A fraction has denominator 36 and lies strictly between 1/3 and 1/2. How many such fractions are already in their lowest terms?

Rewrite the bounds over 36: 1/3 = 12/36 and 1/2 = 18/36, so the numerator runs over 13, 14, 15, 16 and 17 — five fractions in the gap. To be in lowest terms the numerator must share no factor with $36=2^2\times 3^2$, which knocks out the even numerators 14 and 16 and the multiple of 3, namely 15. Only 13/36 and 17/36 survive, so the answer is 2. The trap 5 counts every fraction in the gap without asking whether it cancels; 3 removes only the even numerators and forgets 15; and 1 forgets that 17 is prime and therefore safe.