I am an odd number.
Take away one letter from me and I become even.
ANSWER : Seven
ANSWER : 3 minutes
Answer: In 20 years
Answer: 17
Answer: 10 (you still have them)
Answer: 0
Answer: Not boys = girls
Answer: 3 o’clock
Answer: On a clock
Answer: Turn on A long, off; turn on B; enter: hot = A, lit = B, cold = C.
Answer: 17 minutes.
Answer: Use standard 3-weigh decision tree (partition, compare, resolve).
Answer: Impossible (removed same color → unequal colors).
Answer: First wins; leave multiples of 5 — remove 1 initially.
Answer: Light A both ends, B one end; when A finishes (30'), light B other end → +15'.
Answer: 33 (2^n+1).
Answer: Solve linear system over GF(2) via Gaussian elimination.
Answer: Mix a drop from all four cups into one sample. If blue → at least one poisoned; then use elimination by pair-splitting (logic puzzle). (Alternate strict version uses binary mixing if multiple strips allowed.)
Answer: Minimum worst-case = 14 drops (1+2+…+14 ≥ 100).
Answer: Designate a counter; others turn the bulb on once; counter counts ON occurrences.
Answer: The sum decomposes only into product pairs whose products factor uniquely (classic 4–13 logic puzzle structure).
Answer: Fill 5L → pour into 7L; fill 5L again → top up 7L (needs 2L) → remaining 3L in 5L; empty 7L; pour 3L into 7L; fill 5L again and pour until 7L full (needs 4L) → 1L left.
Answer: Yes — a closed knight’s tour exists.
Answer: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 (single digits). Many multi-digits collapse to them.
Answer: Perfect squares (1,4,9,16…). Only they have odd number of divisors.
Answer: Topological sort + dynamic programming in linear time
Answer: Expected-value reasoning seems paradoxical; resolution: switching expectation is flawed because prior distribution not symmetric.
Answer: Yes (Four-colour theorem — requires computer-assisted proof).