The Elevator With Three Working Buttons

Case 50 · shortest-path challenge · Hard · About 10-15 minutes

The archive elevator has a peculiar test mode. Its floor display works, but only three buttons respond. You start on floor 2 and want to reach floor 17 with as few button presses as possible. Each press changes the floor immediately according to a precise rule.

This is a mathematical machine, not a real elevator instruction. You can solve it with pencil and paper by listing reachable floors in rounds.

Generated story illustration. The complete puzzle evidence is given in the text.

The test-mode rules

• U: add 4 to the current floor.
• D: subtract 3 from the current floor.
• X: double the current floor.
• Floors must stay between 0 and 30, inclusive. A move outside that range is forbidden.
• You may not press the same button twice in succession. You may use it again after a different button.

For example, from floor 2, pressing U then X takes you to 6 and then 12. Pressing X then X is forbidden, even though both resulting floors would be in range. All three buttons are available for the first move, subject to the floor limit.

What is the minimum number of presses needed to reach floor 17? Find a shortest button sequence. For an extra challenge, find every shortest sequence.

To prove a route is shortest, finding a route is only half the job. You must also rule out shorter ones. Track the last button as well as the current floor: being on floor 8 after U gives different next options from being on floor 8 after X.

Hint 1 - reveal when you want a nudge

Search in layers: one press, then two, then three. Do not begin by guessing a long route.

Hint 2 - reveal when you want a nudge

There are three shortest routes, each with five presses. Two begin with X; one begins with U.

Reveal the solution and explanation

The minimum is five presses. One shortest sequence is U, X, U, D, U. Its floor sequence is 2, 6, 12, 16, 13, 17. Every move stays within the allowed range, and adjacent buttons are different.

The other shortest sequences are X, U, X, U, D, giving 2, 4, 8, 16, 20, 17; and X, U, X, D, U, giving 2, 4, 8, 16, 13, 17. These are all three shortest routes.

Here is a reproducible shortestness check. Begin with the state (2, no previous button). At each round, apply every permitted button to every state in the previous round. Discard out-of-range moves and moves repeating the previous button. Keep both the new floor and the button just pressed. Floor 17 appears in none of rounds one through four. In round five, it appears on precisely the three sequences above.

You can audit the final step as well. To land on 17, the preceding floor must be 13 before U or 20 before D. X cannot land there because doubling a whole number gives an even number. The four-press routes that can make those legal final moves are UXUD, XUXD, and XUXU. This narrows the full search to the same three answers.

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The Cabinet That Opens Only Once

Case 47 · code deduction · Hard · About 10-15 minutes

A restoration cabinet has four number wheels. Its owner has left a record of four unsuccessful guesses. Each attempt received two counts, and every count is correct. The final attempt must be your solution.

This is a paper puzzle. There is no need to try any real lock. The cabinet, its record, and its rules are entirely fictional.

Generated story illustration. The complete puzzle evidence is given in the text.

Read the feedback precisely

The code uses four different digits from 0 to 9. Its first digit is not zero. A correct place count means that a digit is both present and in its correct position. A wrong place count means that a digit is present but in another position. Digits counted in the first column are not counted again in the second.

Guess	Correct place	Wrong place
1234	0	2
6307	2	0
8612	0	3
5083	1	1

All counts are exact. A guess with two digits present has exactly two absent digits. There is no extra rule about increasing order, prime numbers, or the appearance of the artwork.

What is the four-digit code? Check your proposed answer against all four rows. If one count is even slightly wrong, keep working.

For an orderly paper search, begin with 6307. Choose which two positions are correct. The other two digits in that guess must be absent altogether. There are six possible pairs of correct positions, so this gives you a small set of branches instead of thousands of unstructured guesses.

Hint 1 - reveal when you want a nudge

The 8612 row requires three of its four digits. Compare that with the two present digits in 6307.

Hint 2 - reveal when you want a nudge

The 5083 row says exactly two of those digits occur. Try the branch in which 6 and 3 begin the code.

