Problem 1

Find all real numbers

\alpha

so that, for every positive integer

n

, the integer

\lfloor\alpha\rfloor+\lfloor2\alpha\rfloor+\lfloor3\alpha\rfloor+\cdots+\lfloor n\alpha\rfloor

is divisible by

n

Step 3 of 4: Case 0 < alpha < 1 fails

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In plain words

For small positive alpha, every floor is zero until the running sum of alpha finally crosses 1, so the total sum is just 1 once, which cannot be divisible by any m at least 2.

0<\alpha<1\implies S(m,\alpha)=1

Detailed analysis

If $0<\alpha<1$ , let $m\ge2$ be the smallest integer with $m\alpha\ge1$ ; then $\lfloor i\alpha\rfloor=0$ for $i=1,\ldots,m-1$ and $\lfloor m\alpha\rfloor=1$ (since $\alpha<1$ forces $m\alpha<m$ , in fact $m\alpha<2$ by minimality, so the floor is exactly $1$ ), giving $S(m,\alpha)=1$ , which is not divisible by $m\ge2$ . This contradicts the required property.