Problem is accessible but requires concentration and determination.
Congratulations to the USA for winning first place, followed by China, Korea, India, and Belarus.
Problem Statement
Determine all real numbers ( \alpha ) such that
For every positive integer ( n ), ( \text{floor}(\alpha) + \text{floor}(2\alpha) + \text{floor}(3\alpha) + ... + \text{floor}(n\alpha) ) is a multiple of ( n ).
The floor function rounds down to the nearest integer.
Try Substituting Whole Numbers for ( \alpha )
Case: ( \alpha = 1 )
For ( n = 1 ): ( 1 ) is a multiple of ( 1 ).
For ( n = 2 ): ( 1 + 2 = 3 ), not a multiple of ( 2 ).
Conclusion: ( \alpha = 1 ) is not a solution.
Case: ( \alpha = 2 )
For ( n = 1 ): ( 2 ) is a multiple of ( 1 ).
For ( n = 2 ): ( 2 + 4 = 6 ) is a multiple of ( 2 ).
For ( n = 3 ): ( 2 + 4 + 6 = 12 ) is a multiple of ( 3 ).
For ( n = 4 ): ( 2 + 4 + 6 + 8 = 20 ) is a multiple of ( 4 ).
For ( n = 5 ): ( 2 + 4 + 6 + 8 + 10 = 30 ) is a multiple of ( 5 ).
Conclusion: ( \alpha = 2 ) works, showing that all even integers are solutions.
Sum Formula
Formula for the sum of integers from 1 to ( n ): ( S = \frac{n(n+1)}{2} ).
If ( \alpha ) is even, ( \alpha/2 ) is an integer, hence the sum is a multiple of ( n ).
Case for Odd Integers
If ( \alpha ) is odd, the sum ( \frac{\alpha n(n + 1)}{2} ) is not divisible by ( 2 ), which means it cannot be a multiple of ( n ) for all ( n ).
Try Substituting Non-Integer Values for ( \alpha )
Case: ( \alpha = 0.5 )
For ( n = 1 ): ( 0 ) is a multiple of ( 1 ).
For ( n = 2 ): ( 0 + 1 = 1 ), not a multiple of ( 2 ).
Conclusion: ( \alpha = 0.5 ) is not a solution.
Case: ( \alpha = \frac{2}{3} )
Fails similarly for ( n = 2 ).
Case: ( \alpha = 1.75 )
Works for ( n = 1, 2, 3, 4 ) but fails for ( n = 5 ).
Case: ( \alpha = -5 )
Works for ( n = 1, 2, 3, 4, 5 ), but fails for ( n = 6 ).
Narrowing Down Solutions
Values between ( 0 ) and ( 1 ) lead to repeat failures for ( n ) increasing.
For ( \alpha ) between ( -1 ) and ( 1 ): only ( 0 ) remains valid.
Valid solutions are constrained to even integers based on the analysis.
Proof Completion
Any real ( \alpha ) can be expressed as ( k + \beta ), with ( k ) as an integer and ( \beta ) as a fraction.
The integer part sums prove to be multiples of ( n ), while the fractional parts fail unless ( \beta = 0 ).
Conclusion
The only solutions for ( \alpha ) are all even integers.
No odd integers or fractions work as valid solutions, completing the proof.