Decimal expansion of rational numbers

The reduced denominator is (10^5), so the decimal terminates exactly after (5) places. The numerator condition indicates no further cancellation.

In the first option, (121=11^2) cancels the denominator's (11), leaving only (2) and (5) in the denominator, so it terminates. No option is non-terminating here, so the options need rechecking.

In (\frac{49}{2\cdot 5^2\cdot 7^2}), (49=7^2) cancels completely, so it terminates. For a non-terminating recurring decimal, a factor other than (2) and (5) must remain in the reduced denominator.

A rational number has a terminating decimal expansion after the fraction is reduced only when the denominator has no prime factors other than 2 and 5. If any other prime factor remains in the lowest terms, its decimal expansion is non-terminating but recurring. Therefore, cancellation must be performed before classifying the decimal; looking only at the original denominator could give a wrong conclusion.

Here, \(14=2\cdot7\), so cancellation with the numerator gives \(\frac{14}{2\cdot5^2\cdot7^2}=\frac{1}{5^2\cdot7}\). The reduced denominator still contains the prime factor 7. Hence the decimal cannot terminate and must be non-terminating recurring. Thus option B is correct. It is not non-recurring because every rational number has either a terminating or recurring decimal expansion.

At most (5) decimal places means the fraction can be written with denominator (10^5). The reduced denominator must divide (10^5).

(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.

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

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

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.

The denominator has only (2) and (5), so the decimal terminates with the larger exponent (8). In exams, use the larger exponent, not the sum.

Since (198=2\cdot 3^2\cdot 11), the reduced denominator is (2\cdot 5^5). The larger exponent is (5), so reduce first and then count places.

For termination, (3^2) and (7^2) must cancel completely, so (n=441). For the least value, cancel only the factors other than (2) and (5).