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 Compass Behind Five Plaques

Case 49 · truth-count deduction · Medium · About 5-10 minutes

A brass compass is hidden in one of five closed drawers. Above the drawers are five statements. A single reliable note says that exactly two statements are true. That is the entire trick: no witness changes their story, and no drawer is secretly empty and full at once.

Read the plaques as statements about the compass’s location. The letter on a plaque names the plaque, not necessarily the drawer it describes.

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

Five plaques, one compass

The drawers are A, B, C, D, and E. Exactly one drawer contains the compass. All other drawers are empty. The five statements are:

• Plaque A: The compass is in drawer A or drawer B.
• Plaque B: The compass is not in drawer C.
• Plaque C: The compass is in drawer D or drawer E.
• Plaque D: The compass is in drawer A.
• Plaque E: The compass is not in drawer E.

The separate note guarantees that exactly two of those five statements are true. Every other statement is false. “Or” has its ordinary inclusive meaning; with only one compass, both named drawers cannot contain it anyway.

Which drawer should you open? Give the drawer and identify its two true plaques. Then show why no other drawer fits the note.

You can solve by testing the five possible locations. Keep the candidate location fixed while evaluating all five plaques. Do not decide that a plaque is true merely because it is attached to the right drawer.

Hint 1 - reveal when you want a nudge

There are five candidate locations, not five independent hidden objects. Test one candidate at a time.

Hint 2 - reveal when you want a nudge

If the compass is in E, both negative statements need careful reading: “not in C” is true, but “not in E” is false.

Reveal the solution and explanation

Open drawer E. Plaques B and C are the two true statements.

With the compass in E, plaque A is false because neither A nor B contains it. Plaque B is true because E is not C. Plaque C is true because the compass is in one of D or E. Plaque D is false because A is empty. Plaque E is false because the compass really is in E. The total is exactly two.

Testing the remaining locations proves uniqueness. In A, plaques A, B, D, and E are true, for a total of four. In B, A, B, and E are true, for three. In C, only E is true, for one. In D, B, C, and E are true, for three. None gives two.

The lesson is a useful one for statement puzzles: evaluate claims against a proposed world. Do not assume that a truthful statement must have been placed beside the hidden object. Here the plaque attached to the correct drawer is itself false.

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The Counterfeit in the Conservator’s Tray

Case 48 · balance deduction · Medium · About 7-12 minutes

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

Eight metal tokens are laid out for an exhibition. Seven have exactly the same weight. One is counterfeit and is either heavier or lighter than a genuine token. The conservator has already made three balance-scale comparisons.

You do not get another weighing. The three recorded results contain enough information to name the token and the direction of its weight difference.

The evidence

The tokens are labeled A, B, C, D, E, F, G, and H. Each listed group contains three tokens. The scales are accurate, and every genuine token has the same weight. The counterfeit differs by a nonzero amount. No token changes weight during the test.

Left pan	Right pan	Observed result
A, B, C	D, E, F	Left is lighter
A, D, G	B, C, E	Left is heavier
B, E, H	A, C, G	Left is heavier

Which token is counterfeit? Is it heavier or lighter?

The picture sets the scene. Token positions or apparent sizes in that illustration are not measurements. Only the labeled comparisons above count as evidence.

A compact notation can help. Write “C light” for the possibility that C is the counterfeit and is too light. For each possibility, predict all three results. A possibility survives only if every prediction matches. You never need to know the counterfeit's precise weight.

Hint 1 - reveal when you want a nudge

After the first weighing, G and H are genuine. The counterfeit is either a light token on the left or a heavy token on the right.

Hint 2 - reveal when you want a nudge

The first two results leave B light, C light, and D heavy. Let the third result separate them.

Reveal the solution and explanation

C is counterfeit, and it is lighter than a genuine token.

The first result permits six possibilities: A light, B light, C light, D heavy, E heavy, or F heavy. A light would make the second left pan lighter, so reject it. E heavy would also make that left pan lighter. F heavy is absent from the second comparison and would leave it balanced. Reject both.

That leaves B light, C light, and D heavy. In the third weighing, B is on the left. A light B would make the left pan lighter, contrary to the record. D is on neither pan. A heavy D would make the two pans balance. A light C, however, sits on the right and makes the left pan heavier. Only C light matches.

It also explains the first two results directly. C is on the first left pan, making it lighter. C is on the second right pan, making the left heavier. The third right pan also contains C, so the left is heavier again. This proves both the token's identity and whether it is heavy or light.

