What can I know from what I see? What can I learn when someone else says “I do not know”? Try each puzzle before opening its solution. Some have a unique answer; others teach us to recognize insufficient information. The answers stay closed until you choose to reveal them.
Three hats
A, B and C can see the others’ red or blue hats, but not their own. At least one hat is red. A says “I do not know my color”; after hearing this, B says the same; C then says “I know mine”. What color is C’s hat?
Show solution and explanation
Red. If A saw two blue hats, A would know their own was red: therefore B or C is red. If B saw C wearing blue, B could infer that their own hat was red. Since B cannot, C wears red.
Consecutive numbers
Anna and Bruno have consecutive integers from 1 to 10 on their foreheads. Each sees only the other’s number. Anna sees 9 and says “I do not know mine.” Bruno says “Neither do I.” Anna now knows. What is her number?
Show solution and explanation
8. Anna initially considers 8 or 10. If she had 10, Bruno would see 10 and know that 9 was the only allowed neighbor. His uncertainty rules out 10.
One is twice the other
Anna and Bruno have integers from 1 to 20 on their foreheads; one is twice the other. Anna sees 8 and cannot identify her number. Bruno cannot identify his either. Anna then deduces hers. What is it?
Show solution and explanation
4. Anna’s initial possibilities are 4 and 16. If hers were 16, Bruno would see 16 and know he had 8, since 32 is outside the range. His uncertainty rules out 16.
The three daughters
Three daughters’ positive integer ages have product 36. A friend sees the house number, equal to the sum of their ages, but still cannot determine them. The father adds, “The eldest daughter plays piano.” What are their ages?
Show solution and explanation
2, 2 and 9. The possible unordered triples have sums 38, 21, 16, 14, 13, 13, 11 and 10. Only sum 13 leaves two possibilities: (1,6,6) and (2,2,9). A single eldest daughter rules out (1,6,6).
Cheryl’s birthday
Cheryl’s birthday is among May 15, 16, 19; June 17, 18; July 14, 16; August 14, 15, 17. She tells Albert the month and Bernard the day. Albert: “I do not know the date, but I know Bernard does not.” Bernard: “I did not know; now I do.” Albert: “Now I know too.” What is the date?
Show solution and explanation
July 16. Albert’s first statement eliminates May and June because days 19 and 18 are unique. Among July and August, Bernard’s new knowledge excludes day 14, which occurs in both months. August would still leave two dates, 15 and 17, for Albert. Therefore his month is July.
Every label is wrong
Three boxes contain apples only, pears only, or both. Their labels APPLES, PEARS and APPLES AND PEARS are all wrong. You may draw one fruit from one box. How can you label all three?
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Draw from the box labeled APPLES AND PEARS: it cannot be mixed. If you draw an apple, it contains apples only; the box labeled PEARS cannot contain pears or the already assigned apples, so it is mixed; the remaining box contains pears only. If you draw a pear, swap the roles of apples and pears.
Three switches
Three switches outside a closed room control one incandescent bulb inside. You may enter only once. How do you find its switch?
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Turn on the first switch for a few minutes, then turn it off; turn on the second and enter. If the bulb is on, it is the second; off but warm, the first; off and cold, the third. The heat test needs a bulb that warms up, not a cool LED.
Knights and knaves
A and B are either knights, who always tell the truth, or knaves, who always lie. A says “B is a knave.” B says “We are the same type.” Which is which?
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A is a knight and B a knave. If A speaks truthfully, B lies when claiming they are alike: consistent. If A lies, B is a knight, yet B’s statement would be false: impossible.
Who is guilty: is it determined?
Exactly one of A, B and C is guilty. A says “B did it”; B says “C did it”; C says “B is lying.” Exactly one statement is true. Is there a unique culprit?
Show solution and explanation
No: both A and C are possible. If A did it, only C speaks truthfully. If B did it, A and C are truthful, so B is excluded. If C did it, only B is truthful. Adding “the culprit lied” still does not distinguish A from C: each lies in their respective scenario.
The lighter coin among nine
Nine coins look identical, but one is lighter. With a two-pan balance and only two weighings, how do you find it?
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Weigh three against three. If equal, the fake is among the three unweighed coins; otherwise it is among the lighter three. Weigh one candidate against another: if equal, the third is fake; otherwise the lighter is fake.
