Decimal expansion of rational numbers

The factors (3^3) and (23) must be removed from the reduced denominator, so the minimum factor is (27\cdot 23=621). Factors (2) and (5) may remain.

Since (175=5^2\cdot 7), 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.

Since (10^7=2^7\cdot 5^7), the denominator lacks (2^4). Thus (N=13\cdot 16=208), so the correct value is not listed.

Key idea: \(10^7=2^7\cdot 5^7\). The given denominator has only \(2^3\), so multiply numerator and denominator by \(2^4\) to raise the power of 2 to 7. Thus \(N=13\cdot 2^4=13\cdot16=208\). Option C (416) would be \(13\cdot2^5\) — one extra factor of 2 — and A (104) corresponds to \(13\cdot2^3\), so both are incorrect. Exam tip: match prime powers of 2 and 5 to form \(10^n\) quickly by comparing exponents.

The reduced denominator is (10^6), so the decimal terminates exactly after (6) places. If the denominator is reduced, do not assume further cancellation.

The denominator must be examined after cancelling common factors. Factor the numerator as 14 = 2 × 7. Cancelling these factors from 2^2 × 5^3 × 7^2 leaves the reduced denominator 2 × 5^3 × 7. The criterion says that a rational number has a terminating decimal only when its reduced denominator has no prime factors other than 2 and 5. The remaining factor 7 prevents termination. Because the number is rational, its non-terminating decimal is recurring rather than non-recurring. Therefore option B is correct. Option A is wrong because the factor 7 remains; option D is also wrong for the same reason, and option C would describe a non-rational decimal rather than this rational number.

Since (81=3^4), the reduced denominator is (2^4\cdot 5^6). The larger exponent is (6), so the decimal terminates after (6) places.

At most (7) decimal places means the fraction can be written with denominator (10^7). Therefore the reduced denominator must divide (10^7).

(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.

For a rational number written in lowest form, the decimal expansion terminates precisely when the denominator has no prime factors other than 2 and 5. In this question, r > 0, so 11^r is a genuine factor of the reduced denominator. Because the fraction is already in lowest form, this factor 11 cannot cancel with the numerator. Consequently the denominator cannot be changed into a power of 10 by multiplying numerator and denominator by suitable factors. The division will continue without ending. Since every rational decimal is either terminating or eventually recurring, this continuing decimal must be non-terminating recurring. Thus option B is correct. Option A and option D ignore the surviving factor 11, whereas option C is associated with irrational numbers, not a rational fraction.

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.