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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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The Night Ferry and the Connection You Cannot Catch

Case 46 · timetable deduction · Medium · About 5-10 minutes

The island archive closes its parcel hatch at 08:30. You are waiting at Quay at 07:58 with a small box that must reach Island as early as possible. The harbor clerk offers a timetable. One tempting connection is too tight.

Every time below is exact. There are no delays, shortcuts, or other services. Your task is to find the earliest possible arrival, rather than merely an arrival before closing.

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

The harbor rules

You may wait at any stop. Changing boats requires at least two full minutes between arrival and departure. This also applies at Quay before your first departure. A boat departs on time; arriving one minute before departure is too late to board.

From	To	Departs	Arrives
Quay	Lock	08:00	08:12
Quay	Glass	08:00	08:08
Glass	Island	08:09	08:20
Glass	Reed	08:12	08:18
Lock	Reed	08:14	08:19
Lock	Island	08:15	08:26
Reed	Island	08:20	08:29

What route gives the earliest legal arrival at Island? Give the arrival time and explain why the apparently faster alternative fails.

A useful way to work is to write the earliest time you are ready to board at each stop. That time is two minutes after arrival. Do not confuse reaching a pier with being ready to leave it.

Hint 1 - reveal when you want a nudge

There are only three listed arrivals at Island. Test the earliest one first.

Hint 2 - reveal when you want a nudge

Arriving at Glass at 08:08 makes you ready to board at 08:10.

Reveal the solution and explanation

Take Quay to Lock at 08:00, then Lock to Island at 08:15. You arrive at 08:26.

The first boat reaches Lock at 08:12. Your two-minute change is complete at 08:14, so the 08:15 departure is legal. You started at Quay at 07:58, which also satisfies the first boarding requirement.

The timetable's 08:20 Island arrival looks better. It belongs to the boat leaving Glass at 08:09. Your only way to Glass arrives at 08:08. You would need to be there by 08:07 to catch that boat, so this connection cannot be made.

The remaining Island arrival is 08:29 from Reed. You can reach it by Quay-Glass-Reed: 08:18 plus two minutes permits boarding at 08:20. Quay-Lock-Reed fails because 08:19 plus two minutes is 08:21. Either way, no Reed route beats 08:26. These are all listed Island arrivals, so the earliest legal choice is unique.

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Three Registers and One Surviving Letter

Case 45 · wordplay / sets · Medium · About 7–10 minutes

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

An observatory clerk keeps three registers for every delivery. Two registers record possible letters; the third lists letters that must be rejected. The surviving characters identify one direction.

There are five deliveries, and their row order is fixed. You need no compass, star knowledge, or interpretation of the accompanying art.

For each row, find the letters occurring in both A and B. Then discard any of those letters that also occurs in C. Exactly one distinct letter will remain in each row.

Row	A	B	C
1	LANTERN	NOODLE	LEAF
2	COPPER	ORCHID	CRISP
3	RAVEN	BRIAR	CABIN
4	THIMBLE	STONE	HONEY
5	SHADOW	HEARTH	WART

Read the five survivors from row 1 to row 5. What direction is recorded?

Repeated occurrences do not matter. Treat a letter as either present or absent in each word. You are not subtracting letter counts, matching positions, finding anagrams, or choosing a shared sound. All strings use their exact printed uppercase spellings.

For a separate miniature example, if A were ABC, B were BCD, and C were B, the common letters would be B and C; rejecting B would leave C. That example is not an extra message row.

Hint 1 - reveal when you want a nudge

Make the shared-letter list before consulting C. A letter found only in A cannot survive, even if it is absent from C.

Hint 2 - reveal when you want a nudge

In row 1, LANTERN and NOODLE share L, E, and N. LEAF removes L and E, leaving N.

Reveal the solution and explanation

The direction is NORTH.

• Row 1: the common letters are L, E, N. C removes L and E, leaving N.
• Row 2: the common letters are C, O, R. CRISP removes C and R, leaving O.
• Row 3: the common letters are A and R. CABIN removes A, leaving R.
• Row 4: the common letters are T and E. HONEY removes E, leaving T.
• Row 5: the common letters are H and A. WART removes A, leaving H.

Each row therefore has one survivor before the five letters are combined. The recognizable direction confirms the extraction but is not needed to choose among multiple surviving letters.

A complete check tests every alphabet letter for three conditions: present in A, present in B, and absent from C. Applying that test independently to all five rows gives the singleton sets N, O, R, T, H. Extra occurrences such as the repeated O in NOODLE or R in BRIAR do not change membership. Reading the rows backward would contradict the stated order and is not an alternative solution.

