Difficulty: Intermediate | Prerequisites: Comfort with fractions, place value, and basic algebra.
Decimals sit at the intersection of number theory and algebra, and these problems go well beyond routine arithmetic. They test whether you can translate between decimal notation and fractions, work with repeating decimals as rational numbers, and apply algebraic reasoning to digit-level puzzles. If you are not yet confident converting repeating decimals to fractions, review that technique before tackling these. The problems come from a problem-solving context rather than a standard lecture, so expect to think sideways.
Three competition-style problems that use decimal place-value shifts, repeating-decimal-to-fraction conversion, and factorial divisibility. Solving them requires setting up equations from the decimal representation, not just plugging numbers in. The core skill throughout is moving fluently between a decimal and its fraction form.
Repeating decimal (recurring decimal)
A decimal in which one or more digits repeat infinitely after a certain point. Written with a bar over the repeating block, e.g. 0.636363... = 0.63 (bar over 63).
In simple terms, it is a decimal that never ends but cycles through the same digits forever.
Repetend
The block of digits that repeats in a repeating decimal. In 0.636363..., the repetend is "63".
Think of it as the "loop" in the decimal.
Standard repeating decimal
As defined in this problem set: a repeating decimal whose repetend is exactly one digit long. For example, 0.333... (repetend is 3) is standard; 0.636363... (repetend is 63, two digits) is not.
Terminating decimal
A decimal that ends after a finite number of digits, e.g. 0.25 or 1.5. A fraction terminates in decimal form if and only if its denominator (in lowest terms) has no prime factors other than 2 and 5.
In simple terms, if you can write it out completely without needing a repeating bar, it terminates.
Reciprocal
The reciprocal of a number x is 1/x. For example, the reciprocal of 4 is 0.25, and the reciprocal of 0.02 is 50.
Factorial (n!)
The product of all positive integers from 1 up to n. So 4! = 4 × 3 × 2 × 1 = 24.
Think of it as "multiply every whole number from 1 to n together."
Problem statement
When the decimal point of a certain positive decimal number is moved four places to the right, the new number is four times the reciprocal of the original number. Find the original number.
Setup
Let the original number be x. Moving the decimal point four places to the right multiplies x by 10,000, giving 10000x. The problem says this equals 4 times the reciprocal of x:
10000x = 4/x
Solution
Multiply both sides by x: 10000x² = 4
Divide both sides by 10000: x² = 4/10000 = 1/2500
Take the positive square root (the problem specifies a positive number): x = 1/50 = 0.02
Verification
Moving the decimal point of 0.02 four places right gives 200. The reciprocal of 0.02 is 50, and 4 × 50 = 200. Confirmed.
Key technique
Translate "move the decimal point k places right" into "multiply by 10^k." Then set up the algebraic equation and solve. The reciprocal of x is 1/x.
Problem statement
A student multiplied 66 by the repeating decimal 1.ababab... (written 1.ab with a bar over ab), where a and b are digits. He misread the repeating decimal as the terminating decimal 1.ab and multiplied 66 × 1.ab instead. His answer was 0.5 less than the correct answer. Find the two-digit number ab.
Setup
The repeating decimal 1.ababab... can be converted to a fraction. The repeating part is 0.ababab... Let that equal y.
100y = ab.ababab...
100y − y = ab
99y = ab
y = ab/99
So 1.ababab... = 1 + ab/99.
The terminating decimal 1.ab = 1 + ab/100.
Equation
Correct answer minus incorrect answer = 0.5:
66 × (1 + ab/99) − 66 × (1 + ab/100) = 0.5
66 × ab × (1/99 − 1/100) = 0.5
66 × ab × (100 − 99)/(99 × 100) = 0.5
66 × ab × 1/9900 = 0.5
66 × ab = 0.5 × 9900 = 4950
ab = 4950/66 = 75
Answer: ab = 75
Verification
1.757575... = 1 + 75/99 = 174/99 = 58/33
66 × 58/33 = 2 × 58 = 116
66 × 1.75 = 115.5
Difference: 116 − 115.5 = 0.5. Confirmed.