Two guards
One of two doors is safe. One guard always speaks truthfully, the other always lies; you do not know which is which. You may ask one guard one question. What do you ask?
Show solution and explanation
Ask, “Which door would the other guard say is safe?” Both point to the wrong door, so choose the other one.
Two ropes
Each of two ropes takes exactly 60 minutes to burn, though neither burns uniformly. How can you measure 45 minutes?
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Light the first rope at both ends and the second at one end at the same time. The first finishes in 30 minutes. Then light the other end of the second rope: its remaining part, which would take another 30 minutes from one end, finishes in 15. Total: 45 minutes.
The 100 prisoners and boxes
There are 100 numbered prisoners and 100 numbered boxes. Each box holds one randomly placed prisoner number. Each prisoner may open at most 50; everyone wins only if all find their own number. They may agree beforehand, cannot communicate during the search, and boxes are closed after each turn. What strategy maximizes success?
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Prisoner k opens box k, then the box numbered by the slip found, following a permutation cycle. Everyone succeeds exactly when every cycle has length at most 50. For a uniform random permutation the probability is 1 − Σ from j=51 to 100 of 1/j ≈ 31.18%, far above independent random choices.
The night bridge
Four people must cross a bridge with one torch. At most two cross together, at the speed of the slower person. Their times are 1, 2, 7 and 10 minutes. What is the minimum?
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17 minutes. The 1- and 2-minute people cross (2); 1 returns (1); 7 and 10 cross (10); 2 returns (2); 1 and 2 cross (2). Total 2+1+10+2+2=17. Taking the two slowest across separately costs more.
Poisoned wine
Exactly one of 1,000 bottles is poisoned. You have 10 test animals; the poison acts within 24 hours, and only one testing round is allowed. How can you identify the bottle?
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Number bottles 0 to 999 and write each number in 10 binary digits, since 2¹⁰=1024. Animal i receives a sample from every bottle whose bit i is 1. After 24 hours, the affected animals spell out the poisoned bottle’s binary number, assuming a mixed sample still contains an effective dose.
Three logicians at a café
A waiter asks three logicians A, B and C, in order, “Do all three of you want coffee?” Each knows their own wish, hears earlier answers and speaks truthfully. A says “I do not know”; B says “I do not know”; C says “Yes.” What does each want?
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All three want coffee. If A did not, A would already know the group answer was no; uncertainty shows A wants it. The same reasoning applies to B. C knows their own wish and, from the previous answers, knows A and B want coffee too; thus C says yes.
Sum and product
Two integers satisfy 1 < x < y and x+y < 100. Sergio knows their sum and Paolo their product. Paolo: “I do not know the numbers.” Sergio: “I knew you did not.” Paolo: “Now I know.” Sergio: “Now I know too.” What are they?
Show solution and explanation
4 and 13; sum 17 and product 52. Paolo’s first statement excludes products with one allowed factorization. Sergio knew that every pair with his sum would leave Paolo uncertain. With that information Paolo has one remaining factorization; Sergio’s final statement selects sum 17. This is Freudenthal’s classic puzzle; the full elimination checks all pairs with x+y<100.
One number is the sum of the other two
A, B and C see two positive integers on the other foreheads; one of the three numbers is the sum of the other two. A sees B=2 and C=3. In order, A, B and C say “I do not know.” A then says “Now I know.” What is A’s number?
Show solution and explanation
5. At first A considers 1 or 5. If A=1, C sees 1 and 2 and considers 1 or 3 for their own number. But if C=1, B sees two 1s and knows at once that B=2. Since B remained uncertain, C would infer C=3. C instead says they do not know, ruling out A=1. Hence A=5.
The blue-eyed island
Everyone on an island sees the others’ eye colors, not their own. Anyone who deduces that their own eyes are blue leaves that night. No one discusses eye color. A visitor publicly announces, “At least one person has blue eyes.” Everyone knows the rules and observes the nightly departures. If exactly 100 residents have blue eyes, what happens?
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All 100 leave on night 100. With one blue-eyed person, the announcement lets that person infer their color and leave on night one. With two, each expects the other to leave on night one if they themselves are not blue; when nobody does, both leave on night two. Induction gives night n for n blue-eyed people. The public announcement creates common knowledge.
