Instructions
Mathematicians are assigned a number called the Erdős number (named after the famous mathematician Paul Erdős).
- Only Paul Erdős himself has an Erdős number of 0.
- Any mathematician who has written a research paper with Erdős has an Erdős number of 1.
- For other mathematicians:
- Suppose mathematician X has co-authored papers with several others. Among them, mathematician Y has the smallest Erdős number (y). Then, X‘s Erdős number is y + 1.
- If no co-authorship chain connects to Erdős, the Erdős number is infinity (∞).
Conference Scenario:
A seven-day mini-conference was organized in memory of Paul Erdős, with eight mathematicians: A, B, C, D, E, F, G, H.
Initial Conditions:
- At the start of the conference:
- A was the only participant with an infinite (∞) Erdős number.
- Nobody had an Erdős number less than F’s (i.e., F had the smallest).
Key Events:
- Day 3:
- F co-authored a paper with A and C.
- This reduced the average Erdős number of the group to 3.
- The Erdős numbers of B, D, E, G, H remained unchanged.
- No other co-authorship among any three members could have reduced the average to 3.
- At the end of Day 3:
- Five members had identical Erdős numbers.
- The other three had distinct Erdős numbers.
- Day 5:
- E co-authored a paper with F.
- This reduced the group’s average Erdős number by 0.5.
- The Erdős numbers of the other six remained unchanged.
- No other papers were written during the conference.
Easy Questions
- Who had an Erdös number of infinity at the beginning of the conference?
A) F B) A C) C D) H - What can be said about the Erdös number of F at the start of the conference?
A) It was the highest B) It was infinite C) It was the lowest among all 8 participants D) It was equal to that of E - If F’s Erdös number is denoted by y, then after co-authoring a paper with F, what will be A’s new Erdös number?
A) y B) y + 1 C) y + 2 D) Cannot be determined - What was the total sum of Erdös numbers of all 8 participants at the end of Day 3?
A) 20 B) 24 C) 26 D) 28 - At the end of Day 3, how many participants had the same Erdös number?
A) 3 B) 4 C) 5 D) 6 - Who were the three participants whose Erdös numbers remained distinct from the rest after Day 3?
A) A, B, C B) E, F, H C) C, D, E D) F, G, H
Solutions
Answer B: Only A had ∞ at the beginning as per the given instruction.
Answer C: F had the lowest Erdös number among all 8, as explicitly stated.
Answer B: Co-authoring with F (Erdös number y) gives A an Erdös number of y + 1.
Answer B: After Day 3, total average was 3, so total = 3 × 8 = 24.
Answer C: Five participants (A, B, C, D, G) shared the same Erdös number y + 1.
Answer B: The three with distinct Erdös numbers were E (6), F (1), and H (7).
CAT Questions
11. How many participants did not change their Erdős number during the conference?
- (1) 2
- (2) 3
- (3) 4
- (4) 5
- (5) Cannot be determined
12. The person with the largest Erdős number at the end must have had:
- (1) 5
- (2) 7
- (3) 9
- (4) 14
- (5) 15
13. How many participants had the same Erdős number at the beginning?
- (1) 2
- (2) 3
- (3) 4
- (4) 5
- (5) Cannot be determined
14. The Erdős number of C at the end was:
- (1) 1
- (2) 2
- (3) 3
- (4) 4
- (5) 5
15. The Erdős number of E at the beginning was:
- (1) 2
- (2) 5
- (3) 6
- (4) 7
- (5) 8
Step-by-Step Solution
Step 1: Assign Initial Variables
- Let F’s initial Erdős number = y (smallest in the group).
- Since A starts with ∞, but after co-authoring with F, A’s number becomes y + 1.
- C also co-authors with F, so if C had a higher number, it reduces to y + 1.
Step 2: After Day 3 (F collaborates with A & C)
- A: ∞ → y + 1
- C: ∞ → y + 1 (assuming C had ∞ initially)
- F: Remains y
- B, D, E, G, H: Unchanged
- Average = 3 → Total sum = 8 × 3 = 24
Equation:
[ (y+1) + b + (y+1) + d + e + y + g + h = 24 ]
Simplify:
[ 3y + b + d + e + g + h + 2 = 24 ]
[ 3y + b + d + e + g + h = 22 ]
Step 3: Five Identical Numbers at End of Day 3
- Likely A, C, and three others have y + 1.
- Suppose B, D, G also have y + 1.
- Then:
[ 3y + (y+1) + (y+1) + e + (y+1) + h = 22 ]
[ 6y + e + h + 3 = 22 ]
[ 6y + e + h = 19 \quad \text{(Equation 1)} ]
Step 4: Day 5 (E collaborates with F)
- E’s number changes from e → y + 1.
- Average drops by 0.5 → New average = 2.5 → New sum = 20.
- Reduction in sum = 4 (since 24 → 20).
- So:
[ e – (y + 1) = 4 ]
[ e = y + 5 \quad \text{(Equation 2)} ]
Step 5: Solve Equations
From Equation 1:
[ 6y + (y + 5) + h = 19 ]
[ 7y + h = 14 ]
Possible integer solutions (since y ≥ 1, and h ≥ y):
- y = 1, h = 7 (only valid solution).
Final Assignments:
- F: y = 1
- A, C, B, D, G: y + 1 = 2
- E: y + 5 = 6 → Later becomes 2 (after Day 5).
- H: 7 (unchanged).
Final Answer Table
| Participant | Initial Erdős No. | Change Event | Final Erdős No. | Changed? |
|---|---|---|---|---|
| A | ∞ | Co-authored with F (Day 3) | 2 | ✅ Yes |
| B | 2 | No change | 2 | ❌ No |
| C | ∞ | Co-authored with F (Day 3) | 2 | ✅ Yes |
| D | 2 | No change | 2 | ❌ No |
| E | 6 | Co-authored with F (Day 5) | 2 | ✅ Yes |
| F | 1 | None | 1 | ❌ No |
| G | 2 | No change | 2 | ❌ No |
| H | 7 | None | 7 | ❌ No |
Answers to CAT Questions
11. 5 participants did not change (B, D, F, G, H) → (4) 5
12. Largest final number = 7 (H) → (2) 7
13. 3 had same initial number (B, D, G = 2) → (2) 3
14. C’s final number = 2 → (2) 2
15. E’s initial number = 6 → (3) 6