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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 Tailor’s Six-Letter Seam

Case 43 · letter ordering · Easy · About 6–9 minutes

A tailor is stitching six different letter patches into a single straight seam. Before the cloth is folded, she copies down relationships between neighboring patches. The six patches later fall loose, but the notes survive.

Rebuild the seam from those relationships. Recognizing an English word can confirm your result, but it is not one of the rules you must assume.

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

The patches are T, I, N, S, E, L. Use each patch exactly once in six positions, read from left to right.

The tailor’s notes say:

• I is immediately before N.
• N is immediately before S.
• E is immediately before L.
• L occupies the last position.
• T occurs somewhere before I.

What six-letter string do the patches form?

“Immediately before” means the positions are adjacent, with no intervening patch. “Somewhere before” need not mean adjacent. The line has a beginning and an end; it is not a necklace whose last patch connects around to the first.

No patch may be turned into another letter, reused, or removed. There are no spaces in the result. All five notes are true at the same time. The arrangement in the supply list is simply the list of available patches and is not an extra order constraint.

Hint 1 - reveal when you want a nudge

The first two notes combine into a single three-letter block. Do not arrange I–N and N–S as though they used two different N patches.

Hint 2 - reveal when you want a nudge

The blocks are INS and EL. L must be last, so EL occupies positions 5 and 6. Consider where T can go in the four remaining positions.

Reveal the solution and explanation

The seam reads TINSEL.

I immediately before N and N immediately before S make the uninterrupted block INS. E immediately before L makes the block EL. Since L must be the sixth and final patch, E must be fifth.

The first four positions must therefore contain T and the intact block INS. Only two block orders are possible: T–INS or INS–T. The last note requires T before I, so only T–INS survives. Adding the fixed final EL gives TINSEL.

Every note can now be checked directly: I is second and N third; N is third and S fourth; E is fifth and L sixth; L is last; and T, in first position, precedes I.

The block argument proves uniqueness without consulting a dictionary. As a separate finite check, the six distinct patches have 720 possible orders. Exactly one satisfies all five notes. Treating the seam as a loop or weakening “immediately” to “somewhere” would admit different possibilities, which is why those conventions are explicitly fixed before the puzzle begins.

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Five Word Tickets in Five Coat Pockets

Case 42 · word logic · Medium · About 8–12 minutes

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

A cloakroom attendant has placed five word tickets in a row of pockets. She remembers relationships between the spellings better than she remembers the pocket numbers. Her four notes are enough to reconstruct the row.

The names are printed words rather than colors you must distinguish. Even if every ticket were gray, the puzzle would be unchanged.

The pockets are numbered 1–5 from left to right. Put one of these tickets in each pocket, using every ticket exactly once:

AMBER · COBALT · CINDER · SILVER · VELVET

All four notes are true:

1. COBALT lies somewhere to the left of the only ticket whose spelling repeats a letter.
2. The only five-letter ticket is immediately to the right of that repeated-letter ticket.
3. SILVER is immediately to the left of the ticket containing N.
4. The ticket containing M lies somewhere to the left of the ticket containing N.

What are the five tickets, in left-to-right order?

“Somewhere to the left” allows a gap. “Immediately to the left” or “immediately to the right” means neighboring pockets. A letter is repeated if it occurs at least twice within the same printed word. There is no requirement that the tickets be sorted alphabetically or by color.

You may move paper slips while solving, but the five written spellings contain all the information needed.

Hint 1 - reveal when you want a nudge

VELVET is the only ticket with repeated letters, and AMBER is the only five-letter ticket. These two must form a neighboring pair in that order.

Hint 2 - reveal when you want a nudge

SILVER and CINDER form the other neighboring pair. AMBER must precede CINDER, so the two pairs cannot be placed in either order freely.

Reveal the solution and explanation

The order is COBALT, VELVET, AMBER, SILVER, CINDER.

VELVET repeats V and E; the other four words have no repeated letter. AMBER alone has five letters. Note 2 therefore joins VELVET–AMBER into one two-pocket block.

CINDER is the only word containing N. Note 3 joins SILVER–CINDER into the other two-pocket block. AMBER is the only word containing M, so note 4 requires AMBER before CINDER. Because the two blocks cannot overlap, VELVET–AMBER must come before SILVER–CINDER.

The remaining ticket is COBALT. Note 1 puts it before VELVET, so it must come before both blocks. The three units have lengths 1,2, and 2 and occupy all five pockets. No gaps or alternative placements remain.

For an exhaustive check, there are 120 ways to order five distinct tickets. Testing the four notes against every order leaves exactly the arrangement shown. The argument does not depend on whether AMBER and COBALT suggest actual colors, and the quantities count letters in the printed labels rather than anything in the accompanying art.