Hats in a line
One hundred prisoners in a line wear red or blue hats. Each sees only the hats ahead. Starting from the back, everyone must say a color aloud; they hear previous answers and may agree on a strategy beforehand. How many can be guaranteed to survive?
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99. The last person communicates the parity of red hats ahead, perhaps using “red” for even and “blue” for odd; their own guess may be wrong. The next compares that parity with the hats visible ahead to deduce their own color. Each later person updates the parity from the earlier spoken colors.
The prisoners’ light bulb
One hundred isolated prisoners are repeatedly taken, one at a time, to a room with an initially unlit bulb. The daily choice is random. They may agree on a plan beforehand but cannot communicate afterward. How can someone announce with certainty that all have visited?
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Choose one counter. Each of the other 99 prisoners turns on the light exactly once in their life, only when finding it off. Whenever the counter finds it on, the counter turns it off and increases a mental count. At count 99, all other prisoners must have visited; the counter has visited too. There is no fixed time limit, but repeated random visits lead to success with probability one.
Five pirates
Five pirates A, B, C, D and E divide 100 coins. A proposes a split; it passes with at least half the votes of those present. Otherwise A dies and B proposes, and so on. Each pirate prioritizes survival, then more gold, then killing another pirate if otherwise indifferent. What does A propose?
Show solution and explanation
A=98, B=0, C=1, D=0, E=1. Work backward: E alone takes 100; with D and E, D’s own vote suffices and D takes 100. With C,D,E, C=99,D=0,E=1 passes. With B,C,D,E, B=99,C=0,D=1,E=0 passes. A can therefore buy C and E, who otherwise get zero, for one coin each; with A’s own vote that is three of five.
The 100 lockers
One hundred lockers start closed. Student 1 opens them all; student 2 toggles every second locker, student 3 every third, up to student 100. Which remain open?
Show solution and explanation
Exactly the perfect squares: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100. Locker n is toggled once per divisor of n. Divisors pair off except the square root of a perfect square, so only perfect squares are toggled an odd number of times.
Twelve coins in three weighings
Twelve coins look identical; one is counterfeit and may be heavier or lighter. Can you identify it and the direction of its weight difference in three two-pan weighings?
Show solution and explanation
Yes. Number them 1–12. First weigh 1,2,3,4 against 5,6,7,8. If equal, weigh 9,10,11 against genuine 1,2,3: if equal compare 12 with 1; otherwise compare 9 with 10 to locate the heavier or lighter coin among 9–11. If the first left pan is heavier, weigh 1,2,5 against 3,6,9 (9 is genuine). If the second left is heavier, possibilities are 1 heavy, 2 heavy or 6 light: compare 1 with 2. If the second right is heavier, possibilities are 3 heavy or 5 light: compare 3 with 9. If equal, possibilities are 4 heavy, 7 light or 8 light: compare 7 with 8. If the first right pan is heavier, reverse “heavy” and “light”: second left heavier → 3 light or 5 heavy (compare 3 with 9); second right heavier → 1 light, 2 light or 6 heavy (compare 1 with 2); second equal → 4 light, 7 heavy or 8 heavy (compare 7 with 8). Every branch ends by weighing three.
The mystery number
A teacher chooses an integer from 1 to 100. Anna privately learns its remainder modulo 3, Bruno modulo 5, and Carlo modulo 7. Each says “I do not know.” The teacher adds that the number exceeds 50. If the three share their remainders, can they always find it? Can you give a numeric answer without knowing those remainders?
Show solution and explanation
Together they can always find it, but the statement gives no particular number. The Chinese remainder theorem determines one class modulo 3·5·7=105. The interval 1–100 contains at most one representative of that class. Knowing only n>50 does not identify it: the three actual remainders are needed. The initial statements add nothing, since every individual remainder belongs to several numbers in 1–100.
Four reading paths
- Knowledge and statements: 1, 2, 3, 4, 5, 8, 9, 16, 17, 18, 19, 25.
- Strategy and information: 13, 15, 20, 21, 22, 24.
- Practical deduction: 6, 7, 10, 11, 12.
- Combinatorics and optimization: 14, 23.