Difference between revisions of "2024 IMO Problems/Problem 1"
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is divisible by \( n \), i.e., \( S_n(\alpha) \equiv 0 \mod n \). | is divisible by \( n \), i.e., \( S_n(\alpha) \equiv 0 \mod n \). | ||
| − | + | Step 1: Break Down \( \alpha \) into Integer and Fractional Parts | |
Let \( \alpha = m + f \), where \( m = \lfloor \alpha \rfloor \in \mathbb{Z} \) and \( f = \{\alpha\} \in [0,1) \) is the fractional part of \( \alpha \). | Let \( \alpha = m + f \), where \( m = \lfloor \alpha \rfloor \in \mathbb{Z} \) and \( f = \{\alpha\} \in [0,1) \) is the fractional part of \( \alpha \). | ||
| − | + | Step 2: Express the Sum in Terms of \( m \) and \( f \) | |
Using this, we have: | Using this, we have: | ||
| Line 32: | Line 32: | ||
</cmath> | </cmath> | ||
| − | + | Step 3: Modulo \( n \) Simplification | |
We are interested in \( S_n(\alpha) \mod n \): | We are interested in \( S_n(\alpha) \mod n \): | ||
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</cmath> | </cmath> | ||
| − | + | Step 4: Analyze the Fractional Part \( f \) | |
Our goal is to find all \( f \in [0,1) \) such that: | Our goal is to find all \( f \in [0,1) \) such that: | ||
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</cmath> | </cmath> | ||
| − | + | Step 5: Test \( f = 0 \) | |
If \( f = 0 \), then \( \lfloor kf \rfloor = 0 \) for all \( k \), so: | If \( f = 0 \), then \( \lfloor kf \rfloor = 0 \) for all \( k \), so: | ||
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which satisfies the condition for all \( n \). | which satisfies the condition for all \( n \). | ||
| − | + | Step 6: Consider \( f \in (0,1) \) | |
For \( f \in (0,1) \), let's test small values of \( n \) and \( f \). It turns out that the sum \( \sum_{k=1}^{n} \lfloor kf \rfloor \) does not satisfy the congruence \( 0 \mod n \) for all \( n \). For example, if \( f = \frac{1}{2} \): | For \( f \in (0,1) \), let's test small values of \( n \) and \( f \). It turns out that the sum \( \sum_{k=1}^{n} \lfloor kf \rfloor \) does not satisfy the congruence \( 0 \mod n \) for all \( n \). For example, if \( f = \frac{1}{2} \): | ||
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Thus, \( f \ne 0 \) fails the condition. | Thus, \( f \ne 0 \) fails the condition. | ||
| − | + | Step 7: Conclude that Only \( f = 0 \) Works | |
Since \( f \ne 0 \) does not satisfy the condition, the only possible value is \( f = 0 \), meaning \( \alpha \) must be an integer. | Since \( f \ne 0 \) does not satisfy the condition, the only possible value is \( f = 0 \), meaning \( \alpha \) must be an integer. | ||
| − | + | Step 8: Verify for Integer \( \alpha \) | |
Let \( \alpha = m \in \mathbb{Z} \). Then: | Let \( \alpha = m \in \mathbb{Z} \). Then: | ||
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This sum is always divisible by \( n \) because \( \frac{n(n+1)}{2} \) times any integer \( m \) is divisible by \( n \). | This sum is always divisible by \( n \) because \( \frac{n(n+1)}{2} \) times any integer \( m \) is divisible by \( n \). | ||
| − | |||
| − | |||
The only real numbers \( \alpha \) satisfying the condition are the integers. | The only real numbers \( \alpha \) satisfying the condition are the integers. | ||
Revision as of 06:55, 14 November 2024
Determine all real numbers 
such that, for every positive integer 
, the integer
![\[\lfloor \alpha \rfloor + \lfloor 2\alpha \rfloor + \dots +\lfloor n\alpha \rfloor\]](http://latex.artofproblemsolving.com/d/e/8/de8ec4b01ecf552960e95a201881ef7e35e7a2ce.png)
is a multiple of . (Note that 
denotes the greatest integer less than or equal to 
. For example, 
and 
.)
Contents
Solution 1
To solve the problem, we need to find all real numbers \( \alpha \) such that, for every positive integer \( n \), the integer
![\[S_n(\alpha) = \lfloor \alpha \rfloor + \lfloor 2\alpha \rfloor + \dots + \lfloor n\alpha \rfloor\]](http://latex.artofproblemsolving.com/2/3/9/239fa75a577e8b736a1ead473f405ecbaa8b4f02.png)
is divisible by \( n \), i.e., \( S_n(\alpha) \equiv 0 \mod n \).
