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90 logic puzzles with worked solutions

Read the prompt, give the group time to reason, then use the worked solution to check the route rather than only the final answer.

90 puzzles · 6 kinds · every one with a worked solution

All question banks

Grid deduction

Each one-to-one matching has exactly one complete assignment.

  1. Ivo, Lea, and Mina present at 9, 10, and 11, one person per time. Ivo presents at 9. Lea is not at 9. Mina is not at 10. Who presents at 10?

    Answer: Lea presents at 10.Worked solution: Ivo uses 9. Mina cannot use 9 or 10, so Mina uses 11. The remaining time, 10, belongs to Lea.Open the method sourcePrint source: https://mathigon.org/world/Logic_and_Paradoxes

  2. Ari, Bo, Cia, and Dev each choose one project: kite, maze, robot, or volcano. Ari chooses neither maze, robot, nor volcano. Bo chooses neither robot nor volcano. Cia does not choose volcano. Which project does Dev choose?

    Answer: Dev chooses the volcano project.Worked solution: Ari must have kite. Bo then must have maze. Cia must have robot, leaving volcano for Dev.Open the method sourcePrint source: https://mathigon.org/world/Logic_and_Paradoxes

  3. Four workshops use Cedar, Elm, Maple, and Oak rooms. Design is not in Elm, Maple, or Oak. Coding is not in Cedar, Maple, or Oak. Writing is not in Cedar, Elm, or Oak. Where is the Music workshop?

    Answer: Music is in Oak.Worked solution: Design is forced into Cedar, Coding into Elm, and Writing into Maple. Oak is the only room left for Music.Open the method sourcePrint source: https://mathigon.org/world/Logic_and_Paradoxes

  4. Four labeled lunch boxes contain apple, crackers, grapes, and yogurt, one snack each. Box Red has crackers. Box Blue has neither apple nor yogurt. Box Green has neither crackers nor yogurt. What is in Box Yellow?

    Answer: Box Yellow contains yogurt.Worked solution: Red has crackers. Blue cannot have apple or yogurt, so Blue has grapes. Green then has apple. Yogurt remains for Yellow.Open the method sourcePrint source: https://mathigon.org/world/Logic_and_Paradoxes

  5. Jae, Kira, Luis, and Noor each visit on Monday, Tuesday, Wednesday, or Thursday. Noor visits Thursday. Jae does not visit Tuesday or Wednesday. Kira does not visit Monday or Wednesday. Which day does Luis visit?

    Answer: Luis visits Wednesday.Worked solution: Noor has Thursday. Jae must have Monday. Kira must then have Tuesday. Wednesday remains for Luis.Open the method sourcePrint source: https://mathigon.org/world/Logic_and_Paradoxes

  6. Eli, Fara, Gus, Hana, and Ivan wear circle, diamond, moon, square, and star badges. Eli does not wear diamond, moon, square, or star. Fara does not wear moon, square, or star. Gus does not wear square or star. Hana does not wear star. Which badge does Ivan wear?

    Answer: Ivan wears the star badge.Worked solution: The exclusions force Eli to circle, Fara to diamond, Gus to moon, and Hana to square. Star remains for Ivan.Open the method sourcePrint source: https://mathigon.org/world/Logic_and_Paradoxes

  7. Paz, Quin, Ravi, and Sol use lockers 12, 18, 24, and 30. Ravi uses 24. Paz does not use 18 or 30. Quin does not use 12 or 30. Which locker does Sol use?

    Answer: Sol uses locker 30.Worked solution: Ravi has 24. Paz must have 12. Quin must have 18. Sol therefore has 30.Open the method sourcePrint source: https://mathigon.org/world/Logic_and_Paradoxes

  8. Uma, Vero, Wes, and Xia play drum, flute, piano, and violin. Xia plays violin. Uma plays neither flute nor piano. Vero plays neither drum nor piano. Which instrument does Wes play?

    Answer: Wes plays piano.Worked solution: Xia has violin. Uma must have drum, and Vero must have flute. Piano remains for Wes.Open the method sourcePrint source: https://mathigon.org/world/Logic_and_Paradoxes

  9. Packets A, B, C, and D contain basil, dill, mint, and thyme. Packet B contains dill. Packet A contains neither dill nor mint nor thyme. Packet C contains neither basil, dill, nor thyme. What is in Packet D?

    Answer: Packet D contains thyme.Worked solution: A must contain basil and B contains dill. C cannot contain basil, dill, or thyme, so C has mint. The remaining herb, thyme, belongs in D.Open the method sourcePrint source: https://mathigon.org/world/Logic_and_Paradoxes

  10. Four parcels arrive by bike, car, train, and van. Parcel North arrives by train. Parcel East arrives by neither car nor van. Parcel South arrives by neither bike nor van. How does Parcel West arrive?