Reveal the solution and explanation

The code is 6381. Check it directly: 1234 has 1 and 3 in wrong places, with no correct positions. 6307 has 6 and 3 correctly placed, while 0 and 7 are absent. 8612 has 8, 6, and 1 in wrong places, while 2 is absent. Finally, 5083 has 8 correctly placed and 3 in the wrong place.

There is a useful short uniqueness check. Apply the rows in order to all four-digit arrangements with distinct digits and a nonzero first digit. There are 4,536 starting arrangements. The 1234 feedback leaves 1,155. Adding 6307 leaves 52. Adding 8612 leaves just three: 6127, 6328, and 6381.

For 6127, the last guess 5083 contains no code digits at all, so its feedback would be zero and zero. For 6328, it contains 3 and 8, but both are misplaced, giving zero and two. For 6381, it gives the required one and one. The last row therefore eliminates the two remaining alternatives.

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A Radio Code With No Spaces Between Letters

Case 44 · code-breaking / prefix code · Hard · About 10–15 minutes

A small radio station has a five-letter emergency alphabet. Its operators can send a message without separators between letters, because none of the letter codes is the beginning of another.

The receiving strip has arrived intact. Your task is to recover the letters first and the three words afterward.

AI-generated atmospheric illustration. It contains no puzzle evidence; use the complete written clues.

The complete codebook is:

Letter	Code
A	00
E	01
G	10
M	110
T	111

The full received strip contains these twenty-four digits:

110010111100111100011101

Start at the left and consume complete codes in order. Read every digit exactly once. There are no missing digits, errors, reversed segments, or other letters outside the five in the codebook.

After decoding the letters, insert spaces so that the message uses each of these three words exactly once: AT, GATE, MEET. Their order is not given.

What message did the station send? A code is a string, not an ordinary decimal number. In particular, the zero at the beginning of 00 or 01 must be preserved. The illustration’s lights and controls are not an additional transmission.

Hint 1 - reveal when you want a nudge

The strip begins 110. Its first two digits 11 are not a complete code, so read the third digit before choosing a letter.

Hint 2 - reveal when you want a nudge

The beginning divides as 110 | 01 | 01 | 111, which gives MEET. Resume at the very next unused digit.

Reveal the solution and explanation

The message is MEET AT GATE.

The complete parsing is 110 | 01 | 01 | 111 | 00 | 111 | 10 | 00 | 111 | 01. Converting each group through the printed codebook gives MEETATGATE. The supplied word list then determines MEET, AT, GATE in that order.

Why is that parsing unique? Codes beginning with 0 are completed after two digits:00 is A and 01 is E. A beginning 10 is G. A beginning 11 must continue to a third digit, becoming 110 for M or 111 for T. None of these complete codes can also be the beginning of a longer complete code.

At each boundary, those cases force both the next letter and the number of digits consumed. Repeating the argument reaches the end without leaving a partial code. No later guess can create a different earlier boundary.

As an independent check, concatenating the codes for the recovered ten letters reproduces all twenty-four digits. Testing the six orders of AT, GATE, and MEET finds only one matching the decoded letters. This code is invented for the puzzle; prior knowledge of Morse code or binary alphabet numbering would add an unnecessary second system.

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The Signwriter Who Erased Every Space

Case 37 · wordplay / segmentation · Hard · About 10–15 minutes

A signwriter has joined six labels into one continuous inscription. The original labels survive in a supply list, but all their boundaries have vanished. Short words inside longer words make the restoration less obvious than it first appears.

You must recover the labels themselves. There is no second hidden message and no need to turn the inscription into a grammatical sentence.

It contains no puzzle evidence; use the complete written clues.

The supply list contains exactly these six labels:

STAR · START · ART · ARTS · TAR · TART

Each label was used once, in some order, with no alteration. The writer then removed the spaces. The complete inscription reads:

TARTSTARSTARTARTARTSTAR

Restore five boundaries so that the result contains all six supplied labels exactly once. Give the labels in their recovered order.

A boundary can occur only between letters. You may not read backward, overlap labels, discard a character, or introduce a word outside the supply list. Two labels may share letters in their spellings, but they cannot share a physical character position in the inscription.

It may help to put six small checkboxes beside the supply list. Crossing out a used label keeps a seemingly promising partial reading from quietly using the same label twice.

