CAT 2025 Lesson : Numerical Reasoning - Case 3

Read the following and answer the questions that follow

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.

1) 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.

2) 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.

3) 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.

4) No other paper was written during the conference.
[CAT 2006]

5) How many participants in the conference did not change their Erdös number during the conference?

(1) 2
(2) 3
(3) 4
(4) 5
(5) Cannot be determined

6) The person having the largest Erdös number at the end of the conference must have had Erdös number (at that time):

(1) 5
(2) 7
(3) 9
(4) 14
(5) 15

7) How many participants had the same Erdös number at the beginning of the conference?

(1) 2
(2) 3
(3) 4
(4) 5
(5) Cannot be determined

8) The Erdös number of C at the end of the conference was:

(1) 1
(2) 2
(3) 3
(4) 4
(5) 5

9) The Erdös number of E at the beginning of the conference was:

(1) 2
(2) 5
(3) 6
(4) 7
(5) 8

Solution

If a mathematician (X) co-authored a paper with another mathematician (Y), his Erdös number will be 1 more than the mathematician (Y) ⇒ X = Y + 1. Thus if mathematician (X) co-authors with several authors, then his Erdös number is calculated by taking the mathematician with smallest Erdös number.

For example, if a mathematician (A) with an infinite Erdös number co-authors papers with B, C, D and E with number 7, 5, 3 and 8. Then A’s Erdös number will be 3 + 1 = 4.

It is given that A had infinite Erdös number and F had the least Erdös number.

From point 1, F co-authored a paper jointly with A and C. Since F has the least Erdös number, the Erdös number of A and C will become F+1. Thus, A and C have same Erdös number.

From point 2, 5 members have same Erdös number. Then A, C and three other mathematicians have the same Erdös number, which will be F+1.
F and the remaining 2 mathematicians have distinct Erdös number.

From point 1, A + B + C + D + E + F + G + H =$8 \times 3$ = 24,
Let A, B, C, D, G be the members with same Erdös number F+1.
⇒ 5F + 5 + F + E + H = 24

F will have the least value, so substituting 1
11 + E + H = 24
⇒ E + H = 13 (This is possible)

Let us substitute F = 2
17 + E + H = 24
⇒ E + H = 7 (This is not possible as E and H need to have distinct Erdös numbers with a value more than 2, as F has the least Erdös number.

Then, F = 1 is the only possible scenario.


From point 3, A + B + C + D + E + F + G + H = $8 \times 2.5$ = 20,
E’s Erdös number will be 2 on day 5, as he co-authored paper with F.
We know F = 1 and A, C, E = 2 and 3 others have 2 and H has a distinct Erdös number.
$6 \times 2$ + 1 + H = 20
H = 20 - 13 = 7. Then E = 6 from beginning till the end of day 4.


5)
From the above table only A, C and E have changed their Erdös number. Therefore, 5 participants have not changed their Erdös number.

Answer: (4) 5

6)
From the table, 7 is the largest Erdös umber at the end of the conference.

Answer: (2) 7

7)
In the table, only one person will have an Erdös number of 7. As there are only 3 distinct Erdös numbers at the end of day 3, then 2 Erdös numbers will be given to the remaining 3 participants. Therefore 3 participants have same Erdös number at the beginning.

Answer: (2) 3

8)
From the table, C has Erdös number of 2 at the end of conference.

Answer: (2) 2

9)
From the table C has Erdös number of 6 at the beginning of conference.

Answer: (3) 6

Answer:
5) (4) 5
6) (2) 7
7) (2) 3
8) (2) 2
9) (3) 6