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The Moss Key and the Moving Letters

Case 41 · cipher · Medium · About 10–15 minutes

A conservatory messenger uses a repeating keyword to change the alphabet shift from letter to letter. The keyword has been left in plain sight. The message is secure only against a visitor who skips the operating instructions.

Recover the instruction exactly. No plant identification, historical cipher knowledge, or interpretation of the illustration is required.

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

Write the alphabet ABCDEFGHIJKLMNOPQRSTUVWXYZ with values A=0 through Z=25.

For encryption, write the keyword MOSS repeatedly underneath the letters of a spaceless message. Add each message-letter value to the key-letter value beneath it. If the total is 26 or more, subtract 26. Convert the result back to a letter.

For example, A encrypted with key B becomes B, because 0+1=1. Z with key C becomes B, because 25+2=27, and 27−26=1.

The complete encrypted message is:

RCDDAKJWPHZJQOV

The key begins at M beneath the first ciphertext letter and continues M–O–S–S–M–O–S–S without resetting. The plaintext has fifteen letters. After decrypting, put spaces at word lengths 6, 3, 6.

What instruction was sent? Every ciphertext character is correct. The example pairs are separate demonstrations and do not use up any part of the repeated keyword.

Hint 1 - reveal when you want a nudge

To undo addition, subtract the key value. When the result is negative, add 26 before converting back to a letter.

Hint 2 - reveal when you want a nudge

The first four key letters are MOSS, with values 12,14,18,18. R−M gives F, C−O wraps to O, and the next two letters each decode to L.

Reveal the solution and explanation

The instruction is FOLLOW RED THREAD.

Repeat the key as MOSSMOSSMOSSMOS. Subtracting its values from RCDDAKJWPHZJQOV, with alphabet wraparound, gives FOLLOWREDTHREAD. The specified lengths divide it into FOLLOW, RED, and THREAD.

The first letter is a direct subtraction: R has value 17 and M has value 12, leaving 5, which is F. The second crosses the alphabet boundary: C has value 2 and O has value 14;2−14=−12, and adding 26 gives 14, which is O. The key does not start over at RED. It remains aligned with the continuous fifteen-character string.

This is a fully keyed cipher. A solver is not expected to infer MOSS by frequency analysis or guess a password from a picture. Once the key alignment is fixed, each position is a reversible addition modulo 26 and has one possible plaintext letter.

A forward check adds the same key values to the recovered message. It recreates all fifteen printed ciphertext letters, including the wraparound positions. The spacing check is separate:6+3+6 accounts for every letter exactly once.

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The Beacon’s Alternating Lantern Rows

Case 40 · cipher / binary · Medium · About 8–12 minutes

A harbor signal keeper has copied six rows from a lantern panel. To make the wiring convenient, alternate rows run in opposite directions. The copy preserves every lamp state, and the decoding instructions explain the reversal.

The signal names a destination. You do not need nautical knowledge or to count any lights in the illustration.

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

Each row contains five bits. A bit is the printed digit 0 or 1. Read odd-numbered rows left to right and even-numbered rows right to left.

1. 00001
2. 10001
3. 01000
4. 11000
5. 00110
6. 00100

After choosing the correct reading direction, assign the five positions weights 16, 8, 4, 2, 1 from left to right. Add the weights whose bits are 1. A 0 contributes nothing.

Convert each total to a letter using A=0, B=1, C=2, …, Z=25. The alphabet is ABCDEFGHIJKLMNOPQRSTUVWXYZ. Totals 26–31 would be invalid, but none should appear after correct decoding.

Read the six resulting letters in row order. What destination is signaled? As a separate example, the already correctly oriented row 00101 totals 4+1=5, giving F. It is not a seventh row of the message.

Hint 1 - reveal when you want a nudge

Correct the direction before applying weights. Reversing an even row can move a 1 from a large weight to a small one.

Hint 2 - reveal when you want a nudge

Row 4 is printed 11000 but must be read as 00011. It totals 2+1=3, so its letter is D.

Reveal the solution and explanation

The destination is BRIDGE.

Row	Oriented bits	Total	Letter
1	00001	1	B
2	10001	17	R
3	01000	8	I
4	00011	3	D
5	00110	6	G
6	00100	4	E

Rows 2 and 6 look unchanged when reversed because their bit patterns are symmetrical. They still obey the alternating rule. It is row 4 that makes ignoring that rule visibly fail:11000 totals 24, which would produce Y instead of D.

The zero-based alphabet is equally important. Using A=1 would shift every interpretation and would contradict the supplied example, where a total of 5 means F.