    Answer: Parcel West arrives by van.Worked solution: North uses train. East must use bike. South must use car. Van remains for West.Open the method sourcePrint source: https://mathigon.org/world/Logic_and_Paradoxes

  11. Clay, glass, paper, and wood models sit on shelves 1, 2, 3, and 4. Glass is on shelf 2. Clay is not on shelves 2, 3, or 4. Paper is not on shelves 1, 2, or 4. Which shelf holds wood?

    Answer: Wood is on shelf 4.Worked solution: Clay is forced to shelf 1, glass is on shelf 2, and paper is forced to shelf 3. Wood must be on shelf 4.Open the method sourcePrint source: https://mathigon.org/world/Logic_and_Paradoxes

  12. Nia, Oren, and Pia each use a telescope at dawn, noon, or dusk. Oren uses it at noon. Nia uses it at neither noon nor dusk. When does Pia use it?

    Answer: Pia uses the telescope at dusk.Worked solution: Oren takes noon. Nia can take neither noon nor dusk, so Nia takes dawn. The remaining slot, dusk, belongs to Pia.Open the method sourcePrint source: https://mathigon.org/world/Logic_and_Paradoxes

  13. Red, Blue, Green, and White recipe cards show soup, rice, salad, and stew. White shows stew. Red shows neither rice, salad, nor stew. Blue shows neither soup, salad, nor stew. What does Green show?

    Answer: Green shows salad.Worked solution: White has stew. Red is excluded from the other three choices, so Red has soup. Blue must then have rice, leaving salad for Green.Open the method sourcePrint source: https://mathigon.org/world/Logic_and_Paradoxes

  14. Kite, Lime, Moss, and Navy envelopes contain a badge, map, pass, and token. Moss contains the pass. Kite contains neither the map, pass, nor token. Lime contains neither the badge, pass, nor token. What is in Navy?

    Answer: Navy contains the token.Worked solution: Moss has the pass. Kite is forced to the badge and Lime is forced to the map. The only insert left for Navy is the token.Open the method sourcePrint source: https://mathigon.org/world/Logic_and_Paradoxes

  15. Aster, Birch, Cedar, Dune, and Elm segments occupy radio slots 1 through 5. Aster has slot 3. Birch has none of slots 2 through 5. Cedar has neither slot 1 nor slots 3 through 5. Dune has neither slots 1 through 3 nor slot 5. Which slot has Elm?

    Answer: Elm has slot 5.Worked solution: Aster has 3. The exclusions force Birch to 1, Cedar to 2, and Dune to 4. Slot 5 remains for Elm.Open the method sourcePrint source: https://mathigon.org/world/Logic_and_Paradoxes

  16. Five lanterns named Arc, Beam, Cove, Drift, and Echo glow amber, blue, green, silver, and violet. Cove glows green. Arc glows neither blue, green, silver, nor violet. Beam glows neither amber, green, silver, nor violet. Drift glows neither amber, blue, green, nor violet. What color is Echo?

    Answer: Echo glows violet.Worked solution: Cove has green. Arc is forced to amber, Beam to blue, and Drift to silver. Violet is the remaining color for Echo.Open the method sourcePrint source: https://mathigon.org/world/Logic_and_Paradoxes

  17. Poster tubes J, K, L, M, and N hold art, code, dance, music, and science posters. Tube L holds dance. Tube J holds neither code, dance, music, nor science. Tube K holds neither art, dance, music, nor science. Tube M holds neither art, code, dance, nor science. Which poster is in Tube N?

    Answer: Tube N holds the science poster.Worked solution: Tube L has dance. The exclusions force J to art, K to code, and M to music. Science remains for Tube N.Open the method sourcePrint source: https://mathigon.org/world/Logic_and_Paradoxes

  18. Folders A through F have coral, gold, gray, lilac, teal, and white tabs. Folder C has gray. A has none of gold, gray, lilac, teal, or white. B has none of coral, gray, lilac, teal, or white. D has none of coral, gold, gray, teal, or white. E has none of coral, gold, gray, lilac, or white. Which tab does F have?

    Answer: Folder F has the white tab.Worked solution: C has gray. The exclusions force A to coral, B to gold, D to lilac, and E to teal. White remains for F.Open the method sourcePrint source: https://mathigon.org/world/Logic_and_Paradoxes

Sequence rules

The rule is stated in each prompt so a finite list cannot support rival answers.

