Decimal expansion of rational numbers

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.

The factor (7^2) makes the decimal recurring, and the larger exponent among (2) and (5) is (3), giving the non-repeating start. In exams, separate recurrence from the initial delay.

Since (63=3^2\cdot 7), the reduced denominator is (2^4\cdot 5^3). The reason directly explains the terminating decimal rule.

(0.\overline{625}) is a fixed recurring decimal, so it is rational but not terminating. A decimal is terminating only when zeros continue after some point.

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

Since (q) has only (2) and (5), the decimal terminates. Not dividing (10^6) means the larger exponent is (7) or (8).

(0.00\overline{72}=\frac{72}{9900}=\frac{2}{275}). The correct denominator is (275), so choose (275) from the options.

Two non-repeating zeros and two repeating digits give (\frac{72}{9900}). Reducing it gives (\frac{2}{275}).

The direct answer is B: 4 decimal places. First cancel common prime factors: \(\frac{2^3\cdot5^2}{2^7\cdot5^5}=\frac{1}{2^4\cdot5^3}\), because three factors of 2 and two factors of 5 cancel. The denominator is \(2^4\cdot5^3=16\cdot125=2000\). Since a terminating decimal has a denominator made only of 2s and 5s, it terminates. To write the denominator as a power of 10, the larger exponent is 4, so multiply by one extra 5: \(\frac1{2000}=0.0005\), which has four places after the decimal point. Option A, 3, is too small; four places are needed. Option B, 4, is correct. Option C, 5, counts an unnecessary zero or misreads the numerator. Option D, 7, ignores cancellation. Exam cue: after reducing, use the larger exponent of 2 and 5.

A decimal terminates when the reduced denominator has only (2) and (5). The other statements are incomplete because other prime factors may also be present.

When (9)'s continue forever at the end, the number equals the next terminating decimal. Thus (0.124999\ldots=0.125).

The factors (3^2) and (17) must be removed from the reduced denominator, so the minimum factor is (3^2\cdot 17=153). Factors (2) and (5) may remain.

Since (189=3^3\cdot 7), the reduced denominator is (2^2\cdot 5). The larger exponent is (2), so the decimal terminates after (2) places.

This decimal does not end, and the number of zeros between the (2)'s keeps changing. Since there is no fixed repeating block, it is non-terminating non-recurring.

Since (10^6=2^6\cdot 5^6), the denominator lacks (5^4). Thus (N=11\cdot 5^4=6875), so the correct option is (6875).

To obtain denominator \(10^6=2^6\cdot5^6\), we must make the power of 5 equal to 6. The given denominator is \(2^6\cdot5^2\), so multiply numerator and denominator by \(5^4=625\). Hence \(N=11\cdot5^4=11\cdot625=6875\). Common mistakes: 1375 equals \(11\cdot125\) (using \(5^3\) instead of \(5^4\)), while 2750 comes from another incorrect multiplier. Exam tip: prime-factorize the denominator and match exponents of 2 and 5 to form \(10^k\).