Key technique
Convert the repeating decimal to a fraction using the standard method (multiply by the appropriate power of 10 to shift the repeating block, then subtract). The difference between the fraction form and the truncated decimal form sets up the equation.
Problem statement
A repeating decimal is called "standard" if its repetend is exactly one digit long (e.g. 0.6555... where the 5 repeats, or 0.333...). Find the sum of all positive integers n such that 1/n! can be expressed as a standard repeating decimal that is not a terminating decimal.
Key background
A fraction p/q in lowest terms is:
Terminating if and only if q has no prime factors other than 2 and 5.
A repeating decimal with a one-digit repetend if and only if, after removing all factors of 2 and 5 from q, what remains is a divisor of 9. The only divisors of 9 are 1, 3, and 9. If the remaining factor is 1, the decimal terminates (not what we want). So q, after stripping out all 2s and 5s, must leave exactly 3 or 9.
Working through the values
n = 1: 1/1! = 1. This is a terminating decimal. Does not qualify.
n = 2: 1/2! = 1/2 = 0.5. Terminating. Does not qualify.
n = 3: 1/3! = 1/6. In lowest terms, 1/6 has denominator 6 = 2 × 3. Strip the 2: remainder is 3, which divides 9. So 1/6 = 0.1666... = 0.16 (bar over the 6). This is a standard repeating decimal. Qualifies.
n = 4: 1/4! = 1/24. Denominator 24 = 2³ × 3. Strip the 2s: remainder is 3, which divides 9. So 1/24 = 0.041666... = 0.0416 (bar over the 6). Qualifies.
n = 5: 1/5! = 1/120. Denominator 120 = 2³ × 3 × 5. Strip 2s and 5: remainder is 3. Qualifies. 1/120 = 0.00833...
n = 6: 1/6! = 1/720. Denominator 720 = 2⁴ × 3² × 5. Strip 2s and 5: remainder is 9 (= 3²), which divides 9. Qualifies.
n = 7: 1/7! = 1/5040. Denominator 5040 = 2⁴ × 3² × 5 × 7. Strip 2s and 5: remainder is 9 × 7 = 63. Does 63 divide 9? No (63 > 9). Does not qualify.
For n ≥ 7: n! includes 7 as a factor (or higher primes), and after stripping 2s and 5s, the remainder will include primes other than 3, so it cannot divide 9. None qualify.
Answer
The qualifying values are n = 3, 4, 5, 6.
Sum = 3 + 4 + 5 + 6 = 18
Key technique
To determine the repeating behaviour of a fraction, factor the denominator, remove all 2s and 5s, and examine what remains. The length of the repetend is the smallest positive integer k such that 10^k ≡ 1 (mod the remaining factor). For a one-digit repetend, that remaining factor must divide 9 (i.e. 10¹ − 1).
Converting a two-digit repeating decimal to a fraction
If 0.ababab... is a repeating decimal with a two-digit repetend, then:
0.ababab... = ab/99
More generally, for a k-digit repetend: 0.(repetend)(repetend)... = repetend / (10^k − 1).
Shifting the decimal point
Moving the decimal point k places to the right = multiplying by 10^k.
Moving the decimal point k places to the left = dividing by 10^k.
Condition for a terminating decimal
A fraction p/q in lowest terms terminates if and only if every prime factor of q is 2 or 5.
Condition for a one-digit repetend (standard repeating decimal)
A non-terminating fraction p/q in lowest terms has a one-digit repetend if and only if, after removing all factors of 2 and 5 from q, the result divides 9 (meaning the result is 3 or 9).
Students often confuse the repetend with the entire decimal. In 0.1666..., the repetend is just 6 (one digit), not 16 or 166. Only the part under the bar repeats.