  1. Start at 4 and add 7 each time: 4, 11, 18, 25, 32. What is the next term?

    Answer: 39.Worked solution: Add 7 to 32. The result is 39.Open the method sourcePrint source: https://nrich.maths.org/patterns-and-sequences-short-problems-0

  2. Start at 2 and multiply by 3 each time: 2, 6, 18, 54. What is the next term?

    Answer: 162.Worked solution: Multiply 54 by 3. The result is 162.Open the method sourcePrint source: https://nrich.maths.org/patterns-and-sequences-short-problems-0

  3. Starting at 4, alternate add 3 and multiply by 2: 4, 7, 14, 17, 34. What is next?

    Answer: 37.Worked solution: The last operation shown was multiply by 2, so the next operation is add 3. Then 34 + 3 = 37.Open the method sourcePrint source: https://nrich.maths.org/patterns-and-sequences-short-problems-0

  4. Start at 3 and add consecutive even numbers 2, 4, 6, 8, and so on: 3, 5, 9, 15, 23, 33. What is next?

    Answer: 45.Worked solution: The next even-number gap is 12, so 33 + 12 = 45.Open the method sourcePrint source: https://nrich.maths.org/patterns-and-sequences-short-problems-0

  5. The nth term is n times n + 1, beginning with n = 1: 2, 6, 12, 20, 30. What is the sixth term?

    Answer: 42.Worked solution: For n = 6, calculate 6 times 7, which is 42.Open the method sourcePrint source: https://nrich.maths.org/patterns-and-sequences-short-problems-0

  6. Begin with 2 and 3, then make each new term the sum of the previous two: 2, 3, 5, 8, 13, 21. What is next?

    Answer: 34.Worked solution: Add the previous two terms: 13 + 21 = 34.Open the method sourcePrint source: https://nrich.maths.org/patterns-and-sequences-short-problems-0

  7. The odd-position terms start at 2 and rise by 4. The even-position terms start at 20 and fall by 3: 2, 20, 6, 17, 10, 14, 14, 11. What is next?

    Answer: 18.Worked solution: The next term is in an odd position. Continue 2, 6, 10, 14 by adding 4 to get 18.Open the method sourcePrint source: https://nrich.maths.org/patterns-and-sequences-short-problems-0

  8. Start at 3. Double each term and subtract 1: 3, 5, 9, 17, 33. What is next?

    Answer: 65.Worked solution: Double 33 to get 66, then subtract 1 to get 65.Open the method sourcePrint source: https://nrich.maths.org/patterns-and-sequences-short-problems-0

  9. Starting at 10, add 5, subtract 2, add 10, subtract 4, add 15, and continue the two growing patterns: 10, 15, 13, 23, 19, 34. What is next?

    Answer: 28.Worked solution: The subtraction amounts are 2, 4, then 6. Subtract 6 from 34 to get 28.Open the method sourcePrint source: https://nrich.maths.org/patterns-and-sequences-short-problems-0

  10. The nth term is n cubed minus 1, beginning with n = 1: 0, 7, 26, 63, 124. What is the sixth term?

    Answer: 215.Worked solution: Six cubed is 216. Subtract 1 to get 215.Open the method sourcePrint source: https://nrich.maths.org/patterns-and-sequences-short-problems-0

Balance-scale deductions

The odd item's direction is known, and the recorded outcomes isolate one item.

  1. Coins A, B, and C contain exactly one heavier coin. A balances B. Which coin is heavier?

    Answer: Coin C.Worked solution: A and B have equal weight, so neither can be the one heavier coin. C is the only candidate left.Open the method sourcePrint source: https://nrich.maths.org/problems/weighing-scales

  2. Among A, B, C, and D, exactly one item is lighter. A and B outweigh C and D. Then C outweighs D. Which item is lighter?

    Answer: D is lighter.Worked solution: The first weighing places the lighter item among C and D. Since C outweighs D, D is the lighter item.Open the method sourcePrint source: https://nrich.maths.org/problems/weighing-scales

  3. Five items A through E include exactly one heavier item. A and B balance C and D. Which item is heavier?

    Answer: E is heavier.Worked solution: All four weighed items balance in equal-size groups, so none is heavier. The unweighed item E must be heavier.Open the method sourcePrint source: https://nrich.maths.org/problems/weighing-scales

  4. Six items A through F include exactly one lighter item. A, B, and C are lighter as a group than D, E, and F. Then A outweighs B. Which item is lighter?

    Answer: B is lighter.Worked solution: The first result puts the lighter item among A, B, and C. A outweighs B, so B is lighter than A. If C were lighter, A and B would balance. Therefore B is the odd item.Open the method sourcePrint source: https://nrich.maths.org/problems/weighing-scales

  5. Seven items A through G include exactly one heavier item. A, B, and C balance D, E, and F. Which item is heavier?

    Answer: G is heavier.Worked solution: The six weighed items balance, so the only unweighed candidate G is heavier.Open the method sourcePrint source: https://nrich.maths.org/problems/weighing-scales

  6. Eight labeled items A through H include exactly one heavier item. A, B, and C are lighter as a group than D, E, and F. D balances E. Which item is heavier?

