Question: The Erdös number of C at the end of the conference was:
Explanation:
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 fifth day,
Among the eight mathematicians, F has the smallest Erdös number.
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.
Now, we can solve all the questions.
From the above explanation, C’s Erdös number was f + 1 = 2 on the third day and thereafter.
Hence, option (b).