2021 AMC Fall 10B Problem 10

P1-10~3 minPrint

Logic Puzzles Number Properties Primes Consider the largest/smallest element or boundary case Test small/specific values or special cases to find or verify the answer

Forty slips of paper numbered $1$ to $40$ are placed in a hat. Alice and Bob each draw one number from the hat without replacement, keeping their numbers hidden from each other. Alice says, "I can't tell who has the larger number." Then Bob says, "I know who has the larger number." Alice says, "You do? Is your number prime?" Bob replies, "Yes." Alice says, "In that case, if I multiply your number by $100$ and add my number, the result is a perfect square. " What is the sum of the two numbers drawn from the hat?

Show solution without answering

Key insight

Alice's doubt rules out 1 and 40. Then Bob can only be sure with 1, 2, 39 or 40; prime means 2, and 200 + 25 = 225.

Solution

Let Alice's number be $a$ and Bob's be $b$ . Go through the conversation one sentence at a time, and after each one ask what it rules out.

1. Alice: "I can't tell who has the larger number." If Alice had $1$ , she would know Bob is larger. If she had $40$ , she would know she is larger. So Alice has something from $2$ to $39$ .
2. Bob: "I know who has the larger number." Bob heard step 1, so he knows Alice is between $2$ and $39$ .
- If Bob has $2$ : Alice cannot also have $2$ (the slips are different), so she has $3$ to $39$ . Bob knows he is smaller.
- If Bob has $39$ : Alice has $2$ to $38$ . Bob knows he is larger.
- If Bob has $1$ or $40$ : he obviously knows.
- If Bob has, say, $20$ : Alice could have $5$ or $30$ , so he cannot know.

So Bob has $1$ , $2$ , $39$ or $40$ .
3. Bob's number is prime. Of $1, 2, 39, 40$ , only $2$ is prime ( $1$ is not prime, $39 = 3 \times 13$ , $40$ is even). So $b = 2$ .
4. $100b + a$ is a perfect square. That is $200 + a$ , with $a$ from $2$ to $39$ , so it is between $202$ and $239$ . Squares near there: $14^2 = 196$ , $15^2 = 225$ , $16^2 = 256$ . Only $225$ fits, so $a = 25$ .
5. The sum is $2 + 25 = 27$ .

The answer is $\boxed{\textbf{(A)}\ 27}$ .

Why this works

In puzzles where people take turns saying what they know, each statement cuts down the possibilities, and the next speaker uses the smaller list. Bob could only be sure with $2$ or $39$ because Alice had already ruled out $1$ and $40$ . After the logic narrows things down, a small arithmetic check finishes the problem.

Concise solution

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

Let Alice hold $a$ and Bob hold $b$ .

Alice cannot tell who is larger, so $a$ is neither $1$ nor $40$ : $2 \le a \le 39$ .

Bob hears this. He now knows who is larger, so his number must beat or lose to every value in $\{2, \dots, 39\}$ other than his own. That happens exactly when $b \in \{1, 2, 39, 40\}$ : with $b = 2$ , Alice's number is at least $3$ ; with $b = 39$ , hers is at most $38$ .

Bob's number is prime, so $b = 2$ .

Finally $100 \cdot 2 + a = 200 + a$ is a perfect square with $2 \le a \le 39$ , i.e. between $202$ and $239$ . The only square in that range is $225 = 15^2$ , so $a = 25$ .

The sum is $2 + 25 = 27$ .

The answer is $\boxed{\textbf{(A)}\ 27}$ .

Review in the AoPS books

Intermediate Counting & Probability, Chapter 2: Sets and Logic (§2.4)
Introduction to Number Theory, Chapter 1: Integers: The Basics
Introduction to Number Theory, Chapter 2: Primes and Composites
Introduction to Algebra, Chapter 20: Special Functions (§20.3)