    Answer: F is heavier.Worked solution: The first result puts the heavier item among D, E, and F. D balances E, so F is the only candidate.Open the method sourcePrint source: https://nrich.maths.org/problems/weighing-scales

  7. One item among A through H is lighter than all the others. A, B, and C are lighter as a group than D, E, and F. A balances B. Which item is lighter?

    Answer: C is lighter.Worked solution: The first result puts the lighter item among A, B, and C. A balances B, leaving C.Open the method sourcePrint source: https://nrich.maths.org/problems/weighing-scales

  8. Of the nine items A through I, exactly one is heavier. A, B, and C balance D, E, and F. Then G is lighter than H. Which item is heavier?

    Answer: H is heavier.Worked solution: The first balance puts the heavier item among G, H, and I. H outweighs G. If I were heavier, G and H would balance. Therefore H is heavier.Open the method sourcePrint source: https://nrich.maths.org/problems/weighing-scales

  9. Within items A through J, exactly one is lighter. A, B, and C balance D, E, and F. G and H outweigh I and J. Then I outweighs J. Which item is lighter?

    Answer: J is lighter.Worked solution: The first balance leaves G, H, I, and J. The second result places the lighter item among I and J. Since I outweighs J, J is lighter.Open the method sourcePrint source: https://nrich.maths.org/problems/weighing-scales

  10. Twelve items labeled A through L include exactly one heavier item. A through D balance E through H. I and J are lighter as a pair than K and L. K outweighs L. Which item is heavier?

    Answer: K is heavier.Worked solution: The first balance leaves I, J, K, and L. The second result puts the heavier item among K and L. The final weighing identifies K.Open the method sourcePrint source: https://nrich.maths.org/problems/weighing-scales

  11. Exactly one of four tiles A, B, C, and D is heavier. A balances B, then C outweighs A. Which tile is heavier?

    Answer: C is heavier.Worked solution: The balance removes A and B from consideration. C outweighs a known normal tile, so C is the unique heavier tile and D is normal.Open the method sourcePrint source: https://nrich.maths.org/problems/weighing-scales

  12. Five counters A through E contain exactly one lighter counter. A and B balance C and D. Which counter is lighter?

    Answer: E is lighter.Worked solution: A balanced comparison shows A, B, C, and D all have normal weight. E is the only counter not cleared, so E must be lighter.Open the method sourcePrint source: https://nrich.maths.org/problems/weighing-scales

  13. Among six blocks A through F, one is heavier. A and B balance C and D. A second weighing shows E outweighs A. Which block is heavier?

    Answer: E is heavier.Worked solution: The first balance clears A through D, leaving E or F. Since E outweighs the known normal block A, E is the heavier block.Open the method sourcePrint source: https://nrich.maths.org/problems/weighing-scales

  14. Seven discs A through G include one lighter disc. A, B, and C balance D, E, and F. Identify the lighter disc.

    Answer: G is lighter.Worked solution: The six weighed discs balance, so none of them is the lighter one. G is the only unweighed disc and must be lighter.Open the method sourcePrint source: https://nrich.maths.org/problems/weighing-scales

  15. A set of nine tokens A through I has one lighter token. A through D balance E through H. Which token is lighter?

    Answer: I is lighter.Worked solution: The eight tokens on the balanced pans all have normal weight. I is the only token outside that cleared set, so I is lighter.Open the method sourcePrint source: https://nrich.maths.org/problems/weighing-scales

  16. Ten beads A through J include one heavier bead. A through D balance E through H. Then J outweighs I. Which bead is heavier?

    Answer: J is heavier.Worked solution: The first balance clears A through H and leaves only I or J. J outweighs I, so J is the heavier bead.Open the method sourcePrint source: https://nrich.maths.org/problems/weighing-scales

  17. Eleven cubes A through K include one heavier cube. A through D balance E through H, and I balances J. Which cube is heavier?

    Answer: K is heavier.Worked solution: The first balance clears A through H. The second balance clears I and J. K is the only cube not shown to be normal, so K is heavier.Open the method sourcePrint source: https://nrich.maths.org/problems/weighing-scales

River crossings

Each question asks for the minimum crossing count, which the checker proves by breadth-first search.

  1. One adult weighs 2 units and two children weigh 1 unit each. A boat holds 2 units, everyone can row, and nobody may swim. What is the minimum number of one-way crossings?

    Answer: 5 crossings.Worked solution: Children 1 and 2 cross; Child 1 returns; Adult crosses; Child 2 returns; both children cross. That is five crossings.Open the method sourcePrint source: https://nrich.maths.org/problems/river-crossing

  2. Two adults weigh 2 units each and two children weigh 1 unit each. A boat holds 2 units, everyone can row, and nobody may swim. What is the minimum number of one-way crossings?

