CAT 2006 — DILR Question 7
Answer the following question based on the information given below.
Mathematicians are assigned a number called Erdös number, (named after the famous mathematician, Paul Erdös). 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, the calculation of his/her Erdös number is illustrated below:
Suppose that a mathematician X has co-authored papers with several other mathematicians. From among them, mathematician Y has the smallest Erdös number. Let the Erdös number of Y be y. Then X has an Erdös number of y + 1. Hence any mathematician with no co-authorship chain connected to Erdös has an Erdös number of infinity.
- In a seven day long mini-conference organized in memory of Paul Erdös, a close group of eight mathematicians, call them A, B, C, D, E, F, G and H, discussed some research problems. At the beginning of the conference, A was the only participant who had an infinite Erdös number. Nobody had an Erdös number less than that of F.
- On the third day of the conference F co-authored a paper jointly with A and C. This reduced the average Erdös number of the group of eight mathematicians to 3. The Erdös numbers of B, D, E, G and H remained unchanged with the writing of this paper. Further, no other co-authorship among any three members would have reduced the average Erdös number of the group of eight to as low as 3.
- At the end of the third day, five members of this group had identical Erdös numbers while the other three had Erdös numbers distinct from each other.
- On the fifth day, 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 with the writing of this paper.
- No other paper was written during the conference.
How many participants in the conference did not change their Erdös number during the conference?
Answer & solution
- A
2
- B
3
- C
4
5
- E
Cannot be determined
Let F and E have Erdös numbers f and e respectively at the beginning of the conference.
On the third day, A’s and C’s Erdös numbers become (f + 1)
The sum of Erdös numbers changed to 8 × 3 = 24
At the end of the third day, five members had identical Erdös numbers while the other three had distinct ones.
On the fifth day, E’s Erdös numbers became f + 1 and this reduced the group’s average by 0.5. This means that E’s Erdös numbers was not f + 1 on the third day.
Thus we have,
At the end of the third day, 5(f + 1) + f + e + y = 24
Hence 6f + 5 + e + y = 24
Hence 6f + e + y = 19
At the end of the fifth day,
6(f + 1) + f + y = 2.5 × 8 = 20
Hence 7f + y = 14
Among the eight mathematicians, F has the smallest Erdös number.
Let f = 2
∴ y = 0
However, only Paul Erdös himself has an Erdös number of 0. So f cannot be equal to 2. Any other value greater than 2, would render y as a negative number, which is also not possible.
So, f = 1
∴ y = 7
∴ e = 6
Now, we can solve all the questions.
From the above explanation, the Erdös numbers of B, D, G, H and F did not change during the conference.
Hence, option (d).