Step 1: Break Down \( \alpha \) into Integer and Fractional Parts
Let \( \alpha = m + f \), where \( m = \lfloor \alpha \rfloor \in \mathbb{Z} \) and \( f = \{\alpha\} \in [0,1) \) is the fractional part of \( \alpha \).
Step 2: Express the Sum in Terms of \( m \) and \( f \)
Using this, we have:
![\[\lfloor k\alpha \rfloor = \lfloor k(m + f) \rfloor = km + \lfloor kf \rfloor.\]](http://latex.artofproblemsolving.com/9/2/4/924279402dac300846da65148ad861b728871cc0.png)
So, the sum becomes:
![\[S_n(\alpha) = m \sum_{k=1}^{n} k + \sum_{k=1}^{n} \lfloor kf \rfloor = m \frac{n(n+1)}{2} + \sum_{k=1}^{n} \lfloor kf \rfloor.\]](http://latex.artofproblemsolving.com/4/3/1/4318f858320e700433de5ce3fcf3dda5b0c49b17.png)
Step 3: Modulo \( n \) Simplification
We are interested in \( S_n(\alpha) \mod n \):
![\[S_n(\alpha) \equiv \left( m \frac{n(n+1)}{2} + \sum_{k=1}^{n} \lfloor kf \rfloor \right) \mod n.\]](http://latex.artofproblemsolving.com/1/1/f/11f20e1aa3d1c107906a4ed4684f600944fe070a.png)
Since \( m \frac{n(n+1)}{2} \) is divisible by \( n \), the expression simplifies to:
![\[S_n(\alpha) \equiv \sum_{k=1}^{n} \lfloor kf \rfloor \mod n.\]](http://latex.artofproblemsolving.com/6/3/a/63a61a6aa595368c0ba9fedd127a83fae095fee8.png)
Step 4: Analyze the Fractional Part \( f \)
Our goal is to find all \( f \in [0,1) \) such that:
![\[\sum_{k=1}^{n} \lfloor kf \rfloor \equiv 0 \mod n \quad \text{for all } n \in \mathbb{N}.\]](http://latex.artofproblemsolving.com/5/8/e/58e5774c521c86305a570f3171b47ad66517393d.png)
Step 5: Test \( f = 0 \)
If \( f = 0 \), then \( \lfloor kf \rfloor = 0 \) for all \( k \), so:
![\[\sum_{k=1}^{n} \lfloor kf \rfloor = 0,\]](http://latex.artofproblemsolving.com/3/9/e/39e6cefbb9553c851733ded557342a001965b063.png)
which satisfies the condition for all \( n \).
Step 6: Consider \( f \in (0,1) \)
For \( f \in (0,1) \), let's test small values of \( n \) and \( f \). It turns out that the sum \( \sum_{k=1}^{n} \lfloor kf \rfloor \) does not satisfy the congruence \( 0 \mod n \) for all \( n \). For example, if \( f = \frac{1}{2} \):
![\[\sum_{k=1}^{2} \lfloor k \cdot \frac{1}{2} \rfloor = \lfloor \frac{1}{2} \rfloor + \lfloor 1 \rfloor = 0 + 1 = 1 \not\equiv 0 \mod 2.\]](http://latex.artofproblemsolving.com/b/1/d/b1dfa0bbddbe18ae283d5399cba279db8d7ec79b.png)
Thus, \( f \ne 0 \) fails the condition.
Step 7: Conclude that Only \( f = 0 \) Works
Since \( f \ne 0 \) does not satisfy the condition, the only possible value is \( f = 0 \), meaning \( \alpha \) must be an integer.
Step 8: Verify for Integer \( \alpha \)
Let \( \alpha = m \in \mathbb{Z} \). Then:
![\[S_n(\alpha) = m \sum_{k=1}^{n} k = m \frac{n(n+1)}{2}.\]](http://latex.artofproblemsolving.com/3/6/c/36caf052c81a14ec3b5b79f86dea683dea2ca309.png)
This sum is always divisible by \( n \) because \( \frac{n(n+1)}{2} \) times any integer \( m \) is divisible by \( n \).
The only real numbers \( \alpha \) satisfying the condition are the integers.
Video Solution(In Chinese)
Video Solution
https://www.youtube.com/watch?v=50W_ntnPX0k
Video Solution
Part 1 (analysis of why there is no irrational solution)
Part 2 (analysis of even integer solutions and why there is no non-integer rational solution)
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution
See Also
| 2024 IMO (Problems) • Resources | ||
| Preceded by First Problem |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 2 |
| All IMO Problems and Solutions | ||