    Answer: 9 crossings.Worked solution: Both children cross; one returns; Adult 1 crosses; the other child returns; both children cross; one returns; Adult 2 crosses; the other child returns; both children cross. Nine crossings are necessary.Open the method sourcePrint source: https://nrich.maths.org/problems/river-crossing

  3. Three adults weigh 2 units each and two children weigh 1 unit each. A boat holds 2 units, everyone can row, and nobody may swim. What is the minimum number of one-way crossings?

    Answer: 13 crossings.Worked solution: Use four crossings to transfer one adult while resetting both children to the starting bank. Repeat for each adult, then send both children across. The total is 4 + 4 + 4 + 1 = 13.Open the method sourcePrint source: https://nrich.maths.org/problems/river-crossing

  4. Ari, Bo, and Cy weigh 1, 2, and 3 units. The boat holds 3 units and all three can row. What is the minimum number of one-way crossings?

    Answer: 5 crossings.Worked solution: Ari and Bo cross; Ari returns; Cy crosses; Bo returns; Ari and Bo cross. The total is five.Open the method sourcePrint source: https://nrich.maths.org/problems/river-crossing

  5. Dax, Emi, Flo, and Gio weigh 1, 2, 3, and 4 units. The boat holds 5 units and all four can row. What is the minimum number of one-way crossings?

    Answer: 5 crossings.Worked solution: Dax and Gio cross; Dax returns; Emi and Flo cross; Emi returns; Dax and Emi cross. The total is five.Open the method sourcePrint source: https://nrich.maths.org/problems/river-crossing

  6. A guide must take a fox, a hen, and grain across. The boat carries the guide plus one item. Without the guide, the fox cannot stay with the hen, and the hen cannot stay with the grain. What is the minimum number of one-way crossings?

    Answer: 7 crossings.Worked solution: Take hen over; return alone; take fox over; bring hen back; take grain over; return alone; take hen over. Seven is the shortest safe route.Open the method sourcePrint source: https://nrich.maths.org/problems/river-crossing

  7. A guide must move A, B, C, and D across in a boat holding two items. When unattended, A cannot stay with B, B cannot stay with C, and C cannot stay with D. What is the minimum number of one-way crossings?

    Answer: 3 crossings.Worked solution: Take A and C over; return alone; take B and D over. Both unattended pairs of items are safe, and all four arrive in three crossings.Open the method sourcePrint source: https://nrich.maths.org/problems/river-crossing

  8. A guide must move cats C1 and C2 plus mice M1 and M2. The boat holds two animals. When unattended, C1 cannot stay with M1 and C2 cannot stay with M2. What is the minimum number of one-way crossings?

    Answer: 3 crossings.Worked solution: Take both cats over; return alone; take both mice over. Each unattended bank contains only one animal type, so the route is safe and uses three crossings.Open the method sourcePrint source: https://nrich.maths.org/problems/river-crossing

  9. A guide must move A, B, C, D, and E in a boat holding two items. When unattended, each neighboring letter pair is unsafe: A-B, B-C, C-D, and D-E. What is the minimum number of one-way crossings?

    Answer: 7 crossings.Worked solution: A shortest safe route is B and D over; return alone; A over; B back; C and E over; D back; B and D over. This uses seven crossings.Open the method sourcePrint source: https://nrich.maths.org/problems/river-crossing

  10. Ada and Bran each weigh 1 unit, and Ciro weighs 2 units. A boat holds 3 units, everyone can row, and nobody may swim. What is the minimum number of one-way crossings?

    Answer: 3 crossings.Worked solution: Ada and Ciro cross, Ada returns, then Ada and Bran cross. All three finish across in three trips, and one trip cannot carry all 4 units.Open the method sourcePrint source: https://nrich.maths.org/problems/river-crossing

  11. Dara and Eli weigh 1 unit each, while Finn and Gail weigh 2 units each. The boat holds 3 units and everyone can row. Find the minimum number of one-way crossings.

    Answer: 5 crossings.Worked solution: Dara takes Finn over and returns. Dara takes Gail over and returns again. Dara and Eli then cross together, for five crossings.Open the method sourcePrint source: https://nrich.maths.org/problems/river-crossing

  12. Hugo, Iris, and Juno weigh 1 unit each; Kai and Luz weigh 2 units each. The boat holds 3 units and all five can row. Determine the fewest one-way crossings.

    Answer: 5 crossings.Worked solution: Hugo takes Kai over and returns, then takes Luz over and returns. Hugo, Iris, and Juno cross together on the fifth trip.Open the method sourcePrint source: https://nrich.maths.org/problems/river-crossing

  13. Mara weighs 1 unit and can row. Niko and Ola each weigh 2 units and cannot row. A boat holds 3 units. How few one-way crossings move everyone across?

    Answer: 3 crossings.Worked solution: Mara rows Niko across, returns alone, and rows Ola across. Both loaded trips meet the 3-unit limit, so three crossings suffice.Open the method sourcePrint source: https://nrich.maths.org/problems/river-crossing

  14. Pax and Quin weigh 1 unit and can row. Rumi and Sela weigh 2 units and cannot row. With a 3-unit boat, what is the minimum number of one-way crossings?

