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.
Under  and , the possibilities for lowest and highest panels are: 20 and 90 or 22 and 99. If it were 22 and 99, that plus 55  would equal 176; the remaining 57 panels must consist of a perfect square  and a prime number , which is impossible without violating condition . Therefore, the lowest number of panels is 20, and the highest is 90.
One knitter made 55 panels ; the remaining two knitters made a perfect square  and a prime number . These have to be 25 and 43 to bring the total to 233 without violating condition .
Ember didn't make the striped or argyle panels  or the polka-dot or checkerboard panels , so they made the harlequin, and there are 20 of them, as deduced above.
By condition , 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 , Casey did not make the striped panels; they made the argyle panels, and Blake made the striped ones.
|Number of panels knitted:||55||90||25||43||20|
|Allocation of 19 lanterns:||5||8||2||3||1|
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|
|Allocation of 12 lanterns:||3||1||0||1||7|
All the shades of gray shown are achromatic (i.e., R=G=B); the number to input is that common value:
|Allocation of 19 lanterns:||2||1||7||5||4|
Take the mean radius of the largest moon of each planet listed, multiply by the factor shown, and round:
|Allocation of 16 lanterns:||1||4||7||3||1|
All the statistics referred to are for the noted Giants slugger Mel Ott. (Source)
|Runs Batted In||1934||135|
|Walks (Bases on Balls)||1942||109|
|Allocation of 20 lanterns:||3||6||5||2||4|
Solving the cryptic gives the party sizes in the unclued shaded entries:
- 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
- reversed WE + FA
- CO + DER
- A NUDE anagram
- C + RAT
- reversed I + YAM
- A + GREET + O
- SHAD + OW
- OIL SO anagram
- reversed E + V + LED
- E[njoy] + WWW
- ER + SE
- VEE[p] + R
- reversed SEAN
|Allocation of 13 lanterns:||1||7||3||0||2|
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|
Solving the puzzle as a Sudoku using all numbers and a Shikaku using only the highlighted numbers gives:
The party sizes are given by the 5-digit numbers in the size-5 areas
|Allocation of 29 lanterns:||2||7||9||3||8|
Taking the tan of each number (in radians) gives (to the precision possible) an integer:
|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:
Translating the numbers into letters gives the answer, FOLKETING (the Danish Parliament, which uses D'Hondt allocation in part to allocate seats to parties).