For each of the 9 parts, we deduce the sizes of the parties, and use D'Hondt allocation to allocate lanterns to them. (D'Hondt is older Dutch for "The Hound")
Each lantern amount is a single digit; we thus get 9 5-digit sequences.
Afghan Hound
Under [1] and [2], the possibilities for lowest and highest panels are: 20 and 90 or 22 and 99. If it were 22 and 99, that plus 55 [6] would equal 176; the remaining 57 panels must consist of a perfect square [4] and a prime number [7], which is impossible without violating condition [1]. Therefore, the lowest number of panels is 20, and the highest is 90.
One knitter made 55 panels [6]; the remaining two knitters made a perfect square [4] and a prime number [7]. These have to be 25 and 43 to bring the total to 233 without violating condition [1].
Ember didn't make the striped or argyle panels [7] or the polka-dot or checkerboard panels [3], so they made the harlequin, and there are 20 of them, as deduced above.
By condition [5], then, Casey made 25, Drew made 43, and Alex made 55. Alex therefore made the polka-dot panels, and Drew made the checkerboard ones [3 and 6].
By condition [4], Casey did not make the striped panels; they made the argyle panels, and Blake made the striped ones.
| Knitter: | A | B | C | D | E |
|---|---|---|---|---|---|
| Pattern: | polka-dot | stripe | argyle | checkerboard | harlequin |
| Number of panels knitted: | 55 | 90 | 25 | 43 | 20 |
| Allocation of 19 lanterns: | 5 | 8 | 2 | 3 | 1 |
Dachshund
Determine how many bags of buns they have to buy, and how many buns are left over after each week's sausage fest.
Each week, they get p packs of s sausages and b bags of 16 buns, so they have 16b - ps buns remaining.
(You can also think of this as 16 minus the remainder of ps/16.)
Andouille: 3 packs of 19 sausages = 57 sausages. 57/16 = 3.5625, so they have to buy 4 bags of buns. 4*16 = 64 total buns. 64 - 57 = 7 leftover buns.
Bratwurst: 7*11 = 77. 77/16 = 4.8125. (5*16) - 77 = 3.
Chorizo: 7*9 = 63. 63/16 = 3.9375. (4*16) - 63 = 1.
Italian sausage: 6*10 = 60. 60/16 = 3.75. (4*16) - 60 = 4.
Kielbasa: 5*13 = 65. 65/16 = 4.0625. (5*16) - 65 = 15.
| Type of Sausage: | And | Brat | Chor | Italian | Kielbasa |
|---|---|---|---|---|---|
| Leftover buns: | 7 | 3 | 1 | 4 | 15 |
| Allocation of 12 lanterns: | 3 | 1 | 0 | 1 | 7 |
Greyhound
All the shades of gray shown are achromatic (i.e., R=G=B); the number to input is that common value:
| RGB value: | 85 | 54 | 240 | 160 | 130 |
| Allocation of 19 lanterns: | 2 | 1 | 7 | 5 | 4 |
Irish Wolfhound
Take the mean radius of the largest moon of each planet listed, multiply by the factor shown, and round:
| Moon | Multiplier | Radius | Result | Rounded |
|---|---|---|---|---|
| Triton | 1 | 1353.4 | 1353.4 | 1353 |
| Titania | 4 | 788.4 | 3153.6 | 3154 |
| Titan | 2 | 2574.73 | 5149.46 | 5149 |
| Ganymede | 1 | 2634.1 | 2634.1 | 2634 |
| Phobos | 80 | 11.267 | 901.36 | 901 |
| Allocation of 16 lanterns: | 1 | 4 | 7 | 3 | 1 |
Otter Hound
All the statistics referred to are for the noted Giants slugger Mel Ott. (Source)
| Stat | Year | Number |
|---|---|---|
| At Bats | 1926 | 60 |
| Runs Batted In | 1934 | 135 |