    Answer: 5 crossings.Worked solution: Pax rows Rumi across and returns, then rows Sela across and returns. Pax and Quin cross together on trip five.Open the method sourcePrint source: https://nrich.maths.org/problems/river-crossing

  15. Tara and Ugo weigh 1 unit and can row. Vera, Wynn, and Xeno weigh 2 units and cannot row. The boat holds 3 units. What is the minimum crossing count?

    Answer: 7 crossings.Worked solution: Tara ferries Vera, Wynn, and Xeno one at a time, returning after the first two. Tara returns once more for Ugo, and the two rowers make the seventh crossing together.Open the method sourcePrint source: https://nrich.maths.org/problems/river-crossing

  16. Yara and Zed weigh 1 unit each, Ami weighs 2 units, and Bex weighs 3 units. A boat holds 4 units and everyone can row. What is the fewest number of one-way crossings?

    Answer: 3 crossings.Worked solution: Yara and Bex cross at the 4-unit limit. Yara returns, then crosses with Zed and Ami, again totaling 4 units. Three crossings are enough.Open the method sourcePrint source: https://nrich.maths.org/problems/river-crossing

  17. Cora and Deni weigh 1 unit each, Evan weighs 2, and Faye and Gino weigh 3 each. Everyone can row a boat holding 4 units. Find the minimum one-way crossings.

    Answer: 5 crossings.Worked solution: Cora takes Faye across and returns, then takes Gino across and returns. Cora, Deni, and Evan make the fifth trip at the 4-unit limit.Open the method sourcePrint source: https://nrich.maths.org/problems/river-crossing

Knights and knaves

Knights always speak truthfully and knaves always speak falsely. Truth-table enumeration leaves one role assignment.

  1. On the island, A says, 'B and I are different types.' B says, 'A is a knave.' Who is who?

    Answer: A=knight; B=knave.Worked solution: If A is truthful, the two differ, so B is a knave and B's claim is false. If A is a knave, the two would be the same, making B a knave whose claim that A is a knave is true, which is impossible.Open the method sourcePrint source: https://mathigon.org/world/Logic_and_Paradoxes

  2. A says, 'B is a knight.' B says, 'A and I are different types.' Who is who?

    Answer: A=knave; B=knave.Worked solution: If A were a knight, B would be a knight, but then B's different-types claim would be false. Therefore A is a knave, so B is not a knight. Both are knaves, making B's different-types claim false as required.Open the method sourcePrint source: https://mathigon.org/world/Logic_and_Paradoxes

  3. A says, 'B and I are both knaves.' A is the only speaker. Who is who?

    Answer: A=knave; B=knight.Worked solution: A cannot be a knight while truthfully calling A a knave. So A is a knave and the whole statement is false. With A already a knave, B must be a knight to make 'both knaves' false.Open the method sourcePrint source: https://mathigon.org/world/Logic_and_Paradoxes

  4. A says, 'A and B are both knaves.' B says, 'C is a knave.' Who is who?

    Answer: A=knave; B=knight; C=knave.Worked solution: A's self-including claim forces A to be a knave and B to be a knight. B then speaks truthfully, so C is a knave.Open the method sourcePrint source: https://mathigon.org/world/Logic_and_Paradoxes

  5. A says, 'A and B are both knaves.' B says, 'C is a knave.' C says, 'D is a knight.' Who is who?

    Answer: A=knave; B=knight; C=knave; D=knave.Worked solution: The first statement forces A knave and B knight. B truthfully makes C a knave. C's statement must be false, so D is a knave.Open the method sourcePrint source: https://mathigon.org/world/Logic_and_Paradoxes

  6. A says, 'Exactly one of A and B is a knight.' B says, 'C is a knight.' C says, 'A is a knave.' Who is who?

    Answer: A=knight; B=knave; C=knave.Worked solution: Testing the eight role rows leaves one consistent row: A knight makes the exactly-one claim true; B knave makes 'C is a knight' false; C knave makes 'A is a knave' false.Open the method sourcePrint source: https://mathigon.org/world/Logic_and_Paradoxes

  7. A says, 'B is a knight or C is a knight.' B says, 'A is a knave.' C says, 'B is a knave.' Who is who?

    Answer: A=knight; B=knave; C=knight.Worked solution: A knight makes the or-claim true through C. B is a knave because its claim about A is false. C is a knight because it truthfully identifies B.Open the method sourcePrint source: https://mathigon.org/world/Logic_and_Paradoxes