Uniqueness follows at each stage: the row number fixes orientation, the five weights fix a total, and each permitted total fixes one letter. Encoding BRIDGE back into five-bit values and reversing its even rows returns all six printed rows exactly. This check uses the actual digits, independent of any expectation about a harbor destination.

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A Letter Hiding in Three Familiar Words

Case 39 · letter riddle · Easy · About 4–7 minutes

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

A lantern keeper has lost a single brass letter from a sign. Instead of naming it, she leaves three statements about other words. The statements concern written characters, so a silent reading is just as useful as saying the words aloud.

Your answer must be one uppercase letter, not a whole word, a sound, or an object from the illustration.

The keeper’s complete riddle is:

I appear exactly once in LANTERN.
I appear exactly twice in CRESCENT.
I do not appear in LIGHT.

Which letter is missing from the sign?

Use the printed spellings exactly. Count occurrences of the same letter, wherever they fall in a word. There are no hidden spaces, silent-letter rules, different fonts, or alternate spellings. All three statements are true of the intended answer.

The possible letters are the ordinary uppercase alphabet, ABCDEFGHIJKLMNOPQRSTUVWXYZ. You may cross out candidates as soon as one statement rules them out.

The riddle does not require every statement to eliminate a new candidate. A final statement can also confirm a result established by earlier ones. Your task is to find the letter satisfying the complete set and show why no second letter does.

Hint 1 - reveal when you want a nudge

Treat “exactly twice” literally. CRESCENT contains two repeated letter types; list them before checking the other words.

Hint 2 - reveal when you want a nudge

C and E each occur twice in CRESCENT. C is absent from LANTERN, while E occurs there once.

Reveal the solution and explanation

The missing letter is E.

CRESCENT is C–R–E–S–C–E–N–T. Its only letters occurring exactly twice are C and E. That one statement reduces the alphabet to two candidates, without requiring a guess about the lantern keeper’s sign.

LANTERN is L–A–N–T–E–R–N. It contains E once and C zero times, so only E survives the first statement. The two N characters are worth noticing: N cannot qualify as a once-occurring letter there, although it occurs in both of the first two words.

Finally, LIGHT contains no E, satisfying the third statement. It contains no C either, but C has already failed the LANTERN requirement. A clue’s failure to distinguish two candidates by itself does not make it untrue or useless as a check.

Counting all 26 possible letters against all three requirements gives exactly one survivor. No pronunciation judgment is needed: the answer follows from finite character counts. Inserting another letter because it makes a more appealing sign would change the stated task.

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The Weaver’s Inward Reading

Case 38 · code-breaking / spatial text · Medium · About 6–10 minutes

A paper weaver has filled a small square with an instruction. Reading its rows produces almost a word, then fragments. The maker’s route, however, is written clearly on the back.

The square below is the complete puzzle. The accompanying illustration is decorative, and no particular thread, color, or object in it needs to be interpreted.

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

Start at the upper-left letter. Read that cell, then move right one cell at a time. If the next step would leave the square or enter a cell already read, turn clockwise and continue. Clockwise direction changes are right → down → left → up → right.

Read every cell exactly once and stop after sixteen letters. Do not skip cells, move diagonally, or turn before an obstacle requires it.

L	O	O	K
E	V	A	U
H	E	S	N
T	R	E	D

For accessible transcription, the rows from top to bottom are LOOK / EVAU / HESN / TRED. Each slash marks a new row, not a character.

Separate the recovered message into words of lengths 4, 5, 3, 4. What instruction does the weaver give?

Hint 1 - reveal when you want a nudge

The first turn occurs after K at the top-right corner. Continue downward through U, N, and D.

Hint 2 - reveal when you want a nudge

The outer boundary reads LOOKUNDERTHE. Four unread cells remain in a smaller central square; keep the same turning rule.

Reveal the solution and explanation

The instruction is LOOK UNDER THE VASE.

The top row gives LOOK. The right edge contributes UND. The bottom row, now read from right to left without repeating its corner, contributes ERT. The left edge upward contributes HE. Those twelve perimeter cells therefore give LOOKUNDERTHE.

From the E at row 2, column 1, another upward step would revisit the already read top-left cell. Turn right into the center. The remaining path visits V at row 2, column 2; A at row 2, column 3; S at row 3, column 3; and E at row 3, column 2. That completes VASE.

The supplied lengths split the sixteen letters unambiguously into four words. Every grid cell occurs once in the account above, with no duplicated corner letters. The movement is uniquely determined because a turn is permitted only when the current forward step is blocked by the boundary or prior path.

