Instructions

Mathematicians are assigned a number called Erdos number (named after the famous mathematician, Paul Erdos). Only Paul Erdos himself has an Erdos number of zero. Any mathematician who has written a research paper with Erdos has an Erdos number of 1.For other mathematicians, the calculation of his/her Erdos 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 Erdos number. Let the Erdos number of Y be y. Then X has an Erdos number of y+1. Hence any mathematician with no co-authorship chain connected to Erdos has an Erdos number of infinity. :

In a seven day long mini-conference organized in memory of Paul Erdos, 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 Erdos number. Nobody had an Erdos 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 Erdos number of the group of eight mathematicians to 3. The Erdos 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 Erdos 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 Erdos numbers while the other three had Erdos numbers distinct from each other.

• On the fifth day, E co-authored a paper with F which reduced the group's average Erdos number by 0.5. The Erdos numbers of the remaining six were unchanged with the writing of this paper.

• No other paper was written during the conference.

Question 35

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

Solution

Since at the end of the 3rd day 5 people had identical erdos no.(f+1) so : 5*(f+1) +f+f+5+x = 24 ; Only f=1 and x = 7 satisfies the equation. So out of 5 people who had identical erdos no. at the end of day 3, 2 of them had different nos. at the beginning. So there were 5-2 = 3 participants who had the same Erdos number at the beginning of the conference.

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