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Problem #41 · September 6, 2026

Common fractions

How many irreducible fractions with denominator 24 lie between 16\frac{1}{6} and 56\frac{5}{6}?
  1. A15
  2. B8
  3. C4
  4. D5
  5. E6

WORKED EXAMPLE

Solution

16=424\frac{1}{6}=\frac{4}{24} and 56=2024\frac{5}{6}=\frac{20}{24}, so the numerator must lie in the interval 4<n<204<n<20. Since 24=23⋅324=2^3\cdot3, the irreducibility condition is that nn is divisible by neither 2 nor 3. Such numerators: 5, 7, 11, 13, 17, 195,\,7,\,11,\,13,\,17,\,19 — 6 in total.
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