This is a route through printed characters, so the instruction remains available if the illustration is omitted or the puzzle is read aloud using row-and-column coordinates. Finding a vase in the artwork is neither necessary nor a substitute for decoding the square.

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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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A Dictionary Filed by the Far End

Case 36 · wordplay / ordering · Easy · About 5–8 minutes

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

A candlelit reading room has an eccentric catalogue. Its librarian files words according to their far ends, as though she met every word walking backward. A visitor who uses the ordinary first-letter order will retrieve the wrong instruction.

Six catalogue entries carry a small message. Their spelling, rather than their meaning, supplies the entire key.

The entries are LOVE, NORDIC, CAMERA, ELF, ABSORB, DEED.

1. Reverse the letters of each entry to make its filing key.
2. Sort the six keys into ordinary alphabetical order.
3. Keep each original entry attached to its own key.
4. Read the first letter of the original entry at each position in the sorted list.

What six-letter message do you obtain?

Use ABCDEFGHIJKLMNOPQRSTUVWXYZ as the alphabet. Compare keys from left to right and use the first differing letter. If one key were the beginning of another, the shorter key would come first, although that tie rule will not be needed here.

Do not reverse the extracted message or replace any entry with a synonym. The capitalized spelling NORDIC is simply one supplied string; knowledge of geography is unnecessary. The words begin in a deliberately unsorted order, and each is used once.

Hint 1 - reveal when you want a nudge

Write a second column for reversed keys. LOVE belongs beside EVOL, not beside another word beginning with L.

Hint 2 - reveal when you want a nudge

CAMERA has key AREMAC and ABSORB has key BROSBA. Their keys put them before entries whose reversed keys begin with C, D, E, or F.

Reveal the solution and explanation

The message is CANDLE.

Filing key	Original entry	First letter
AREMAC	CAMERA	C
BROSBA	ABSORB	A
CIDRON	NORDIC	N
DEED	DEED	D
EVOL	LOVE	L
FLE	ELF	E

Here the keys begin with six different letters, so the first comparison already fixes the complete order. DEED happens to be unchanged by reversal. That does not exempt it from the operation or make it a special starting point.

Uniqueness comes from the sorting rule, not from recognizing a useful object in the resulting letters. Every original entry has exactly one reversed key; every key is distinct; and ordinary alphabetical comparison orders those keys unambiguously. Reading the required original initials is then deterministic.

As a check, move each original entry back beside its key and reverse the key again. All six printed entries return. Taking initials from the keys instead would give ABCDEF, which records their filing order but ignores the explicitly named source of the message.

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The Nameplate Seller’s One False Description

Case 35 · word riddle / deduction · Medium · About 6–9 minutes

A market stall has six wooden nameplates. The seller will give you one, but her description contains exactly one false statement. She promises that all the others are true.

This is a riddle about the printed names, not the properties of wood, plants, masonry, or musical instruments. Every fact needed for the answer is visible in the words.

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

The possible nameplates are:

RAVEN · CEDAR · MAPLE · STONE · BRICK · FLUTE

The seller makes these four statements about the correct word:

1. It contains the letter A.
2. In ordinary alphabetical order it falls strictly after BRICK and strictly before FLUTE.
3. Its last letter is R.
4. Its third letter is not D.

Exactly one of these four statements is false for the correct nameplate. It is not enough to satisfy one statement, or to find a word with at least one false statement.

Which nameplate should you take, and which statement is the false one?

Compare the written uppercase spellings exactly as shown. “Strictly” excludes the endpoints BRICK and FLUTE themselves. Alphabetical order means comparing letters from left to right, using the first position where the words differ. The familiar alphabet is ABCDEFGHIJKLMNOPQRSTUVWXYZ.

Hint 1 - reveal when you want a nudge

Make four true/false boxes for each candidate. You are looking for a row with three true boxes and one false box.

Hint 2 - reveal when you want a nudge

CEDAR lies between BRICK and FLUTE. Its third letter is D, so check carefully how statement 4 is phrased.

Reveal the solution and explanation

Take CEDAR. Statement 4 is the single false statement.

CEDAR contains A; it follows BRICK and precedes FLUTE alphabetically; and it ends in R. But its third letter is D, so the claim that its third letter is not D is false.

Checking every rival establishes that the answer is unique. RAVEN and MAPLE each satisfy statements 1 and 4, but fail 2 and 3. STONE fails 1,2, and 3, while satisfying 4. BRICK fails 1 and 3 and also fails 2 because the alphabetical comparison is strict. FLUTE fails those same three statements. Both satisfy 4. Thus the candidates have, respectively, two, one, two, three, three, and three false statements in their printed order.

