You will find here three puzzles, suitable for advanced high school and university students.
As always, the difficulty is noted with coffee cups — the more cups the harder the problem.
Good luck!
🧩 Problem 1: Logic/ Algebra
Difficulty: ☕☕
Three children’s ages are whole numbers >0. Their product is 36. A neighbour knows their sum but still can’t determine the ages. The neighbour then learns “the youngest child likes chocolate ice cream” (i.e., there is a youngest child), and then figures the ages. What are the ages?
Hint: List triple factors of 36 and their sums; find ambiguity.
🧩 Problem 2: Probability/ Combinatorics
Difficulty: ☕☕☕
An absent-minded assistant placed three letters randomly into three envelopes. What is the probability that at least one of the recipients gets his letter ?
Hint. Count how many arrangements give no one their own letter (this is easier). Then use the complement: Pr(at least one gets correct)=1−Pr(none get correct).
🧩 Problem 3: Algebra
Difficulty: ☕☕☕
Solve for real x:
Hint: Let the infinite radical equal y and form an equation for x; then translate back to x.
Stay tuned for more puzzles like these ones!
With love and curiosity,
Sybille ☕
Solutions
Solution of Problem 1
Answer: 1, 6 and 6 (the eldest are twins).
Triples whose product = 36:
(1,1,36) → sum = 1+1+36 = 38.
(1,2,18) → sum = 1+2+18 = 21.
(1,3,12) → sum = 1+3+12 = 16.
(1,4,9) → sum = 1+4+9 = 14.
(1,6,6) → sum = 1+6+6 = 13.
(2,2,9) → sum = 2+2+9 = 13.
(2,3,6) → sum = 2+3+6 = 11.
(3,3,4) → sum = 3+3+4 = 10.
The neighbour knows the sum but still can’t determine ages → the sum must be 13 (ambiguous between (1,6,6) and (2,2,9)). Then the neighbour learns “there is a youngest child” (i.e., not all same youngest age — distinguishes a unique younger child), which eliminates (2,2,9) because that has two youngest twins (2 and 2). So the ages are 1, 6, and 6.
Solution of Problem 2
The answer is: 2/3.
Number of total arrangements: 6 (3 choices for where to put letter A, 2 choices for which envelope for letter B, and only one choice remaining for which envelope for letter C).
Number of arrangements where no one gets their own letter: 2 (2 choices for letter A, and only one remaining choice for letters B and C).
Hence:
P(no one gets their own letter)=2/6=1/3
Finally:
P(at least one gets its own letter) = 1 - 1/3 = 2/3.
