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2021 AMC Fall 10B Problem 10

2021 AMC Fall 10B Problem 10AMC 10 Step by Step

Name: Date: September 13, 2026

Logic Puzzles · difficulty 2 of 5 · about 3 min · position P1-10

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?

A)
$27$
B)
$37$
C)
$47$
D)
$57$
E)
$67$

Answer

Key insight

Alice's uncertainty rules out 1 and 40; Bob then knows only if he holds 1, 2, 39, or 40; prime forces 2, and 200 + a = 225.

Solution

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 $\boxed{\textbf{(A)}\ 27}$ .

Why this works

Each statement in a "common knowledge" puzzle shrinks the set of possibilities, and later speakers reason using the earlier shrinkage. Bob's certainty depends on Alice's admission, which is why $2$ and $39$ become informative. Then a small arithmetic condition pins down the last unknown.

The trap

Concluding Bob must hold 1 or 40 and getting stuck, forgetting that once 1 and 40 are excluded for Alice, the numbers 2 and 39 also let Bob know.

Common mistakes

Concluding Bob must hold 1 or 40 and getting stuck, forgetting that once 1 and 40 are excluded for Alice, the numbers 2 and 39 also let Bob know.
Forgetting that $2$ is prime and hunting for a different prime, or accepting $a = 25$ without checking it is allowed by Alice's first statement.

Techniques

Consider the largest/smallest element or boundary case · Test small/specific values or special cases to find or verify the answer