Answer: 6

Connect remainders to repeating length: Each unique remainder produces a unique quotient digit. Once a remainder repeats, the cycle restarts. There are 6 unique remainders before the first one ($3$) reappears. Why the correct answer works: Option C is correct. Six unique remainders means six quotient digits in the repeating block: $10 \div 7 = 1.\overline{428571}$. Why distractors fail: Option A (3) undercounts the remainders. Option B (7) equals the divisor, but one remainder (0) would mean termination, so at most 6 remainders are possible. Option D (5) miscounts by one.

Answer: $0.58\overline{3}$

Perform long division: $7 \div 12 = 0.58333\ldots$ The digits $5$ and $8$ appear once, then $3$ repeats indefinitely. Apply bar notation correctly: The bar goes only over the repeating portion: $0.58\overline{3}$. Option A is correct. Why distractors fail: Option B ($0.\overline{583}$) implies all three digits repeat, giving $0.583583\ldots$, which is incorrect. Option C ($0.5\overline{83}$) implies $83$ repeats, giving $0.58383\ldots$. Option D treats it as a terminating decimal.

Answer: The reasoning is flawed; $\frac{6}{15}$ simplifies to $\frac{2}{5}$, which terminates since the denominator is only 5

Simplify the fraction first: $\frac{6}{15} = \frac{2}{5}$ after dividing numerator and denominator by $\gcd(6,15) = 3$. Check the simplified denominator: The reduced denominator is $5 = 5^1$, which has only the prime factor 5. Therefore the decimal terminates: $\frac{2}{5} = 0.4$. Why the colleague's error matters: The rule about prime factors applies to the denominator after the fraction is fully reduced. Checking the unreduced denominator can give a wrong conclusion. Why distractors fail: Option A accepts the flawed reasoning. Option C makes the error even broader. Option D gives the right conclusion but the wrong explanation—divisibility of both by 3 is about simplification, not a general termination rule.

Answer: A terminating decimal

Key rule for terminating decimals: A fraction in lowest terms produces a terminating decimal if and only if the denominator has no prime factors other than 2 and 5. Why the correct answer works: Option B is correct because denominators of the form $2^a \times 5^b$ can always be converted to a power of 10, yielding a finite decimal. Why distractors fail: Option A is wrong because repeating decimals occur when the denominator has prime factors other than 2 and 5. Option C describes irrational numbers, not fractions. Option D is incorrect because all fractions produce well-defined decimals.

Answer: 3

Perform the division: $5 \div 6 = 0.8333\ldots = 0.8\overline{3}$. The digit $8$ appears once, then the digit $3$ repeats. Why the correct answer works: Option B is correct. Only the digit $3$ repeats; the $8$ is a non-repeating part of the decimal. Why distractors fail: Option A (83) incorrectly includes the non-repeating digit. Option C (8) is the non-repeating part. Option D (56) has no relation to the decimal expansion.

Answer: 1

Trace the long division: $1 \div 3 = 0.3$ remainder $1$. Bring down a zero: $10 \div 3 = 3$ remainder $1$. The remainder $1$ reappears at every step. Why the correct answer works: Option B is correct because each division step produces a remainder of $1$, which causes the digit $3$ to repeat indefinitely. Why distractors fail: Option A (remainder 0) would mean the decimal terminates. Option C (remainder 2) occurs in other divisions like $2 \div 3$ but not $1 \div 3$. Option D (remainder 3) is impossible when dividing by 3.

Answer: $0.83$

Expand the decimal: $0.8\overline{3} = 0.83333\ldots$ The hundredths digit is $3$, and the digit to its right (thousandths) is also $3$. Apply rounding rule: Since the thousandths digit ($3$) is less than $5$, the hundredths digit stays at $3$. The rounded value is $0.83$. Why distractors fail: Option B ($0.84$) incorrectly rounds up. Option C ($0.80$) rounds to the nearest tenth, not hundredth. Option D ($0.90$) is far too aggressive a rounding.

Answer: Yes, because the hundredths digit is 5, which rounds up

Expand and identify rounding digits: $0.\overline{45} = 0.454545\ldots$ The tenths digit is $4$, and the hundredths digit is $5$. Apply the rounding rule: Since the hundredths digit is $5$ (i.e., $\geq 5$), the tenths digit rounds up from $4$ to $5$. The result is $0.5$. Why distractors fail: Option B invents a false rule about repeating decimals rounding down. Option C is wrong because $0.\overline{45} = \frac{45}{99} = \frac{5}{11} \approx 0.4545$, not $0.5$ exactly. Option D incorrectly identifies the tenths digit as $5$ (it is $4$).

Answer: $0.\overline{45}$

Identify the repeating block: The decimal $0.454545\ldots$ has the two-digit block $45$ repeating from the start. The entire repeating unit is $45$. Why the correct answer works: Option C places a single bar over $45$, meaning $0.\overline{45} = 0.454545\ldots$, which matches the expansion. Why distractors fail: Option A uses separate bars on individual digits, which is non-standard notation. Option B implies only the $5$ repeats ($0.4555\ldots$). Option D redundantly shows $45$ before the bar, suggesting a non-repeating prefix that does not exist.