The end words can be tempting because they appear in a clue, but an endpoint is not inside an open interval. Likewise, changing the negative wording of statement 4 would create a different puzzle. Keeping the truth table literal is more dependable than guessing which object sounds most at home in a market story.

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The Drawer With a Growing Alphabet Shift

Case 34 · cipher · Medium · About 7–10 minutes

A clockmaker dislikes a cipher wheel that stays still. Her encrypted note advances the alphabet setting after every letter, and the mechanism remembers how many letters have already passed.

The note identifies a drawer. Its full operating instructions survived, so the challenge is careful reversal rather than guessing a secret rule.

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

Use the alphabet ABCDEFGHIJKLMNOPQRSTUVWXYZ, with wraparound after Z.

The clockmaker first removes all spaces from an instruction. She encrypts its first letter by shifting it forward one place, the second by two places, the third by three places, and so on. Letter number fourteen is shifted fourteen places. No position resets at a word boundary.

Her complete ciphertext is:

UCNIYNPZMNCMJSG

There are fifteen letters. Recover them, then restore the spaces using the supplied word lengths: 4, 5, 6.

A miniature example of the method is CAT → DCW: C moves one place to D, A moves two places to C, and T moves three places to W. This example is separate from the drawer instruction; do not count its letters when setting the shift.

What instruction did the clockmaker write? No extra substitution, reversed reading order, or interpretation of the picture is involved.

Hint 1 - reveal when you want a nudge

Number the ciphertext letters 1 through 15. Decryption moves backward by the corresponding number, not forward again.

Hint 2 - reveal when you want a nudge

The first four calculations are U−1=T, C−2=A, N−3=K, and I−4=E. Keep increasing the shift after this first word.

Reveal the solution and explanation

The instruction is TAKE THIRD DRAWER.

Subtracting the position from each encrypted letter yields TAKETHIRDDRAWER. The supplied lengths divide this into TAKE, THIRD, and DRAWER. The spaces are fixed by the instructions rather than supplied by an assumption about sensible English.

Here are useful checkpoints. Position 5 is Y; moving backward five places gives T, the beginning of THIRD. Position 9 is M; moving backward nine places gives D, its final letter. Position 10 is N; moving backward ten places gives D, the beginning of DRAWER. The final G moves backward fifteen places, wrapping around the alphabet, to R.

A common near-solution comes from restarting the count at the second word. That would use a shift of one at position 5, contradicting the rule that spaces were removed before encryption.

Each position has one specified shift, and shifting an alphabet is reversible. Consequently every ciphertext character has exactly one predecessor. As a separate check, encrypting all fifteen recovered characters with shifts 1–15 reproduces UCNIYNPZMNCMJSG character for character. There is no competing key to select by taste or context.

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The Conservatory’s Misfiled Route Cards

Case 33 · code-breaking / path · Medium · About 8–12 minutes

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

The keeper of a winter garden uses four-letter route cards. A card’s outside letters connect it to other cards; its second letter carries a fragment of the destination password.

Several cards lead somewhere useful eventually. Tonight, however, the gate accepts a journey of one exact length.

Choose and order exactly five different cards from this tray:

VJWE · KTWS · ARPM · KSYP · LFXA · EVXA · QUMV · ESVK · AIZQ · MOQE

• The first card must begin with L.
• The last card must end with S.
• The last letter of each card must equal the first letter of the following card.
• No card may be used twice. Cards cannot be reversed or altered.

For example, a card ending in Q may be followed by a card beginning with Q. The cards need not spell ordinary words.

Once you have a valid five-card route, read the second letter of each card in route order. What five-letter password opens the gate?

All tray entries are distinct. You must obey the exact route length even if a longer journey also reaches an S-ending card. There are no invisible cards between the printed ones.

Hint 1 - reveal when you want a nudge

Only LFXA can start. After it, the two A-starting cards create the first real choice.

Hint 2 - reveal when you want a nudge

AIZQ forces QUMV and then VJWE. After four cards that route ends at E, with only one card remaining. Can any E-starting card finish at S?

Reveal the solution and explanation

The route is LFXA → ARPM → MOQE → ESVK → KTWS. Its second letters give FROST.

LFXA is forced by the starting rule. If AIZQ follows it, QUMV and VJWE are forced next. Four cards have then been used and the route reaches E. Neither ESVK nor EVXA ends in S, so that branch cannot finish in one more card.

Taking ARPM second instead forces MOQE third. At E, choosing EVXA would leave the route at A after four cards. Neither available A-starting card ends in S. Therefore ESVK must be fourth. It leads to K, where KTWS completes the route and KSYP does not.

