Decimal expansion of rational numbers

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).

(0.254545\ldots=\frac{252}{990}=\frac{14}{55}), so the denominator is (55). In such questions, reduce the final fraction fully.

The non-repeating part (2) and repeating part (54) give (\frac{252}{990}), which reduces to (\frac{14}{55}). In exams, identify repeating and non-repeating digits separately.

Since (147=3\cdot 7^2), the reduced denominator is (2\cdot 5^4). The larger exponent is (4), so the decimal terminates after (4) places.

For exactly (7) places, the larger exponent of (2) and (5) must be (7). Only (2^7\cdot 5^3) satisfies this.

A rational number has a terminating decimal exactly when its denominator (in lowest terms) is of the form \(2^m5^n\). The number of decimal places needed equals max(m,n). Here 29 is coprime to 2 and 5, so the denominator remains \(2^a5^b\) after simplification. Given \(a>b\), the maximum exponent is \(a\); for exactly 8 decimal places we must have \(a=8\). Option 7 is wrong because it would give only 7 decimal digits; \(a+b\) is irrelevant since the decimal length depends on the maximum exponent, not the sum. Exam tip: always reduce the fraction first and then take the larger of the exponents of 2 and 5 to get the number of decimal places.

(0.00096=\frac{96}{100000}), and reducing by (32) gives (\frac{3}{3125}). Even for small decimals, check the greatest common factor carefully.

(0.00096=\frac{96}{100000}=\frac{3}{3125}), so the denominator is (3125). Convert the decimal to a fraction and reduce fully.

The factor (13) makes the decimal recurring, and the larger exponent among (2) and (5) is (5), giving the initial non-repeating part. Understand recurrence and delay separately.

Since (121=11^2), the reduced denominator is (2^3\cdot 5^2). Therefore the reason correctly explains the terminating decimal rule.

(0.04\overline{6}) has a fixed repeating digit, so it is rational but not terminating. A decimal is terminating only when zeros continue after some point.

After cancellation, the denominator becomes (2^6\cdot 5^3). The larger exponent is (6), so the decimal terminates after (6) places.

Since (q) has only (2) and (5), the decimal terminates. Not dividing (10^7) means the larger exponent is (8) or (9).

(0.00\overline{63}=\frac{63}{9900}=\frac{7}{1100}), so the denominator is (1100). In recurring decimals, the first denominator formed may not be final.

Two non-repeating zeros and two repeating digits give (\frac{63}{9900}). Reducing it gives (\frac{7}{1100}).

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

(10^k) has only prime factors (2) and (5), so any divisor gives a terminating decimal. The other statements are not always true because extra factors may occur.

When (9)'s continue forever at the end, the number equals the next terminating decimal. Thus (0.37999\ldots=0.38).