Q22
If Devon wrote more than one two-author papers, then how many two-author papers did Chintan write?
Answer: 3
Totals: 10 single, 4 two-author, 3 three-author, 2 four-author papers; author totals A 5, B 8, C 12, D 10. Every author is on both four-author papers, so for each author: singles + two-author + three-author = total - 2.
- Arman: 5 - 2 = 3 with at least one of each type, so 1 single, 1 two-author, 1 three-author.
- Three-author papers: 3 papers give 9 slots, each author in at most 3. Arman has 1, so B + C + D = 8. Since C and D each exceed B, B = 2 and C = D = 3. So C and D are on all three: the papers are A-C-D, B-C-D, B-C-D.
- Singles: four distinct numbers, each at least 1, adding to 10, so they are 1, 2, 3, 4. Arman has 1.
- Brajen: 8 - 2 - 2 = 4 = singles + two-author, and these are equal, so 2 and 2.
- Chintan and Devon have singles 3 and 4 in some order. Chintan: singles + two-author = 12 - 2 - 3 = 7; Devon: 10 - 2 - 3 = 5.
| Author | Single | Two-author | Three-author | Four-author | Total |
|---|---|---|---|---|---|
| Arman | 1 | 1 | 1 | 2 | 5 |
| Brajen | 2 | 2 | 2 | 2 | 8 |
| Chintan | 3 or 4 | 4 or 3 | 3 | 2 | 12 |
| Devon | 4 or 3 | 1 or 2 | 3 | 2 | 10 |
Both cases are realisable: (C single 3, two-author 4; D single 4, two-author 1) with two-author papers A-C, B-C, B-C, C-D; or (C single 4, two-author 3; D single 3, two-author 2) with A-C, B-C, B-D, C-D.
If Devon wrote more than one two-author paper, Devon has 2 two-author papers and 3 singles, so Chintan has 4 singles and \(7-4 = 3\) two-author papers.
Hence, the answer is 3.