Decimal expansion of rational numbers

Since (62=2\cdot 31), the factor (31) cancels and the reduced denominator is (2^3\cdot 5^3). If an extra prime appears, check cancellation first.

(0.\overline{36}=\frac{36}{99}) and (0.\overline{63}=\frac{63}{99}), so their sum is (1). The sum of two recurring decimals can be terminating.

Since (17) remains, the decimal is non-terminating recurring. The larger exponent in (2^4\cdot 5^4) gives (4) initial non-repeating digits.

For a terminating decimal, the reduced denominator (q) can contain only (2) and (5). In (q^4), powers increase but no new prime factor appears.

For a rational number, the decimal expansion terminates exactly when the denominator in lowest terms contains only the primes 2 and 5. Factors 2 and 5 can be combined to make a power of 10, so division ends. A remaining prime such as 17 prevents this and produces a repeating pattern of digits instead.

After cancellation, \(2^5\) removes part of \(2^9\), and one factor 17 removes part of \(17^2\). The denominator becomes \(2^4\cdot 5^2\cdot 17\). Because 17 remains, the denominator is not of the required form. Thus the decimal expansion is non-terminating recurring, so option B follows. Option A would be possible only if every factor other than 2 and 5 had cancelled.

(0.015625=\frac{15625}{1000000}=\frac{1}{64}). Convert a terminating decimal to a fraction and reduce the denominator.

(0.00015625=\frac{15625}{100000000}), and reducing by (15625) gives (\frac{1}{6400}). Do not forget to cancel common factors in large denominators.

(448=2^6\cdot 7), so (6) non-repeating digits appear before the recurring part. For comparison, check the larger power of (2) and (5).

Core idea: \(10^8=2^8\cdot 5^8\). The given denominator has \(2^8\) but only \(5^5\), so it is short by \(5^3\). Multiply numerator and denominator by \(5^3=125\) to obtain denominator \(10^8\). Thus \(N=11\times125=1375\).
Why other options are wrong: 275 equals \(11\times25\) (wrong if you multiply by \(5^2\) instead of \(5^3\)). 2750 is simply twice the correct N (a mistake from an extra factor 2). 6875 equals \(11\times625\) (would result from multiplying by \(5^4\)).
Exam tip: Compare prime-power factors of the denominator with \(10^n\); multiply numerator by the missing power of 2 or 5 to convert to denominator \(10^n\).

For exactly (3) places, the larger exponent must be (3). Since (16=2^4), it terminates after (4) places.

When (9)'s continue forever at the end, the number equals the next terminating decimal. Thus (0.00999\ldots=0.01).

Because 7^0 = 1, the denominator is effectively 2^5 × 5^5. These powers combine to give 10^5, since 2^5 × 5^5 = (2 × 5)^5 = 10^5. Thus the fraction has a denominator of 100000 after reduction and its decimal expansion terminates within five places. Moreover, because the fraction is in lowest form, the numerator is coprime to both 2 and 5. It therefore cannot supply a factor that would cancel the final power of 10; the fifth decimal digit cannot become an unnecessary trailing zero. Hence the expansion terminates exactly after five places, making option A correct. Option B incorrectly adds exponents, while options C and D contradict the fact that the denominator contains only 2 and 5.

The factors (3), (7), and (13) must be removed from the reduced denominator, so the minimum factor is (3\cdot 7\cdot 13=273). Factors (2) and (5) may remain.

Since (750=2\cdot 3\cdot 5^3), the reduced denominator is (2^5\cdot 5^2). The larger exponent is (5), so the decimal terminates after (5) places.

(0.124545\ldots=\frac{1245-12}{9900}=\frac{1233}{9900}=\frac{137}{1100}). Always reduce the final fraction in mixed recurring decimals.

The factors (7^3) and (11) must be removed from the reduced denominator, so (n=7^3\cdot 11=3773). For the least value, do not cancel (2) and (5).

Since (19^2) remains, the decimal is non-terminating recurring, and the larger exponent among (2) and (5) is (4). In such questions, separate recurrence from the initial delay.

For exactly (6) places, the larger exponent of (2) and (5) must be (6). Since (3125=5^5), it gives only (5) decimal places.

Core idea: write denominators with the same prime factors. Since \(10^9=2^9\cdot5^9\) and the given denominator is \(2^5\cdot5^9\), multiply denominator by \(2^4\) to get \(2^9\). Multiply the numerator by the same factor: \(N=23\times2^4=23\times16=368\). The closest distractor 184 corresponds to multiplying by \(2^3=8\) (one factor of 2 short), hence incorrect; 736 and 1472 result from using larger powers of 2. Exam tip: compare exponents of 2 and 5 in the denominator and multiply numerator/denominator to equalize them to powers of 10.

(0.\overline{216}=\frac{216}{999}=\frac{8}{37}). For a purely recurring decimal, first use a denominator of (9)'s and then reduce fully.