| Walks (Bases on Balls) | 1942 | 109 |
| Home Runs | 1929 | 42 |
| Runs | 1941 | 89 |
| Allocation of 20 lanterns: | 3 | 6 | 5 | 2 | 4 |
Pharaoh Hound
Solving the cryptic gives the party sizes in the unclued shaded entries:
Across
- A[narchi]C + DC
- A + CE + D
- A[d]A + A[v]A
- P[ursue] + R[outine] + I[nvestigation] + OR
- hidden in bED I THink
- LEEWARD INN anagram
- &lit: W in (DO + SE[arching])
- R[oulad]E + reversed WAS
Down
- reversed WE + FA
- CO + DER
- A NUDE anagram
- C + RAT
- [c]ACTI
- reversed I + YAM
- A + GREET + O
- AN(TEN)NA
- O(PEN)ED
- SHAD + OW
- OIL SO anagram
- reversed E + V + LED
- E[njoy] + WWW
- ER + SE
- VEE[p] + R
- reversed SEAN
| 1A | 2C | 3D | 4C | ||||||||||||||||
| 5A | F | O | U | R | 6M | ||||||||||||||
| 7A | C | E | D | 8A | A | A | 9A | ||||||||||||
| 10O | G | T | W | E | N | T | Y | N | 11S | ||||||||||
| 12P | R | I | 13O | R | 14E | 15D | I | T | H | ||||||||||
| E | E | 16E | L | 17E | 18V | E | 19N | E | A | ||||||||||
| 20N | E | W | I | R | E | L | A | N | D | ||||||||||
| E | T | W | O | S | E | V | E | N | O | ||||||||||
| 21D | O | W | S | E | 22R | E | S | A | W |
| Allocation of 13 lanterns: | 1 | 7 | 3 | 0 | 2 |
Plott Hound
Plotting each of these on the interval [0, 1] yields a figure which becomes a digit when you flip the axes:
| (4t + ln (1 - t^100 + e^-4)) / 8 | 7 |
| (1/2) * (1 - cos(4pi * t))^(1/4) | 3 |
| sqrt(|4t^3 - 6t^2 + 2t|) | 3 |
| 25(1 - 98t/99 - |1 - 100t/99|) | 7 |
| cos^2 t - (cos 2t) / 2 | 1 |
| Allocation of 12 lanterns: | 4 | 2 | 2 | 4 | 0 |
Shikoku
Solving the puzzle as a Sudoku using all numbers and a Shikaku using only the highlighted numbers gives:
| 1 | 8 | 5 | 6 | 3 | 2 | 4 | 7 | 9 |
| 7 | 9 | 3 | 1 | 4 | 8 | 2 | 6 | 5 |
| 2 | 6 | 4 | 7 | 5 | 9 | 1 | 3 | 8 |
| 4 | 2 | 6 | 5 | 8 | 7 | 3 | 9 | 1 |
| 9 | 7 | 8 | 3 | 1 | 6 | 5 | 2 | 4 |
| 5 | 3 | 1 | 9 | 2 | 4 | 6 | 8 | 7 |
| 6 | 1 | 7 | 2 | 9 | 5 | 8 | 4 | 3 |
| 8 | 5 | 9 | 4 | 6 | 3 | 7 | 1 | 2 |
| 3 | 4 | 2 | 8 | 7 | 1 | 9 | 5 | 6 |
The party sizes are given by the 5-digit numbers in the size-5 areas
| Allocation of 29 lanterns: | 2 | 7 | 9 | 3 | 8 |
Trigg Hound
Taking the tan of each number (in radians) gives (to the precision possible) an integer:
| Value | Tangent |
|---|---|
| 1.53235 | 26 |
| 1.53082 | 25 |
| 1.54699 | 42 |
| 1.47113 | 10 |
| 1.54858 | 45 |
| Allocation of 25 lanterns: | 4 | 4 | 8 | 1 | 8 |
The Grand Finale
Each 5-digit sequence is the zip code for a US location ending in -ville. These start with the letters A through I, giving an ordering.
Allocating 99 lanterns gives them numbers of lanterns as shown:
| ZIP Code | -Ville | Lanterns | Letter |
|---|---|---|---|
| 17302 | Airville, PA | 6 | F |
| 44818 | Bloomville, OH | 15 | O |
| 36524 | Coffeeville, AL | 12 | L |
| 31017 | Danville, GA | 11 | K |
| 14731 | Ellicottville, NY | 5 | E |
| 58231 | Fordville, ND | 20 | T |
| 27938 | Gatesville, NC | 9 | I |
| 42240 | Hopkinsville, KY | 14 | N |
| 21754 | Ijamsville, MD | 7 | G |
Translating the numbers into letters gives the answer, FOLKETING (the Danish Parliament, which uses D'Hondt allocation in part to allocate seats to parties).