On how many days did Alden and Brim have the same number of new cases?
Answer and solution
Correct answer: A
Solution
Answer: A — 5
1. The daily city total is at least 5 on Day 1: each district has at least 1 and Dune must exceed Cova. Any daily total is at most 11: three districts can have 3 each and Cova at most 2.
2. The five strictly increasing integer city totals sum to 12+12+5+14=43. With Day 3=Day 2+1, the only possible sequence between 5 and 11 is 5,8,9,10,11. This can be checked by listing increasing five-number subsets of 5,…,11.
3. Day 1 total 5 forces Alden=Brim=Cova=1 and Dune=2. Dune total 14 then forces 3 on each later day.
4. Alden has 11 remaining cases over four days. On Day 2 it can have at most 2, because only Dune can have 3 then. The maximum possible remaining sum is 2+3+3+3=11, so all these maxima are forced. The same argument applies to Brim.
5. Subtract Alden, Brim and Dune from daily totals to obtain Cova=1,1,0,1,2. This has one maximum of 2, total 5, and is below Dune on every day.
6. Final grid (Alden, Brim, Cova, Dune): Day 1 (1,1,1,2); Day 2 (2,2,1,3); Day 3 (3,3,0,3); Day 4 (3,3,1,3); Day 5 (3,3,2,3). Independent exhaustive enumeration of all 0–3 district sequences gives exactly one valid grid.
7. Alden and Brim match on all five days. Answer A (5).
Verification: 7. Alden and Brim match on all five days. Answer A (5).
1. The daily city total is at least 5 on Day 1: each district has at least 1 and Dune must exceed Cova. Any daily total is at most 11: three districts can have 3 each and Cova at most 2.
2. The five strictly increasing integer city totals sum to 12+12+5+14=43. With Day 3=Day 2+1, the only possible sequence between 5 and 11 is 5,8,9,10,11. This can be checked by listing increasing five-number subsets of 5,…,11.
3. Day 1 total 5 forces Alden=Brim=Cova=1 and Dune=2. Dune total 14 then forces 3 on each later day.
4. Alden has 11 remaining cases over four days. On Day 2 it can have at most 2, because only Dune can have 3 then. The maximum possible remaining sum is 2+3+3+3=11, so all these maxima are forced. The same argument applies to Brim.
5. Subtract Alden, Brim and Dune from daily totals to obtain Cova=1,1,0,1,2. This has one maximum of 2, total 5, and is below Dune on every day.
6. Final grid (Alden, Brim, Cova, Dune): Day 1 (1,1,1,2); Day 2 (2,2,1,3); Day 3 (3,3,0,3); Day 4 (3,3,1,3); Day 5 (3,3,2,3). Independent exhaustive enumeration of all 0–3 district sequences gives exactly one valid grid.
7. Alden and Brim match on all five days. Answer A (5).
G Strategy shortcut
Method: Use total 43 and daily bounds 5–11 to force 5,8,9,10,11. Day 1 is 1,1,1,2; Alden/Brim each need 11 more, forcing 2,3,3,3. Obtain Cova by subtraction.Verification: 7. Alden and Brim match on all five days. Answer A (5).
G Strategy: Method: Use total 43 and daily bounds 5–11 to force 5,8,9,10,11. Day 1 is 1,1,1,2; Alden/Brim each need 11 more, forcing 2,3,3,3. Obtain Cova by subtraction. Verification: 7. Alden and Brim match on all five days. Answer A (5).