# Solution to A Collection of Conundrums and Riddles

#### by Anderson Wang, Colin Lu, Josh Alman, Scott Kominers, and Zach Wissner-Gross

The headings of “Letters Round” and “Numbers Round” combined with the colors and formatting hint at the British game show Countdown, which contains a letters round where contestants try to make the longest word from a set of 9 letters, and a numbers round where contestants use 6 numbers (which are either 1–10 or 25, 50, 75, or 100) and the four basic arithmetic operations to get as close as possible to a target number.

## Letters Round

Each of these puzzles has a unique longest word, which is always either 8 or 9 letters long:

EHMATNIBITHIAMINEB
ECUHSNLIMMUNCHIESL
REMOAUGMDRUMMAGEDO
SEICYCPTACITYSCAPE
MEIITYMSLMYELITISM
VAEELITLAALLEVIATE
PIIRSBECRCRISPIERB
ATSRPOMKEPOSTMARKE
HPAGTECTIPATHETICG

The unused letters spell out BLOOMBERG.

## Numbers round

In each puzzle, the target number can be exactly achieved, and there is only one possible set of numbers that can be used to make the target number (in fact, in almost all cases there is only one sequence of operations that makes the target number—the lone exception is 641, where there are a few ways to use all 6 numbers). In each puzzle, either 5 or 6 numbers are used.

PuzzleAnswer (order is given by parentheses from inner to outer)Unused number (if 5 numbers)
758, 50 75 9 3 7 5(((50/5) + 75)*9) - 73
815, 100 25 2 3 6 4(((100*4) + 6)*2) + 325
989, 75 25 9 2 4 8(75 - (8*4))*(25 - 2)9
971, 50 100 2 3 8 6((((100*6) + 50)*3) - 8)/2
937, 50 25 6 9 7 8(((25*6) - 9)*7) - 508
759, 25 50 8 6 5 4((50 - 5)*(25 - 8)) - 64
902, 50 75 9 6 4 5((75 - 6)*(9 + 4)) + 550
641, 25 75 4 3 7 6((75 + 6 + 4 + 3)*7) + 25

Plugging the six unused numbers into the blanks at the bottom gives ((3*25) + (9*8))*4 - 50, which evaluates to 538.

Note that there are many online solvers for the Countdown letters and numbers rounds that we can use (the numbers rounds in particular are very difficult to do by hand, given that they are in some sense the highest difficulty possible).

At this point, we’ve extracted BLOOMBERG and 538, which are both news websites/organizations. Perhaps with help from the title, we can realize that both of these websites have weekly puzzle columns, Kominer’s Conundrums on Bloomberg Opinion and The Riddler on FiveThirtyEight. If we view these columns released the week of the hunt, we can quickly verify that they are both relevant to this puzzle: both the Kominer’s Conundrums and the Riddler Classic mention the name Barbara Yew, the keynote speaker for the fictional Mystery Hunt conference. The Riddler also coyly includes the words “mystery” and “hunt”, while the Conundrums column references the Mystery Hunt directly.

We can now solve these puzzles (if you’re doing them a week after the hunt the solutions will be available online, but we will also give solutions below).

## Kominer’s Conundrums

This puzzle consists of two lists of somewhat vague crossword clues. From solving some of the clearer ones, we can see that the answers to these clues are exactly one letter apart.

Actress in "Breaking Bad"BRANDTActor in "The Godfather"BRANDOT/O
Greek underworld figureCHARONMultiple Oscar-winning directorCUARONH/U
Part of an operaOVERTUREWhat the Supreme Court might do to a caseOVERTURNE/N
Popular lemon varietyMEYERUnit of lengthMETERY/T
Popular board gameSCRABBLEWrite quicklySCRIBBLEA/I
Cut shortCURTAILPart of a stageCURTAINL/N
Alternative form, in geneticsALLELEClaim or assertALLEGEL/G
ModestyHUMILITYClimate measureHUMIDITYL/D
ChooseDECIDESolve, as in a cipherDECODEI/O
Card gamePOKERTV series that featured Kendrick LamarPOWERK/W
Narrow naval vesselCANOEJapanese corporationCANONE/N
Upper body garmentTUNICNorth African capitalTUNISC/S

Reading all the changed letters for the first clues, followed by the changed letters for the second clues, spells out THEY ALL LIKE COUNTING DOWNS, referencing the Countdown game show.

