Decimal expansion of rational numbers

Since (169=13^2), the reduced denominator is (2^3\cdot 5^4). Therefore the reason correctly explains the terminating decimal rule.

After cancellation, the denominator becomes (2^7\cdot 5^3). The larger exponent is (7), so the decimal terminates after (7) places.

Since (q) has only (2) and (5), the decimal terminates. Not dividing (10^8) means the larger exponent is (9) or (10).

Two non-repeating zeros and two repeating digits give (\frac{54}{9900}). Reducing it gives (\frac{3}{550}).

After cancellation, the denominator becomes (2^5\cdot 5^4). The larger exponent is (5), so the decimal terminates after (5) places.

When (9)'s continue forever at the end, the number equals the next terminating decimal. Thus (0.46999\ldots=0.47).

The factors (3^4) and (19) must be removed from the reduced denominator, so the minimum factor is (81\cdot 19=1539). Factors (2) and (5) may remain.

Since (245=5\cdot 7^2), the reduced denominator is (2^2\cdot 5\cdot 7). Since (7) remains, the decimal is non-terminating recurring.

This decimal does not end, and the number of zeros keeps changing. Since there is no fixed repeating block, it is non-terminating non-recurring.

Reason: \(10^8=2^8\cdot5^8\). The given denominator has \(5^8\) but only \(2^4\). To convert the denominator to \(2^8\cdot5^8\) multiply numerator and denominator by \(2^4=16\). Thus \(N=37\times16=592\). About distractors: C (1184) corresponds to multiplying by 32 (\(2^5\)) mistakenly; A (296) is half of the correct value (\(37\times8\)) and often comes from dividing instead of multiplying; D (23125) would result from incorrectly multiplying by \(5^4=625\) (\(37\times625\)). Exam tip: factorise \(10^n\) as \(2^n\cdot5^n\) and balance the powers of 2 and 5 to find the factor for the numerator quickly.

The reduced denominator is (10^7), so the decimal terminates exactly after (7) places. If the denominator is reduced, do not assume further cancellation.

Reduce the fraction first. The numerator is 22 = 2 × 11. Cancelling these factors from the denominator 2^2 × 5^4 × 11^2 leaves 2 × 5^4 × 11. A rational number has a terminating decimal only if every prime factor in its reduced denominator is 2 or 5. Although the factors 2 and 5 would support termination, the factor 11 remains after cancellation, so the denominator is not a power of 10 and the decimal cannot terminate. The number is rational, therefore its infinite decimal expansion is eventually periodic, or recurring. Hence option B is correct. Option A and option D incorrectly assume that the factors 2 and 5 alone decide the result before reduction; option C incorrectly treats a rational decimal as non-recurring.

Since (243=3^5), the reduced denominator is (2^5\cdot 5^4). The larger exponent is (5), so the decimal terminates after (5) places.

At most (9) decimal places means the fraction can be written with denominator (10^9). Therefore the reduced denominator must divide (10^9).

(0.\overline{063}=\frac{63}{999}), and reducing by (9) gives (\frac{7}{111}). An initial zero inside the repeating block is also counted as a digit.

Both (7^0) and (19^0) equal (1), so the effective denominator is (2^5\cdot 5^2). The larger exponent is (5), so the decimal terminates after (5) places.

The factor (41) makes the decimal recurring, and the larger exponent of (2) and (5) is (7), giving the non-repeating start. In mixed denominators, the larger exponent gives the delay.

(0.046875=\frac{46875}{1000000}), and reducing gives (\frac{3}{64}). Convert the decimal to a fraction and reduce fully.

The relevant theorem states that a rational number p/q in lowest terms has a terminating decimal if and only if the reduced denominator q is of the form 2^a × 5^b. Here q also contains 13^r, and r is positive, so a factor 13 remains in the denominator. Lowest form guarantees that this factor cannot be cancelled by the numerator. Therefore q cannot be transformed into a power of 10, and the decimal expansion does not end. Since p/q is rational, its non-terminating decimal must eventually repeat; it cannot be non-terminating non-recurring. Thus option B is correct. Options A and D would require the absence of the factor 13, while option C does not describe the decimal behaviour of a rational number.

Since (320=2^6\cdot 5), the reduced denominator is (2\cdot 5^2\cdot 11). Since (11) remains, the decimal is non-terminating recurring.