Decimal expansion of rational numbers

Step 1: The numerator (3^2\cdot 5) cancels from the denominator. Step 2: The reduced denominator is (2^6\cdot 5^3). The larger exponent is (6), so the decimal terminates after (6) places. Step 3: Look for the larger exponent only after cancellation.

Step 1: The block (357) repeats in a fixed way. Step 2: A fixed recurring decimal can always be written as a rational number. Step 3: Identify rationality when a repeating block is fixed.

Step 1: A decimal terminates when the reduced denominator contains only (2) and (5). Step 2: (q=2^4\cdot 5^3) satisfies this condition. The other options contain (3), (7), or (11). Step 3: Check the prime factors of the denominator carefully.

Step 1: (27=3^3). Step 2: The full factor (3^3) cancels from the denominator, leaving (2^3\cdot 5^2). Step 3: Decide the decimal type from the denominator left after cancellation.

Step 1: (0.00\overline{45}=0.00454545\ldots). Step 2: It equals (\frac{45}{9900}=\frac{1}{220}). So the denominator is (220). Step 3: Choose the denominator only after reducing.

Step 1: (0.00\overline{45}) has two non-repeating zeros and two repeating digits. Step 2: Its fraction form is (\frac{45}{9900}), which reduces to (\frac{1}{220}). Step 3: The first denominator formed from a recurring decimal may not be the final denominator.

Step 1: (\frac{18}{999}=\frac{2}{111}). Step 2: (111=3\cdot 37), which has factors other than (2) and (5). Therefore the decimal is non-terminating recurring. Step 3: Fractions from recurring decimals often have denominators made from (9)'s.

Step 1: For exactly (2) places, the larger exponent must be (2). Step 2: (4=2^2), (20=2^2\cdot 5), and (25=5^2) give exactly (2) places. (50=2\cdot 5^2) also gives exactly (2) places, so none of the listed choices is impossible. Step 3: If all options seem possible, check the question or options for an error.

Step 1: (45=3^2\cdot 5). Step 2: After cancellation, the denominator becomes (2^5\cdot 5^3). The larger exponent is (5), so the decimal terminates after (5) places. Step 3: Always reduce the fraction before counting decimal places.

Step 1: Let (x=0.027777\ldots). Step 2: (100x=2.7777\ldots) and (1000x=27.7777\ldots), so (900x=25) and (x=\frac{25}{900}=\frac{1}{36}). Step 3: For a mixed recurring decimal, separate the non-repeating and repeating parts before multiplying.

The denominator has only (2) and (5), so the decimal terminates with the larger exponent (7). In exams, do not add the exponents.

Since (84=2^2\cdot 3\cdot 7), the reduced denominator is (2\cdot 5^2). The larger exponent is (2), so reduce first and then count places.

For a terminating decimal, (3^4) and (13) must cancel completely, so (n=3^4\cdot 13=1053). For the least value, cancel only the unwanted prime factors.

Taking (x=0.31818\ldots), subtracting (10x) from (1000x) gives (\frac{315}{990}=\frac{7}{22}). The reduced denominator is (22), so none of the listed denominators is correct.

(0.3\overline{18}=0.3181818\ldots=\frac{315}{990}=\frac{7}{22}). Always reduce the final fraction in mixed recurring decimals.

Even after (125=5^3) cancels, (11) remains in the denominator. If a reduced denominator has a prime other than (2) and (5), the decimal is non-terminating recurring.

For exactly (6) places, the larger exponent of (2) and (5) must be (6). Only (2^6\cdot 5^2) has larger exponent (6).

When the denominator is of the form \(2^a5^b\) and the numerator is 17 (which is coprime to 2 and 5), the fraction is already in lowest terms. A terminating decimal requires only factors 2 and 5 in the denominator, and the number of decimal places required equals \(\max(a,b)\). Given the decimal terminates exactly after 9 places and \(a<b\), the larger exponent is \(b\), so \(b=9\). Why other choices fail: 8 is too small, 10 is too large, and \(a+9\) is not implied by the condition \(a<b\). Exam tip: first reduce the fraction if possible; then the termination length equals the larger exponent of 2 or 5 in the reduced denominator.

(0.00072=\frac{72}{100000}), and reducing by (8) gives (\frac{9}{12500}). So the correct denominator is (12500); check the common factor carefully in small decimals.

(0.00072=\frac{72}{100000}), and reducing by (8) gives (\frac{9}{12500}). First write the denominator as a power of (10), then reduce.