  8. A says, 'B and C are the same type.' B says, 'A is a knight.' C says, 'B is a knave.' Who is who?

    Answer: A=knave; B=knave; C=knight.Worked solution: The consistent row has A knave, so A's same-type claim is false. B is a knave and falsely calls A a knight. C is a knight and truthfully calls B a knave.Open the method sourcePrint source: https://mathigon.org/world/Logic_and_Paradoxes

  9. A says, 'Exactly one of B and C is a knight.' B says, 'A is a knave.' C says, 'B is a knave.' Who is who?

    Answer: A=knight; B=knave; C=knight.Worked solution: A and C as knights and B as knave makes every statement fit: exactly one of B and C is a knight, B falsely calls A a knave, and C truthfully calls B a knave.Open the method sourcePrint source: https://mathigon.org/world/Logic_and_Paradoxes

  10. A says, 'Exactly one of B and C is a knave.' B says, 'A is a knight.' C says, 'B is a knave.' D says, 'C is a knave.' Who is who?

    Answer: A=knight; B=knight; C=knave; D=knight.Worked solution: A, B, and D speak truthfully while C lies. Exactly one of B and C is a knave, B correctly identifies A, C falsely identifies B, and D correctly identifies C.Open the method sourcePrint source: https://mathigon.org/world/Logic_and_Paradoxes

  11. A says, 'It is not true that B and C are both knights.' B says, 'A and C are different types.' C says, 'A is a knight.' Who is who?

    Answer: A=knight; B=knave; C=knight.Worked solution: With A and C knights and B knave, A truthfully denies that B and C are both knights, B falsely says A and C differ, and C truthfully identifies A.Open the method sourcePrint source: https://mathigon.org/world/Logic_and_Paradoxes

  12. Two islanders speak. A says, 'B is a knave.' B says, 'A and B are different types.' Determine both roles.

    Answer: A=knave; B=knight.Worked solution: If A were a knight, B would be a knave but B's different-types statement would be true. Therefore A is a knave, A's claim is false, and B is a knight.Open the method sourcePrint source: https://mathigon.org/world/Logic_and_Paradoxes

  13. A says, 'Exactly one of A, B, and C is a knight.' B says, 'A is a knight.' C says, 'B is a knight.' Assign every role.

    Answer: A=knave; B=knave; C=knave.Worked solution: If A were truthful, B and then C would also be knights, contradicting A's count of one. With A a knave, B and C both speak falsely, leaving zero knights and making A's count false.Open the method sourcePrint source: https://mathigon.org/world/Logic_and_Paradoxes

  14. Three islanders speak. A says, 'B is a knight.' B says, 'C is a knave.' C says, 'A and B are the same type.' Identify A, B, and C.

    Answer: A=knave; B=knave; C=knight.Worked solution: A's statement makes A and B the same type. C therefore tells the truth, so C is a knight. B's claim that C is a knave is false, making B and A knaves.Open the method sourcePrint source: https://mathigon.org/world/Logic_and_Paradoxes

  15. A says, 'B and C are different types.' B says, 'A is a knave.' C says, 'B is a knight.' Find the three roles.

    Answer: A=knave; B=knight; C=knight.Worked solution: C has the same role as B because C claims B is a knight. A's different-types claim is therefore false, so A is a knave. B then truthfully identifies A.Open the method sourcePrint source: https://mathigon.org/world/Logic_and_Paradoxes

  16. A says, 'B is a knave.' B says, 'C is a knight.' C says, 'D is a knave.' D says, 'A and C are the same type.' Who is who?

    Answer: A=knave; B=knight; C=knight; D=knave.Worked solution: The first three statements force A opposite B, B equal to C, and C opposite D. Thus A and D match while A and C differ. D's same-type claim is false, fixing D and A as knaves and B and C as knights.Open the method sourcePrint source: https://mathigon.org/world/Logic_and_Paradoxes

  17. A says, 'Exactly two of A, B, C, and D are knights.' B says, 'A is a knight.' C says, 'B is a knave.' D says, 'C is a knight.' Resolve the roles.

    Answer: A=knight; B=knight; C=knave; D=knave.Worked solution: B must match A, while C and D must both oppose A. Either branch contains two knights, so A's count is true and A must be a knight. The chain then fixes B as knight and C and D as knaves.Open the method sourcePrint source: https://mathigon.org/world/Logic_and_Paradoxes

  18. A says, 'Exactly four of A, B, C, D, and E are knights.' B says, 'A is a knight.' C says, 'B is a knight.' D says, 'C is a knave.' E says, 'D is a knight.' Assign all five roles.

    Answer: A=knave; B=knave; C=knave; D=knight; E=knight.Worked solution: B and C match A, while D and E oppose A. This creates either three or two knights, never four, so A's statement is false. A, B, and C are knaves, while D and E are knights.Open the method sourcePrint source: https://mathigon.org/world/Logic_and_Paradoxes

Cryptarithms

Different letters represent different base-10 digits, leading letters are nonzero, and exhaustive column search verifies one mapping.