Answer: $\frac{7}{15}$ is already in lowest terms; $15 = 3 \times 5$, which contains the factor 3, so it repeats

Check if the fraction is in lowest terms: $\gcd(7, 15) = 1$, so $\frac{7}{15}$ is already fully reduced. Factor the denominator: $15 = 3 \times 5$. The factor $3$ is neither $2$ nor $5$, so the fraction produces a repeating decimal. Why distractors fail: Option B focuses only on the factor 5 and ignores the factor 3. Option C incorrectly simplifies the fraction. Option D invents a false rule about odd denominators.

Answer: Stored value: $3.1429$; repeating block: $142857$

Compute the division: $22 \div 7 = 3.\overline{142857} = 3.142857142857\ldots$ Identify the repeating block: The sequence $142857$ repeats every 6 digits. Round to 4 decimal places: The first 5 decimal digits are $14285$. The fourth digit is $8$, and the fifth digit is $5$. Since $5 \geq 5$, round up: $3.1429$. Why distractors fail: Option A has the correct repeating block but fails to round up. Option C rounds correctly but gives an incomplete repeating block. Option D gets both the rounding and the repeating block wrong.

Answer: $31.33$ kg

Compute the product: $\frac{2}{3} \times 47 = \frac{94}{3} = 31.\overline{3} = 31.3333\ldots$ Round to the nearest hundredth: The hundredths digit is $3$. The thousandths digit is also $3$, which is less than $5$, so the hundredths digit stays. Result: $31.33$. Why distractors fail: Option B ($31.34$) rounds up incorrectly. Option C ($31.30$) rounds to the nearest tenth. Option D ($31.00$) rounds to the nearest whole number.

Answer: No, $0.1\overline{6} = 0.1666\ldots$ while $0.\overline{16} = 0.161616\ldots$

Expand each notation: $0.1\overline{6} = 0.16666\ldots$ (only the 6 repeats). $0.\overline{16} = 0.161616\ldots$ (the block 16 repeats). Why the correct answer works: Option B correctly identifies that the two expansions differ. $0.1\overline{6} \approx 0.1667$ while $0.\overline{16} \approx 0.1616$, so they are different numbers. Why distractors fail: Option A is wrong because $\frac{1}{6} = 0.1\overline{6}$, but $0.\overline{16} = \frac{16}{99}$, a different value. Option C incorrectly claims bar position is irrelevant. Option D is wrong because $0.17 \neq 0.16$ when rounded to two decimal places.

Answer: $\frac{9}{40}$

Check the denominator's prime factors: A fraction in lowest terms terminates if and only if the denominator's prime factors are limited to 2 and 5. Factor each denominator: $7$ is prime (not 2 or 5). $12 = 2^2 \times 3$ (contains 3). $40 = 2^3 \times 5$ (only 2 and 5). $9 = 3^2$ (contains 3). Why the correct answer works: Option C ($\frac{9}{40}$) has denominator $40 = 2^3 \times 5$, containing only factors of 2 and 5, so it terminates: $9 \div 40 = 0.225$. Why distractors fail: Options A, B, and D all have denominators with prime factors other than 2 and 5, so they produce repeating decimals.

Answer: Because the denominator 7 has a prime factor other than 2 or 5

Recall the terminating decimal rule: A fraction in lowest terms terminates if and only if the denominator's prime factorization contains only the primes 2 and/or 5. Why the correct answer works: Option C correctly identifies the reason: 7 itself is a prime factor that is neither 2 nor 5, so $\frac{1}{7}$ must repeat. Why distractors fail: Option A is irrelevant—the numerator's parity does not determine termination. Option B is close but imprecise; being prime and greater than 5 is not the general rule (e.g., $\frac{1}{2}$ terminates despite 2 being prime). Option D is a consequence, not the fundamental reason.

Answer: The digit 3 repeats infinitely

Understand bar notation: A bar (vinculum) placed over one or more digits in a decimal indicates that those digits repeat infinitely. So $0.\overline{3}$ means $0.333\ldots$ continuing forever. Why the correct answer works: Option B correctly states that the digit 3 repeats infinitely, which is exactly what bar notation communicates. Why distractors fail: Option A incorrectly limits the repetition to exactly three times. Option C confuses bar notation with termination. Option D confuses bar notation with a rounding instruction.

Answer: $0.667$

Expand the repeating decimal: $0.\overline{6} = 0.66666\ldots$ The thousandths digit is $6$, and the ten-thousandths digit is also $6$. Apply the rounding rule: Since $6 \geq 5$, the thousandths digit rounds up from $6$ to $7$, giving $0.667$. Why distractors fail: Option A ($0.666$) fails to round up. Option C ($0.670$) rounds to the nearest hundredth after rounding the thousandths. Option D ($0.700$) over-rounds to the nearest tenth.

Answer: 12

Apply the pigeonhole principle to remainders: When dividing by $n$, the possible remainders are $0, 1, 2, \ldots, n-1$. A remainder of $0$ means the decimal terminates. So at most $n - 1$ non-zero remainders can appear before one must repeat. Why the correct answer works: For divisor $13$, there are at most $13 - 1 = 12$ distinct non-zero remainders. Option C is correct. Why distractors fail: Option A (6) is about half of 13 and has no special significance here. Option B (13) forgets to exclude the remainder $0$. Option D (14) exceeds the total number of possible remainders.