Discussion

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 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, C’s Erdös number was f + 1 = 2 on the third day and thereafter.

Hence, option (b).

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