TITLE: $\lfloor a n\rfloor \lfloor b n\rfloor \lfloor c n\rfloor = \lfloor d n\rfloor \lfloor e n\rfloor \lfloor f n\rfloor$ for all $n$
QUESTION [10 upvotes]: Simple task, but I know only extremely overkill solution :)

Let $a,b,c,d,e,f$ be positive irrational numbers.
Suppose that for any positive integer number $n$,
$$\lfloor a  n\rfloor \cdot \lfloor b  n\rfloor \cdot \lfloor c  n\rfloor  
= \lfloor d n\rfloor \cdot \lfloor e  n\rfloor \cdot \lfloor f  n\rfloor.$$
Prove that the sets $\{a,b,c\}$ and  $\{d,e,f\}$ are equal.

REPLY [2 votes]: So, the overkill  proof;
One of the equidistribution theorems:
 https://en.wikipedia.org/wiki/Equidistribution_theorem
a is irrational, then sequence $a*p_n$ mod 1 is uniformly distributed.
Lemma: 
If a > 0 is irrational, then sequence { $\lfloor a*n\rfloor$ } contains infinitely many prime numbers.
Proof of lemma: 
Case (a>1):
$\lfloor a*n\rfloor = p_m$ <=> $p_m < a*n < p_m + 1$  <=> $\frac{p_m}{a} < n < \frac{p_m+1}{a}$  <=> n =  $\lfloor \frac{p_m+1}{a}\rfloor$, {$\frac{p_m}{a}$} $ \in (1-\frac{1}{a}, 1)$  (where are infinitely many such m by thm.)
Case (a<1) is trivial.
Proof of task:
Without loss of generality $a \ge b \ge c, d \ge e \ge f, a\ge d$
Let a > d. 
Fix n what $n*a > n*d + 2, n*c > 1, \lfloor n*a \rfloor  =p_m$
$\lfloor n*d \rfloor \lfloor n*e \rfloor \lfloor n*f \rfloor  > 0 $ and not divisible to $p_m$ -  it is the absurd.
So, a = d and we need to prove what if $\lfloor n*e \rfloor \lfloor n*f \rfloor = \lfloor n*b \rfloor \lfloor n*c \rfloor$, then sets {b, c}  and {e,f} are equals.
We can do it similary or by method of @marco2013 
P.S. Maybe, we can generalize lemma to rational not-integer numbers (with the Dirichlet thm.)