Erdös Numbers at Mini-Conference

CAT 2006 Slot 1 · DILR · Hard · Data Interpretation

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Passage / data set

Mathematicians are assigned a number called Erdös number. Only Paul Erdös himself has an Erdös number of zero. Any mathematician who has written a research paper with Erdös has an Erdös number of 1. For other mathematicians, if X co-authored papers with several mathematicians and Y has the smallest Erdös number yy among them, then X has Erdös number y+1y+1. Any mathematician with no co-authorship chain connected to Erdös has an Erdös number of infinity.

In a seven-day mini-conference, eight mathematicians A, B, C, D, E, F, G, and H discussed research problems. At the beginning:

  • A was the only participant with an infinite Erdös number.
  • Nobody had an Erdös number less than that of F.

Events:

  1. On day 3, F co-authored a paper jointly with A and C. This reduced the average Erdös number of the 8 mathematicians to 3. The Erdös numbers of B, D, E, G, and H remained unchanged. No other co-authorship among any three members would have reduced the average Erdös number to as low as 3.
  2. At the end of day 3, five members of this group had identical Erdös numbers while the other three had Erdös numbers distinct from each other.
  3. On day 5, E co-authored a paper with F which reduced the group's average Erdös number by 0.5. The Erdös numbers of the remaining six were unchanged.
  4. No other paper was written during the conference.

Question 1 of 5

How many participants in the conference did not change their Erdös number during the conference?

  1. A.

    2

  2. B.

    3

  3. C.

    4

  4. D.

    5

  5. E.

    Cannot be determined

Answer

Option E — Cannot be determined

Explanation

On Day 3, only A and C changed their Erdös numbers. On Day 5, only E changed its Erdös number. The remaining 5 participants (B, D, F, G, H) did not change their Erdös numbers throughout the conference.

Question 2 of 5

The person having the largest Erdös number at the end of the conference must have had Erdös number (at that time):

  1. A.

    5

  2. B.

    7

  3. C.

    9

  4. D.

    14

  5. E.

    15

Answer

7

Explanation

At the end of day 3, Erdös number of F = 1. A, C became 1+1=21+1=2. 5 people had identical Erdös numbers = 2 (A, C and 3 others among B, D, G, H). Let F=1, six people have Erdös number 2 (A, C, E became 2 on day 5, and three others), total sum = 1+6×2+x=201 + 6 \times 2 + x = 20. 13+x=20  ⟹  x=713 + x = 20 \implies x = 7. So the 8th person had Erdös number 7.

Question 3 of 5

How many participants had the same Erdös number at the beginning of the conference?

  1. A.

    2

  2. B.

    3

  3. C.

    4

  4. D.

    5

  5. E.

    Cannot be determined

Answer

Option C — 4

Explanation

At the end of day 3, 5 people had the Erdös number 2. A and C changed to 2 on day 3. Thus, 3 people already had the Erdös number 2 at the beginning of the conference.

Question 4 of 5

The Erdös number of C at the end of the conference was:

  1. A.

    1

  2. B.

    2

  3. C.

    3

  4. D.

    4

  5. E.

    5

Answer

Option B — 2

Explanation

F had Erdös number 1. When C co-authored with F, C's Erdös number became 1+1=21 + 1 = 2.

Question 5 of 5

The Erdös number of E at the beginning of the conference was:

  1. A.

    2

  2. B.

    5

  3. C.

    6

  4. D.

    7

  5. E.

    8

Answer

6

Explanation

On day 5, E co-authored with F (whose Erdös number is 1), so E's Erdös number became 2. This reduced the sum of Erdös numbers by 0.5×8=40.5 \times 8 = 4. Thus, E's initial Erdös number was 2+4=62 + 4 = 6.

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