Students sometimes assume that any fraction with an odd denominator produces a repeating decimal. That is wrong: 1/5 = 0.2, which terminates, because 5 is one of the two allowed prime factors for termination.
When converting repeating decimals to fractions, a common error is using the wrong power of 10. For a two-digit repeating block, multiply by 100 (not 10). For a one-digit block, multiply by 10.
Students occasionally forget that "non-terminating" is part of the condition in Problem 3. The fraction 1/2 = 0.5 has a trivial one-digit representation, but it terminates, so it does not count as a repeating decimal at all.
⚠️ Converting between repeating decimals and fractions is a perennial exam topic. Know the general formula cold: a repeating block of k digits over (10^k − 1).
⚠️ Problems that combine decimals with reciprocals or factorials test whether you can set up an equation from a verbal description. Practise translating words into algebra before the exam.
⚠️ The prime factorisation of a denominator determines everything about a decimal's behaviour: whether it terminates, how long the repetend is, and whether it qualifies as "standard." This fact is tested directly and indirectly across number theory and algebra courses.
True or false: 0.125 is a repeating decimal. (Answer: False. It terminates.)
Fill in the blank: 0.272727... as a fraction is ___/99. (Answer: 27/99, which simplifies to 3/11.)
True or false: Moving the decimal point three places to the right is the same as multiplying by 1000. (Answer: True.)
Fill in the blank: A fraction in lowest terms terminates if and only if its denominator has no prime factors other than ___ and ___. (Answer: 2 and 5.)
True or false: 1/7! has a one-digit repeating block. (Answer: False. 7! includes 7 as a prime factor, so the stripped denominator does not divide 9.)
Q: Express 0.363636... as a fraction in lowest terms.
A: 0.363636... = 36/99 = 4/11.
Q: A positive number x satisfies the condition that moving its decimal point three places to the right gives 8 times the reciprocal of x. What is x?
A: 1000x = 8/x, so 1000x² = 8, x² = 8/1000 = 1/125, and x = 1/(5√5) = √5/25. Alternatively, x² = 0.008, so x ≈ 0.0894. (Exact form: x = 2√2 / (10√10) = √5/25.)
Q: For which values of n (n a positive integer) is 1/n a standard repeating decimal (one-digit repetend, non-terminating)?
A: We need the denominator (after removing factors of 2 and 5) to be 3 or 9. The qualifying denominators are: 3, 6, 9, 12, 15, 18, 24, 30, 36, 45, 48, 72, 90, 96, 120, 144, 180, 240, 360, 720, and so on. Any n of the form 2^a × 3^b × 5^c where b = 1 or 2, a ≥ 0, c ≥ 0, and the fraction is non-terminating (so b ≥ 1).
Q: What is the length of the repetend of 1/7?
A: 1/7 = 0.142857142857... The repetend is 142857, which is 6 digits long. (Because the smallest k with 10^k ≡ 1 mod 7 is k = 6.)
Q: A student multiplied 33 by the repeating decimal 2.ababab... but misread it as the terminating decimal 2.ab. The error was 0.25. Find ab.
A: 33 × ab × (1/99 − 1/100) = 0.25. So 33 × ab / 9900 = 0.25, giving ab = 0.25 × 9900 / 33 = 75. (Same answer; the structure scales.)
This material connects directly to number theory, particularly the structure of rational numbers and modular arithmetic. The idea that the repetend length of 1/n depends on the multiplicative order of 10 modulo n comes up again in topics like Fermat's little theorem and primitive roots.
The factorial-based reasoning in Problem 3 bridges into combinatorics and the study of prime factorisations of n!, which reappears in topics like Legendre's formula for the largest power of a prime dividing n!.
The algebra of setting up equations from word problems (Problems 1 and 2) is a skill tested across every level of mathematics, from pre-algebra competition problems through to university entrance exams.
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