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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.
The person having the largest Erdos number at the end of the conference must have had Erdos number (at that time):
Let us consider the Erdos number of A,B,C,D,E,F,G,H be a,b,c,d,e,f,g,h where f is the min, a is infinity.
At the end of 3rd day, F co authored with A and C. Since F has min Erdos number ,the values of c,a will change to f+1 and the Erdos number of F will remain the same. [Because according to Erdos principle if a person co-authors with some one who has higher Erdos number then the Erdos number of co-authors will be min Erdos value + 1]
Average of the mathematicians is 3
Sum of the Erdos number of eight mathematicians=24
Erdos number at the third day:f+1,b,f+1,d,e,f,g,h
At the end of the fifth day, F co-authors with E thereby changing the average to 2.5 and the Erdos number of rest of the mathematicians remain unchanged.
Sum of the Erdos numbers of eight mathematicians=20
So here the difference of 4[24-20] arose, which means e will be f+5 initially and changed to f+1 after co-authoring with F.
So the Erdos number at the third day:f+1,b,f+1,d,f+5,f,g,h
At the end of the third day, five mathematicians had the same Erdos number and the rest had distinct Erdos number from each other.
It cannot be f+5 because then there will be two mathematicians with the same Erdos number f+1.
So five mathematicians will have f+1, one with f+5,one with f, one with some different value say x
5(f+1)+f+5+f+x=24
7f+x=14
The only value which satisfies the above equation is f=1,x=7
Erdos number at the end of fifth day,f+1,b,f+1,d,f+1,f,g,h
On tabulating, we get
Hence the person having the largest Erdos number at the end of the conference must have had Erdos number 7 . Hence option B.
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