Hint 1 - reveal when you want a nudge

At the beginning, TAR is a possible prefix, but it leaves the next unconsumed character as T. Test what label could start there rather than accepting the first prefix you see.

Hint 2 - reveal when you want a nudge

Starting with TART leaves STARSTARTARTARTSTAR. After choosing STAR next, keep START available as its own separate label, not as permission to reuse STAR.

Reveal the solution and explanation

The unique restoration is TART | STAR | START | ART | ARTS | TAR.

Putting these six labels together recreates TARTSTARSTARTARTARTSTAR exactly. Each supplied label appears once, and the boundaries divide rather than overlap the characters.

A successful reconstruction alone does not prove there are no alternatives. For that check, the six distinct labels can be arranged in 6 × 5 × 4 × 3 × 2 × 1 = 720 orders. Comparing the concatenation of every order with the printed inscription leaves exactly the order above. This enumeration uses only the supplied strings and the stated once-only rule.

A hand solution can use the same principle without writing out 720 rows. At each boundary, try only unused labels that match the next letters exactly. Abandon a branch as soon as no unused label matches. Continue until every character has been consumed and every label crossed out.

The supply list is important. Without its fixed membership and usage counts, words such as STAR and START would invite arbitrary alternative segmentations. The puzzle’s answer is a constrained reconstruction, not a judgment about which phrase sounds nicest.

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The Passage Added to the Plan

Case 27 · logic · Hard · About 6–10 minutes

Passage connections are defined only by the list in the text.

A copied museum plan lists six passages. Exactly one was invented; the other five are genuine. A maintenance walk provides the key to finding the addition.

The walker started in the Dome, finished at the Exit, and used every genuine passage exactly once. Rooms could be revisited, but no passage could be used twice.

Rooms and listed passages

The room abbreviations are A: Atrium, B: Bronze Room, C: Courtyard, D: Dome, and E: Exit.

The copied plan lists these undirected passages:

A–B, A–C, B–C, B–D, C–D, D–E.

Guaranteed rules

• Exactly one listed passage does not exist. Every other listed passage exists.
• There are no additional passages, hidden doors, one-way restrictions, or ways outside the building.
• The recorded walk from D to E is accurate and traverses each genuine passage once.
• Passing through a room does not consume a passage unless the walker uses that connecting route.

Which listed passage must be the invented one? Supply one complete D-to-E walk through the five remaining passages to confirm that your proposed deletion works.

You can draw a dot for each room and a line for each listed passage. The decorative gallery illustration is not a floor plan and supplies no extra connections. The question asks for a unique false passage, not a unique order for the maintenance walk.

Hint 1 - reveal when you want a nudge

At a room other than the start or finish, every arrival must be paired with a departure. Count the passages meeting each room.

Hint 2 - reveal when you want a nudge

Only D and E should have odd counts. On the copied plan, B and C also have odd counts. One false line must fix both.

Reveal the solution and explanation

B–C is the invented passage. One valid walk is D–B–A–C–D–E.

The copied plan gives passage counts A=2, B=3, C=3, D=3, E=1. In a walk using each genuine passage once, intermediate rooms have even counts: each arrival pairs with a departure. The two different endpoints, D and E, have odd counts.

B and C therefore need to change from odd to even, while the other rooms retain their parity. Removing a passage changes the count at its two endpoints by one. It must connect B and C, so the removed entry is B–C.

The example walk uses all five surviving passages exactly once. Reversing the loop gives another valid order, D–C–A–B–D–E. Multiple walks do not weaken the unique identification of the false passage.

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The Freight Office’s Forged Weight

Case 24 · investigation · Hard · About 6–10 minutes

All weights and parcel identities are specified in the text.

Three sealed museum parcels, A, B, and C, have reached the station freight office. Someone altered exactly one line of the weighing record. The parcels themselves were not changed.

Vale has four recorded totals and a trustworthy ordering note. Determine the false line and recover the three individual weights.

The four recorded lines

1. A and B together: 9 units.
2. B and C together: 11 units.
3. A and C together: 8 units.
4. A, B, and C together: 15 units.