## The Riddler

The puzzle asks us to fill in a 3×8 grid with digits, where the product of each row and column is given. The unique solution is the following (a logical path is in the appendix):

First DigitSecond DigitThird DigitTarget Number
776294
983216
953135
72798
827112
74384
577245
85140
8,890,560156,80055,566

The next step is to realize that both the Conundrums and Riddler puzzles can be applied to the original puzzle - in particular, the Conundrums has twelve pairs of clues and we have twelve letters rounds, and the Riddler gives us eight 3-digit numbers and we have eight numbers rounds, each with a 3-digit goal. Additionally, some of the original puzzles did not contribute to extracting BLOOMBERG or 538, strongly implying that we somehow want to use them now.

## Letters round (redux)

Since the Conundrums required us to change a letter in a word to get another word, we can apply the exact same letter change to the corresponding Countdown letters puzzle to get a new puzzle. This is confirmed by the letter T appearing in the first puzzle, the letter H appearing in the second puzzle, and so on down to the letter C appearing in the twelfth and final puzzle.

We can solve all the new puzzles, and find that they once again have a unique longest word, which is always 8 or 9 letters long.

Original PuzzleChanged lettersModified PuzzleAnswerUnusued letter (if 8 letters)
EHMATNIBIT/OEHMAONIBIBOHEMIANI
ECUHSNLIMH/UECUUSNLIMMINUSCULE
REMOAUGMDE/NRNMOAUGMDGOURMANDM
SEICYCPTAY/TSEICTCPTAECSTATICP
MEIITYMSLL/NMEIITYMSNIMMENSITY
MTLLDAERIL/GMTLGDAERIMIGRATEDL
VAEELITLAL/DVAEELITDAVALIDATEE
PIIRSBECRI/OPIORSBECRPROSCRIBE
ATSRPOMKEK/WATSRPOMWETAPEWORMS
HPAGTECTIC/SHPAGTESTISPAGHETTI

Performing the same extraction of taking the unused letter in the puzzles with 8-letter answers gives us IMPALED.

## Numbers round (redux)

Because The Riddler gives us eight 3-digit numbers, we can try to re-solve the Countdown numbers puzzles using these numbers as targets. Doing so, we once again find that each puzzle has a solution that uniquely uses either 5 or 6 of the numbers:

Original PuzzleModified PuzzleAnswer (order is given by parentheses from inner to outer)Unused number (if 5 numbers)
758, 50 75 9 3 7 5776, 50 75 9 3 7 5(((50×3) + 7)×5) − 975
815, 100 25 2 3 6 4983, 100 25 2 3 6 4((6+4) × (100−2)) + 325
989, 75 25 9 2 4 8953, 75 25 9 2 4 8(((25+4) × (75−9)) − 8) / 2
971, 50 100 2 3 8 6727, 50 100 2 3 8 6((100−6) × 8) − (50/2)3
937, 50 25 6 9 7 8827, 50 25 6 9 7 8((((25 − 8) × 6) + 9) × 7) + 50
759, 25 50 8 6 5 4743, 25 50 8 6 5 4((50−6) × (25−8)) − 54
902, 50 75 9 6 4 5577, 50 75 9 6 4 5((9+4) × (50−6)) + 575
641, 25 75 4 3 7 6851, 25 75 4 3 7 6(((75 × 3) − 6) × 4) − 257

Once again, there are six unused numbers, so we can plug them into the blanks at the bottom to get ((75×25) + (3×4))×75 − 7, which evaluates to 141518. This can be split into the small two-digit numbers 14, 15, 18, which we convert to letters alphanumerically to get NOR.