, while options C and D introduce an extra factor of 5 and do not represent the prime factorisation of 80. Reduction before factorisation is important.

(0.0375=\frac{375}{10000}), and dividing by (125) gives (\frac{3}{80}). Convert the decimal to a fraction and reduce fully.

The governing rule is that a rational number p/q in lowest form has a terminating decimal expansion only when the prime factors of q are 2 and/or 5. Here q contains 7^r, and r > 0, so at least one factor 7 remains in the reduced denominator. No cancellation with p is possible because p/q is already in lowest form. A denominator containing another prime factor cannot be converted into a power of 10, so the division continues indefinitely. Since the number is rational, its repeating remainder pattern must eventually recur. Therefore option B, non-terminating recurring, is correct. Option A would apply only if no factor other than 2 or 5 remained; option C describes an irrational decimal; option D incorrectly makes termination depend on m and n being equal.

In (0.58\overline{23}), the block (23) repeats regularly, so it is rational. A fixed repeating block is a strong sign of rationality.

First reduce the fraction before applying the decimal-expansion rule. Since 72 = 2^3 × 3^2, the numerator cancels completely with the factors 2^3 × 3^2 in the denominator. Thus 72/(2^3 × 3^2 × 5^5) = 1/5^5 = 1/3125. To express this with a denominator that is a power of 10, multiply numerator and denominator by 2^5: 1/5^5 = 2^5/10^5 = 32/100000 = 0.00032. The decimal therefore ends after five digits to the right of the decimal point. Option C is correct. Options A and B use smaller exponents without justification, while option D is false because the reduced denominator contains only the prime factor 5, so the decimal is terminating.

Since (38=2\cdot 19), the factor (19) cancels and the reduced denominator is (2\cdot 5^3). Even if an extra prime appears, check cancellation first.

(0.\overline{54}=\frac{54}{99}) and (0.\overline{45}=\frac{45}{99}), so their sum is (1). The sum of two recurring decimals can sometimes be terminating.

Since (9=3^2) remains, the decimal is non-terminating recurring. The larger exponent in (2^2\cdot 5^2) gives (2) initial non-repeating digits.

For a non-terminating recurring decimal, the reduced denominator has at least one prime factor other than (2) and (5). Factors (2) or (5) may also be present, but they are not enough alone.

For a rational number, the decimal expansion terminates only when, in lowest terms, the denominator contains powers of 2 and 5 alone. Any remaining prime factor other than 2 or 5 makes the decimal expansion non-terminating recurring. Thus the important step is to cancel common factors completely before applying this rule.

Cancel the common factors in \(\frac{2^5\cdot7}{2^8\cdot5^2\cdot7^2}\). The result is \(\frac{1}{2^3\cdot5^2\cdot7}\), because \(2^5\) leaves \(2^3\) below and one factor 7 remains below. Since 7 is still a factor of the reduced denominator, the decimal is non-terminating recurring. Therefore option B is correct. It cannot terminate after three places, because the factor 7 prevents termination.

(0.0625=\frac{625}{10000}=\frac{1}{16}). Convert a terminating decimal to a fraction and always reduce the denominator.

(0.000625=\frac{625}{1000000}), and reducing by (625) gives (\frac{1}{1600}). Do not fear large denominators; cancel common factors.

(112=2^4\cdot 7), so (4) non-repeating digits appear before the recurring part. For comparison, check the larger power of (2) and (5).

To form (10^k=2^k5^k), both powers must reach at least the larger exponent. Therefore the minimum (k=\max(r,s)).

(\sqrt{11}) is irrational, so its decimal is non-terminating non-recurring. Rational numbers are either terminating or non-terminating recurring.

Since \(10^6=2^6\cdot5^6\), compare this with the given denominator \(2^4\cdot5^6\): the factor \(2^2\) is missing. Multiply numerator and denominator by \(2^2=4\) to obtain denominator \(10^6\), so \(N=3\times4=12\). The distractor 24 would result from mistakenly multiplying by \(2^3=8\). Exam tip: to express a fraction with denominator \(10^n\), match the prime powers of 2 and 5 in the denominator to exponent \(n\).

Since (242=2\cdot 11^2), the reduced denominator becomes (2^2\cdot 5^4). The larger exponent is (4), so reduce first and then count decimal places.