Verified conditions

• Each parcel weighs a whole number of units, from 1 through 9 inclusive.
• A is lighter than B, and B is lighter than C. No two weights are equal.
• Exactly three recorded lines are correct. The remaining line gives an incorrect total.
• Every weighing uses the same unit and the same unchanged parcels. No container weight or hidden packing adjustment needs to be added.

Which numbered line was altered? Give A’s, B’s, and C’s weights, then calculate all four totals to check your answer.

You may solve by algebra or systematically test increasing triples. A contradiction among the totals does not tell you which line is false by itself. The trustworthy light-to-heavy ordering is an essential additional constraint.

The illustration is atmospheric. The sizes or number of drawn parcels and the empty balance do not supply measured evidence.

Hint 1 - reveal when you want a nudge

Temporarily suppose the first two pair totals and the grand total are correct. Subtract each pair total from the grand total.

Hint 2 - reveal when you want a nudge

Those three lines give C=15−9 and A=15−11. Use either pair total to find B, then test the remaining line.

Reveal the solution and explanation

Line 3 is false. A weighs 4 units, B weighs 5, and C weighs 6. Their totals are A+B=9, B+C=11, A+C=10, and A+B+C=15.

To prove uniqueness, consider the alternatives. If line 1 were false, the other lines would give A=4, B=7, C=4, violating the strict order. If line 2 were false, they would give A=2, B=7, C=6, again violating the order.

If line 4 were false, adding the three pair totals gives twice the combined weight: 28. The total would be 14, producing A=3, B=6, C=5, which also fails B<C.

Only treating line 3 as false yields an increasing triple in the allowed range and exactly three correct lines. Its recorded 8 should be 10.

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The Keeper’s Three Bowls

Case 14 · strategy · Hard · About 10–15 minutes

A keeper sets three bowls on a table and challenges you to a small game. The rules fit in a sentence, but the first move matters. A careless opening hands the keeper a winning reply.

Your task is to find a move that wins against perfect opposition, not merely against one friendly response.

All binding clues are in the text; the artwork does not show a required arrangement.

The bowls are labeled A, B, C and begin with 1, 2, 4 stones, respectively.

You move first. Players alternate turns. On each turn, remove either one or two stones from exactly one bowl. You may not remove more stones than that bowl contains. Removed stones do not return.

The player who takes the final stone wins. There are no passes, transfers, or moves affecting two bowls at once. An empty bowl stays empty.

Which first move guarantees a win, whatever legal replies the keeper chooses? Identify both the bowl and the number removed. Explain why your plan works.

There are five legal opening moves. A convincing solution must survive every possible reply after its chosen opening. To establish uniqueness, you should also be able to explain why each other opening gives the keeper a winning strategy.

Hint 1 - reveal when you want a nudge

Try removing both stones from B. The remaining bowls contain 1 and 4. There are only three possible replies.

Hint 2 - reveal when you want a nudge

After that opening, answer a removal from the 1-stone bowl by leaving 3 in the other. If the keeper removes 2 from the 4-stone bowl, leave two separate single stones.

Reveal the solution and explanation

Remove both stones from bowl B. This is the only winning opening.

The position becomes (1,0,4). If the keeper removes A's stone, remove one from C, leaving a single bowl of three. If the keeper removes one from C, remove A's stone, again leaving a single bowl of three. From three, answer a removal of one by taking two, or a removal of two by taking one.

If the keeper instead removes two from C, the position is (1,0,2). Remove one more from C, leaving two separate single stones. The keeper takes one, and you take the last.

The other openings allow these keeper replies:

• You remove one from A: the keeper removes two from C, leaving (0,2,2).
• You remove one from B: the keeper removes one from C, leaving (1,1,3).
• You remove one from C: the keeper removes one from B, leaving (1,1,3).
• You remove two from C: the keeper removes one from A, leaving (0,2,2).

At (0,2,2), the keeper mirrors your removal in the other equal bowl. At (1,1,3), the keeper mirrors a move on the single stones, or completes a total removal of three from the last bowl. Each paired response finishes the affected part. The keeper therefore takes the final stone in every case.

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