2021 AMC 10B Problem 17

P16-20~3 minPrint

Logic Puzzles Basic Counting Organized listing / direct enumeration Consider the largest/smallest element or boundary case

Ravon, Oscar, Aditi, Tyrone, and Kim play a card game. Each person is given $2$ cards out of a set of $10$ cards numbered $1,2,3, \dots,10.$ The score of a player is the sum of the numbers of their cards. The scores of the players are as follows: Ravon-- $11,$ Oscar-- $4,$ Aditi-- $7,$ Tyrone-- $16,$ Kim-- $17.$ Which of the following statements is true?

Show solution without answering

Key insight

Start with the smallest score. 4 can only be 1 + 3. Then 7 must be 2 + 5, and 11 must be 4 + 7.

Solution

The smallest scores can be made in the fewest ways, so start there and work up. Each card goes to only one player.

Oscar, 4. Two different cards adding to $4$ : only $1 + 3$ . Oscar has cards $1$ and $3$ .
Aditi, 7. The ways are $1 + 6$ , $2 + 5$ , $3 + 4$ . Cards $1$ and $3$ are taken, so Aditi has $2$ and $5$ .
Ravon, 11. The ways are $1 + 10$ , $2 + 9$ , $3 + 8$ , $4 + 7$ , $5 + 6$ . Cards $1, 2, 3, 5$ are taken, so Ravon has $4$ and $7$ .
Tyrone, 16. The cards left are $6, 8, 9, 10$ . Only $6 + 10 = 16$ works. Tyrone has $6$ and $10$ .
Kim, 17. The last two cards are $8$ and $9$ , and $8 + 9 = 17$ . It checks out.

Now test the statements. Aditi has $2$ and $5$ , so (B) and (D) are false. Ravon has $4$ and $7$ , so (A) is false and (C) is true. Tyrone has $6$ and $10$ , so (E) is false.

The answer is $\boxed{\textbf{(C)}\ \text{Ravon was given card 4.}}$

Why this works

Very small (or very large) totals can be made in only a few ways. Each forced choice uses up cards, which cuts down the choices for the next player. Working from the most limited player outward turns a guessing game into a chain of forced steps.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Attack the smallest score first, since it has the fewest options.

Oscar has $4$ : two distinct cards summing to $4$ must be $\{1, 3\}$ .
Aditi has $7$ : the pairs are $\{1,6\}, \{2,5\}, \{3,4\}$ . Cards $1$ and $3$ are gone, so Aditi has $\{2, 5\}$ .
Ravon has $11$ : the pairs are $\{1,10\},\{2,9\},\{3,8\},\{4,7\},\{5,6\}$ . Cards $1,2,3,5$ are gone, so Ravon has $\{4, 7\}$ .
Tyrone has $16$ from the remaining $\{6, 8, 9, 10\}$ : only $\{6, 10\}$ works ( $8 + 9 = 17$ is Kim's).
Kim has $\{8, 9\}$ , summing to $17$ . Consistent.

Ravon holds card $4$ , so statement (C) is true; the others are false (Aditi has $2$ and $5$ , Tyrone has $6$ and $10$ ).

The answer is $\boxed{\textbf{(C)}\ \text{Ravon was given card 4.}}$

Review in the AoPS books

Intermediate Counting & Probability, Chapter 2: Sets and Logic (§2.4)
Introduction to Counting & Probability, Chapter 1: Counting Is Arithmetic
Prealgebra, Chapter 15: Problem-Solving Strategies (§15.2)
Introduction to Counting & Probability, Chapter 2: Basic Counting Techniques