Difference between revisions of "2024 USAMO Problems/Problem 4"

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(Solution 1)
 
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Exactly all positive integers <math>m</math> and <math>n</math> with <math>m \leq n + 1</math>; these are all possible ordered pairs <math>(m, n)</math>.
 
Exactly all positive integers <math>m</math> and <math>n</math> with <math>m \leq n + 1</math>; these are all possible ordered pairs <math>(m, n)</math>.
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~[https://artofproblemsolving.com/wiki/index.php/User:Athmyx Athmyx]
  
 
==Video Solution==
 
==Video Solution==
 
https://youtu.be/ZcdBpaLC5p0 [video contains problem 1 and problem 4]
 
https://youtu.be/ZcdBpaLC5p0 [video contains problem 1 and problem 4]

Latest revision as of 01:50, 15 November 2024

Let $m$ and $n$ be positive integers. A circular necklace contains $m n$ beads, each either red or blue. It turned out that no matter how the necklace was cut into $m$ blocks of $n$ consecutive beads, each block had a distinct number of red beads. Determine, with proof, all possible values of the ordered pair $(m, n)$.

Solution 1

We need to determine all possible positive integer pairs $(m, n)$ such that there exists a circular necklace of $mn$ beads, each colored red or blue, satisfying the following condition:

No matter how the necklace is cut into $m$ blocks of $n$ consecutive beads, each block has a distinct number of red beads.

Necessary Condition:

1. Maximum Possible Distinct Counts: In a block of $n$ beads, the number of red beads can range from $0$ to $n$. Therefore, there are $n + 1$ possible distinct counts of red beads in a block. Since we have $m$ blocks, the maximum number of distinct counts must be at least $m$. Thus, we must have: \[m \leq n + 1\]

Sufficient Construction:

We will show that for all positive integers $m$ and $n$ satisfying $m \leq n + 1$, such a necklace exists.

1. Construct Blocks: Create $m$ blocks, each containing $n$ beads. Assign to each block a unique number of red beads, ranging from $0$ to $m - 1$.

2. Design the Necklace: Arrange these $m$ blocks in a fixed order to form the necklace. Since the necklace is circular, cutting it at different points results in cyclic permutations of the blocks.

3. Verification: In any cut, the sequence of blocks (and thus the counts of red beads) is a cyclic shift of the original sequence. Therefore, in each partition, the blocks will have distinct numbers of red beads.

Example:

Let's construct a necklace for $m = 3$ and $n = 2$:

Blocks: Block 1: $0$ red beads (BB) Block 2: $1$ red bead (RB) Block 3: $2$ red beads (RR)

Necklace Arrangement: Place the blocks in order: BB-RB-RR.

Verification: Any cut of the necklace into $3$ blocks of $2$ beads will have blocks with red bead counts of $0$, $1$, and $2$.

Conclusion:

All ordered pairs $(m, n)$ where $m \leq n + 1$ satisfy the condition. Therefore, the possible values of $(m, n)$ are all positive integers such that $m \leq n + 1$.

Final Answer:

Exactly all positive integers $m$ and $n$ with $m \leq n + 1$; these are all possible ordered pairs $(m, n)$.

~Athmyx

Video Solution

https://youtu.be/ZcdBpaLC5p0 [video contains problem 1 and problem 4]