  1. Solve SEND + MORE = MONEY. Different letters are different digits, and a leading letter is not zero.

    Answer: 9567 + 1085 = 10652. S=9, E=5, N=6, D=7, M=1, O=0, R=8, Y=2.Worked solution: Column-by-column carry constraints and distinct digits leave one mapping: S9 E5 N6 D7 M1 O0 R8 Y2.Open the method sourcePrint source: https://nrich.maths.org/sites/default/files/thumbnails/content-01-06-six2-Two%252520and%252520Two.pdf

  2. Solve I + BB = ILL. Different letters are different digits, and a leading letter is not zero.

    Answer: 1 + 99 = 100. I=1, B=9, L=0.Worked solution: The hundreds carry forces I=1. The tens and ones columns then force B=9 and L=0.Open the method sourcePrint source: https://nrich.maths.org/sites/default/files/thumbnails/content-01-06-six2-Two%252520and%252520Two.pdf

  3. Solve CROSS + ROADS = DANGER. Different letters are different digits, and a leading letter is not zero.

    Answer: 96233 + 62513 = 158746. C=9, R=6, O=2, S=3, A=5, D=1, N=8, G=7, E=4.Worked solution: Exhaustive column search leaves the mapping C9 R6 O2 S3 A5 D1 N8 G7 E4.Open the method sourcePrint source: https://nrich.maths.org/sites/default/files/thumbnails/content-01-06-six2-Two%252520and%252520Two.pdf

  4. Solve FORTY + TEN + TEN = SIXTY. Different letters are different digits, and a leading letter is not zero.

    Answer: 29786 + 850 + 850 = 31486. F=2, O=9, R=7, T=8, Y=6, E=5, N=0, S=3, I=1, X=4.Worked solution: Exhaustive column search leaves F2 O9 R7 T8 Y6 E5 N0 S3 I1 X4.Open the method sourcePrint source: https://nrich.maths.org/sites/default/files/thumbnails/content-01-06-six2-Two%252520and%252520Two.pdf

  5. Solve DONALD + GERALD = ROBERT. Different letters are different digits, and a leading letter is not zero.

    Answer: 526485 + 197485 = 723970. D=5, O=2, N=6, A=4, L=8, G=1, E=9, R=7, B=3, T=0.Worked solution: Exhaustive column search leaves D5 O2 N6 A4 L8 G1 E9 R7 B3 T0.Open the method sourcePrint source: https://nrich.maths.org/sites/default/files/thumbnails/content-01-06-six2-Two%252520and%252520Two.pdf

  6. Solve BASE + BALL = GAMES. Different letters are different digits, and a leading letter is not zero.

    Answer: 7483 + 7455 = 14938. B=7, A=4, S=8, E=3, L=5, G=1, M=9.Worked solution: Exhaustive column search leaves B7 A4 S8 E3 L5 G1 M9.Open the method sourcePrint source: https://nrich.maths.org/sites/default/files/thumbnails/content-01-06-six2-Two%252520and%252520Two.pdf

  7. Solve EAT + THAT = APPLE. Different letters are different digits, and a leading letter is not zero.

    Answer: 819 + 9219 = 10038. E=8, A=1, T=9, H=2, P=0, L=3.Worked solution: Exhaustive column search leaves E8 A1 T9 H2 P0 L3.Open the method sourcePrint source: https://nrich.maths.org/sites/default/files/thumbnails/content-01-06-six2-Two%252520and%252520Two.pdf

  8. Solve COCA + COLA = OASIS. Different letters are different digits, and a leading letter is not zero.

    Answer: 8186 + 8106 = 16292. C=8, O=1, A=6, L=0, S=2, I=9.Worked solution: Exhaustive column search leaves C8 O1 A6 L0 S2 I9.Open the method sourcePrint source: https://nrich.maths.org/sites/default/files/thumbnails/content-01-06-six2-Two%252520and%252520Two.pdf

  9. Solve TO + GO = OUT. Different letters are different digits, and a leading letter is not zero.

    Answer: 21 + 81 = 102. T=2, O=1, G=8, U=0.Worked solution: Exhaustive column search is used to test whether this proposed mapping is the only valid one.Open the method sourcePrint source: https://nrich.maths.org/sites/default/files/thumbnails/content-01-06-six2-Two%252520and%252520Two.pdf

  10. Solve NO + NO + TOO = LATE. Different letters are different digits, and a leading letter is not zero.

    Answer: 74 + 74 + 944 = 1092. N=7, O=4, T=9, L=1, A=0, E=2.Worked solution: Exhaustive column search is used to test whether this proposed mapping is the only valid one.Open the method sourcePrint source: https://nrich.maths.org/sites/default/files/thumbnails/content-01-06-six2-Two%252520and%252520Two.pdf