by Greg Pliska and Guy Jacobson

Solvers will naturally begin by identifying all the famous people shown here.

This should be duck soup; most of these should be familiar, and image search should help to resolve any that they don’t know.

Solvers should next observe that the number of digits in the large number labeling an image (including the ?) matches the length of that person’s full name. They can then extract the letter from the name in the position corresponding to the ? in the digit sequence (people are listed in row-major order):

# | Person | Letter |
---|---|---|

1 | James Earl Jones | R |

2 | Blair Underwood | O |

3 | Franklin Pierce | L |

4 | Bruce Campbell | L |

5 | Maria Tallchief | A |

6 | Nancy Lopez | P |

7 | Walt Disney | A |

8 | Isaac Hayes | S |

9 | Mary-Kate and Ashley Olsen | S |

10 | Paula Poundstone | P |

11 | Ethel Rosenberg | H |

12 | Cindy Crawford | R |

13 | Carl Friedrich Gauss | A |

14 | Jesse Owens | S |

15 | Frederic Chopin | E |

So solvers get the message ROLL A PASSPHRASE. What does this mean? This is a clue that the puzzle uses Diceware, a system for generating passphrases by rolling dice. For each desired word/component of your passphrase, you roll five ordinary d6 dice, obtaining a five-digit number, and then look up this number in a provided list of 6^{5} = 7776 unique words. This puzzle uses the original Diceware word list; also see the Diceware Passphrase home page for more detailed information.

Once they find the Diceware word list, solvers should observe that all the famous names are in fact constructable out of words on this list (thanks to the inclusion of many names and name-y bits). The word “factors” in the flavortext, plus the use of the capital Greek letter Π in the title/flavortext suggest that multiplication may be involved (in mathematics, Π is used to denote a product, just as capital Σ denotes a sum). If you break the names into their Diceware-word pieces and then multiply all the associated five-digit numbers, you in fact get the very numbers that label each image, but with the ?s replaced by actual digits:

Diceware words | Diceware rolls | Product | Digits |
---|---|---|---|

JAMES × EARL × JONES | 34346 × 24121 × 34561 | 28632401428826 | 1 |

BLAIR × UNDER × WOOD | 14266 × 62352 × 64354 | 57243760273728 | 2 |

FRANK × LIN × PIERCE | 26315 × 36463 × 45524 | 43681363519780 | 3 |

BRUCE × CAMP × BELL | 15264 × 15662 × 13624 | 3257018399232 | 3 |

MARIA × TALL × CHIEF | 41543 × 56412 × 16451 | 38553308651916 | 3 |

NANCY × LOPEZ | 43315 × 41122 | 1781199430 | 4 |

WALT × DISNEY | 63431 × 23166 | 1469442546 | 4 |

ISAAC × HAYES | 34234 × 32436 | 1110414024 | 1 |

MARY × KATE × AND × ASHLEY × OLSEN | 41566 × 35214 × 12223 × 12534 × 44332 | 9941191043180804791776 | 7 |

PAULA × POUND × STONE | 45223 × 46313 × 55544 | 116332064507656 | 2 |

ETHEL × ROSEN × BERG | 24615 × 52263 × 13655 | 17566525887975 | 5 |

CINDY × CRAW × FORD | 16565 × 21553 × 26224 | 9362635269680 | 8 |

CARL × FRIED × RICH × GAUSS | 16154 × 26346 × 51642 × 26644 | 585594844192307232 | 7 |

JESSE × OWENS | 34441 × 44545 | 1534174345 | 5 |

FRED × ERIC × CHOPIN | 26325 × 24526 × 16524 | 10668670201800 | 8 |

Solvers now need to figure out what to do with this string of digits: 123334417258758.

It certainly looks like it could be the numeric label for another Diceware-word decomposed name, and in fact it is.

Solvers need to factor this number (“factor” is in the flavortext) to obtain: 2 × 3^{3} × 19 × 1193 × 2917 × 34543.

Now, 34543 is the Diceware number that maps to JOHN, which is very promising. But aren’t there multiple ways to multiply the other factors to get pairs of numbers? Yes, but there’s only one way to get three five-digit numbers whose digits all in the range 1-6, as required by Diceware (see the appendix).

This unique factorization is 34543 × 55423 × 64422 (listing the factors in ascending order), leading to JOHN × STALL × WORTH. So the puzzle’s answer is **JOHN STALLWORTH**.

By logic. First, factor the number into a product of primes. You can just type it into Wolfram Alpha to learn that:

123334417258758 = 2 × 3^{3} × 19 × 1193 × 2917 × 34543

Note that 34543 is a possible Diceware number. Now, 2 × 34543 = 69086, which is not a Diceware number, and no other multiple of 34543 is only five digits. So 34543 is necessarily one of the Diceware numbers in any factorization. Also, 2917 must be a factor of another one of the Diceware numbers, which all lie between 11111 and 66666. Dividing these numbers by 2917, we learn that the factor divisible by 2917 must lie between 3.8 and 22.8 times that number. Given the other factors, the other possible multiplicands in that range are 6, 9, 18, and 19. Multiplying these out, 6 × 2917 = 17502, 9 × 2917 = **26253**, 18 × 2917 = 52506, and 19 × 2917 = **55423**. Of these, the two bolded numbers are Diceware numbers. Dividing the original number by 26253 × 34543 yields 136002, which is too long, but dividing by 55423 × 34543 yields 64422, which *is* a Diceware number, proving that the factorization into Diceware numbers is indeed unique.

Or, if you’d rather write code than think, here’s a brute-force Python 3 program to verify uniqueness:

import re rolls = [i for i in range(11111, 66667) if re.match('^[1-6]*$', str(i))] def dicefactor(l, i, n): if n == 1: print(' x '.join([str(x) for x in l])) else: for j in range(i, len(rolls)): rj = rolls[j] if n % rj == 0: dicefactor(l + [rj], j, n // rj) dicefactor([], 0, 123334417258758)

Its output is: