2024 USAMO Problems/Problem 4: Difference between revisions

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]

@@ Line 3: / Line 3: @@
 ==Solution 1==
-To solve this problem, 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:
+We need to determine all possible positive integer pairs <math>(m, n)</math> such that there exists a circular necklace of <math>mn</math> 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.**
+No matter how the necklace is cut into <math>m</math> blocks of <math>n</math> consecutive beads, each block has a distinct number of red beads.
 Necessary Condition:
-. **Maximum Possible Distinct Counts:**
+. Maximum Possible Distinct Counts:
-   - In a block of \(n\) beads, the number of red beads can range from \(0\) to \(n\).
+In a block of <math>n</math> beads, the number of red beads can range from <math>0</math> to <math>n</math>.
-   - Therefore, there are \(n + 1\) possible distinct counts of red beads in a block.
+Therefore, there are <math>n + 1</math> possible distinct counts of red beads in a block.
-   - Since we have \(m\) blocks, the maximum number of distinct counts must be at least \(m\).
+Since we have <math>m</math> blocks, the maximum number of distinct counts must be at least <math>m</math>.
-   - **Thus, we must have:**
+Thus, we must have:
-     <cmath> m \leq n + 1 </cmath>
+<cmath>m \leq n + 1</cmath>
 Sufficient Construction:
-We will show that for all positive integers \(m\) and \(n\) satisfying \(m \leq n + 1\), such a necklace exists.
+We will show that for all positive integers <math>m</math> and <math>n</math> satisfying <math>m \leq n + 1</math>, such a necklace exists.
-. **Construct Blocks:**
+. Construct Blocks:
-   - Create \(m\) blocks, each containing \(n\) beads.
+Create <math>m</math> blocks, each containing <math>n</math> beads. Assign to each block a unique number of red beads, ranging from <math>0</math> to <math>m - 1</math>.
-   - Assign to each block a unique number of red beads, ranging from \(0\) to \(m - 1\).
-. **Design the Necklace:**
+. Design the Necklace:
-   - Arrange these \(m\) blocks in a fixed order to form the necklace.
+Arrange these <math>m</math> 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.
-   - Since the necklace is circular, cutting it at different points results in cyclic permutations of the blocks.
-. **Verification:**
+. Verification:
-   - In any cut, the sequence of blocks (and thus the counts of red beads) is a cyclic shift of the original sequence.
+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.
-   - Therefore, in each partition, the blocks will have distinct numbers of red beads.
 Example:
-Let's construct a necklace for \(m = 3\) and \(n = 2\):
+Let's construct a necklace for <math>m = 3</math> and <math>n = 2</math>:
-- **Blocks:**
+Blocks:
-  - Block 1: \(0\) red beads (BB)
+Block 1: <math>0</math> red beads (BB)
-  - Block 2: \(1\) red bead (RB)
+Block 2: <math>1</math> red bead (RB)
-  - Block 3: \(2\) red beads (RR)
+Block 3: <math>2</math> red beads (RR)
-- **Necklace Arrangement:**
+Necklace Arrangement:
-  - Place the blocks in order: **BB-RB-RR**
+Place the blocks in order: BB-RB-RR.
-- **Verification:**
+Verification:
-  - Any cut of the necklace into \(3\) blocks of \(2\) beads will have blocks with red bead counts of \(0\), \(1\), and \(2\).
+Any cut of the necklace into <math>3</math> blocks of <math>2</math> beads will have blocks with red bead counts of <math>0</math>, <math>1</math>, and <math>2</math>.
 Conclusion:
-- **All ordered pairs \((m, n)\) where \(m \leq n + 1\) satisfy the condition.**
+All ordered pairs <math>(m, n)</math> where <math>m \leq n + 1</math> satisfy the condition.
-- **Therefore, the possible values of \((m, n)\) are all positive integers such that \(m \leq n + 1\).**
+Therefore, the possible values of <math>(m, n)</math> are all positive integers such that <math>m \leq n + 1</math>.
 Final Answer:
-**Exactly all positive integers \(m\) and \(n\) with \(m \leq n + 1\); these are all possible ordered pairs \((m, n)\).**
+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>.
+~[https://artofproblemsolving.com/wiki/index.php/User:Athmyx Athmyx]
 ==Video Solution==
 https://youtu.be/ZcdBpaLC5p0 [video contains problem 1 and problem 4]

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2024 USAMO Problems/Problem 4: Difference between revisions

Latest revision as of 01:50, 15 November 2024

Solution 1

Video Solution