That branch analysis considers every choice reachable under the rules. It establishes uniqueness without assuming the password must be a recognizable word. In particular, the longer route through Q and V is not a rival solution: it needs six cards to reach KTWS. Counting cards rather than connections is essential. Five cards contain four joins.

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The Postmaster’s Backward Stamp

Case 32 · cipher · Medium · About 6–10 minutes

A postmaster uses two small operations to disguise short delivery instructions. He has written down his complete procedure, but left the wheel setting on yesterday’s test slip.

The sealed parcel must go to one destination. Recover the instruction without guessing what would be a sensible place to hide it.

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

Here is the ordinary alphabet:

ABCDEFGHIJKLMNOPQRSTUVWXYZ

To encrypt, move every letter forward by the same number of alphabet positions, wrapping Z back to A. The setting is a whole number from 1 through 9. Then reverse the letters inside each word. Leave the word order and spaces unchanged.

The test slip records an original word and its final encrypted form:

INK → NQL

The parcel’s encrypted instruction is:

QHSR HKW HXOE ARE

Find the setting, then recover the instruction. The test and the parcel use exactly the same procedure and setting. There are no missing letters, extra reversals, or changes to the spaces.

You can make a paper alphabet wheel if useful. Its appearance is optional; an ordinary written alphabet contains everything the method needs.

Hint 1 - reveal when you want a nudge

Undo the final operation first. Reverse NQL before comparing it with INK.

Hint 2 - reveal when you want a nudge

NQL reversed is LQN. I becomes L, N becomes Q, and K becomes N: each move is three places forward.

Reveal the solution and explanation

The setting is 3, and the instruction is OPEN THE BLUE BOX.

The calibration resolves the setting rather than leaving it to the plausibility of the final sentence. Reversing NQL gives LQN. Comparing corresponding letters with INK gives a shift of three in every position. Of the allowed settings 1–9, only 3 maps I to L; the other two letters independently confirm it.

Now reverse each encrypted word: QHSR becomes RSHQ, HKW becomes WKH, HXOE becomes EOXH, and ARE becomes ERA. Move every letter three places backward. The four results are OPEN, THE, BLUE, and BOX.

Notice that the words themselves never exchange positions. Reversing the entire line would undo an operation the postmaster never performed.

For a check in the original direction, shift OPEN forward three places to RSHQ, then reverse it to QHSR. Repeating that check for all four words recreates the complete printed ciphertext. A fixed shift is a one-to-one letter mapping and word reversal is reversible, so the calibrated procedure permits only one plaintext, whether or not a solver recognizes the sentence early.

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The Archive That Keeps the Missing Letters

Case 31 · wordplay · Easy · About 5–8 minutes

The night archivist has left five long labels on the desk and five short labels in a tray. Her note says that a label can lose a letter without losing its place in the collection. The missing letters have a second job.

You need one five-letter password. The illustration establishes the setting, but nothing in its cards or furnishings is a clue.

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

Match each long label to a different short label. Delete exactly one letter from the long label to obtain its partner. Keep the remaining letters in their original order: rearrangement, substitution, and pronunciation tricks are forbidden.

The long labels, in the order you must use, are:

1. THREAT
2. COAST
3. PLANT
4. CLAMP
5. CHART

The tray contains CART, PLAN, TREAT, LAMP, COST. Use each tray label exactly once.

Write down the letter deleted from each long label. Read those five letters from the first numbered row to the last. What password does the archivist want?

Work only with the supplied spellings. You do not need to find additional words in a dictionary, decide whether an unusual abbreviation counts, or infer an ordering from word meanings.

Hint 1 - reveal when you want a nudge

The short-label tray is a closed set. Try each available short label against THREAT while preserving letter order.

Hint 2 - reveal when you want a nudge

THREAT becomes TREAT when H is removed. Keep H, rather than a letter from the surviving word, as the first password character.

Reveal the solution and explanation

The password is HATCH.

Long label	Short partner	Deleted letter
THREAT	TREAT	H
COAST	COST	A
PLANT	PLAN	T
CLAMP	LAMP	C
CHART	CART	H

The numbered list supplies the reading order, so sorting the tray alphabetically would change the intended operation. The unused order of the tray is merely how its contents were listed.

There is no choice between two valid matching arrangements. Comparing every one-letter deletion of each long label with the five supplied short labels gives exactly the partner shown in that row. Once a partner is fixed, the deleted character is fixed too. Even the repeated H is legitimate: the rule forbids reusing a tray label, not repeating a password letter.

A final forward check is useful. Insert the recorded character back into its original position in each short label. All five original long labels return without any other change. The result therefore satisfies both the matching rule and the extraction rule.

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