IPMAT Indore 2023 — LRDI

The Setup: A round-robin chess tournament features $5$ players where every player plays everyone else exactly once. A win awards $1$ point, a loss awards $0$ points, and no draws occur. We must identify which sequence of scores is mathematically impossible.

Step 1: Establish the total points in the system.
The number of players is $n = 5$. The total number of matches played in the tournament is $\binom{5}{2} = 10$.
Since each match awards exactly $1$ point to the winner and $0$ to the loser, exactly $10$ points are distributed across the 5 players. Every valid score sequence must sum to $10$.

Step 2: Apply Landau's Theorem for score sequences.
Landau's Theorem states that a sequence of $n$ scores $s_1 \le s_2 \le \dots \le s_n$ represents a valid round-robin tournament if and only if:
1. The sum of all scores equals $\binom{n}{2}$.
2. For any subset of $k$ players (sorted from lowest to highest), their combined score must be at least $\binom{k}{2}$, representing the matches they played against each other.

$$\sum_{i=1}^{k} s_i \ge \binom{k}{2} \text{ for all } k=1, 2, \dots, n$$

Step 3: Evaluate the given options against Landau's criteria.
Let's test the suspected invalid sequence $4, 4, 1, 1, 0$.
Sort the sequence in ascending order: $s = (0, 1, 1, 4, 4)$.
* Check $k=1$: $s_1 = 0 \ge \binom{1}{2} = 0$. (Passes)
* Check $k=2$: $s_1 + s_2 = 0 + 1 = 1 \ge \binom{2}{2} = 1$. (Passes)
* Check $k=3$: $s_1 + s_2 + s_3 = 0 + 1 + 1 = 2 \ge \binom{3}{2} = 3$.
Here, $2 \ngtr 3$. The lowest three players must have generated at least $3$ points purely from playing against each other, making a combined score of $2$ physically impossible.

Final Answer: 4, 4, 1, 1, 0

The Setup: A pharmaceutical company tested five drugs (A, B, C, D, E) on three organisms (P, Q, R). We must deduce the complete relationship matrix using the provided table and logical constraints.

Step 1: Establish the initial matrix and apply direct implication rules.
From the provided table, A works on R, B works on Q, and C works on Q.
Rule 3 states that if A works on an organism, B also works on it ($A \implies B$). Since A works on R, B must also work on R.
Rule 4 states that if C works on an organism, D also works on it ($C \implies D$). Since C works on Q, D must also work on Q.

Step 2: Apply column capacity constraints.
Rule 2 states each organism can be treated with at most 3 drugs.
Currently, Organism Q is treated by B, C, and D, reaching its absolute maximum capacity. Therefore, neither A nor E can work on Q.
Since A is restricted from Q and must act as a strict subset of B's placements (which are only Q and R), A must solely work on R.

Step 3: Apply disjoint rules to isolate P and R.
Rule 2 dictates Organism P needs at least 2 drugs. The only available candidates for P are C, D, and E (A and B are locked into Q and R).
Rule 5 states D and E cannot work on the same organism. Thus, P cannot be treated by both D and E.
To secure 2 drugs for P without pairing D and E, P must logically be treated by C and D (choosing C automatically forces D via Rule 4).

Step 4: Finalize Drug E's placement.
Drug E cannot work on Q (capacity full) and cannot work on P (disjoint with D).
Rule 1 mandates each drug works on at least 1 organism. Thus, E is forced to work on Organism R.
The finalized matrix is: P={C,D}, Q={B,C,D}, R={A,B,E}. Drug E strictly works only on organism R.

Final Answer: Only R

The Setup: Using the fully deduced matrix from the pharmaceutical constraints, we need to determine the exact organisms treated by Drug D.

Step 1: Reference the derived matrix.
As mathematically established by the constraint deduction: P={C,D}, Q={B,C,D}, R={A,B,E}.

Step 2: Isolate Drug D.
Evaluating the assignments, Drug D successfully operates on both Organism P and Organism Q.

Final Answer: P & Q

The Setup: We need to identify which organisms are treated by exactly three of the five drugs, relying on the completely deduced matrix.

Step 1: Count the drugs assigned to each organism.
According to the derived matrix:
Organism P is treated by C and D (Total: 2).
Organism Q is treated by B, C, and D (Total: 3).
Organism R is treated by A, B, and E (Total: 3).

Step 2: Select the qualifying organisms.
Both Q and R meet the criteria of being treated by exactly three drugs.

Final Answer: Q and R

The Setup: We must list the specific combination of drugs that successfully treat Organism R based on the logical constraints.

Step 1: Reference Organism R in the derived matrix.
As established in the base matrix deduction: P={C,D}, Q={B,C,D}, R={A,B,E}.

Step 2: Isolate Organism R.
The column for Organism R strictly contains treatments from Drugs A, B, and E.

Final Answer: A, B and E

The Setup: We must list the specific combination of drugs that successfully treat Organism P based on the logical constraints.

Step 1: Reference Organism P in the derived matrix.
As established in the base matrix deduction: P={C,D}, Q={B,C,D}, R={A,B,E}.

Step 2: Isolate Organism P.
To satisfy the minimum 2-drug requirement while avoiding the D/E conflict constraint, Organism P is treated exclusively by Drugs C and D.

Final Answer: Only C and D