where is the integer part of .
I haven't solved it yet, but here's an approach.
. . Maybe it will inspire someone . . .
.where is the integer part of
We have: .
Since the right side must be an integer,
. . That is: .
Substitute into : .
. . which simplifies to: .
And that's it for now . . .
Soroban made a mistake since x is not necessarily an integer. Instead, as he pointed out, it is true that (7x-1)/3 is an integer. Thus, x=(3k+1)/7 for some integer k.
You have an inequality .
In this case, you get (4x+1)/5 - 1 < (7x-1)/3 <= (4x+1)/5 which means -7/23<x<=8/23.
Now, substituting x=(3k+1)/7, you get k=-1, 0.
At the right-hand side of , we do not know that is an integer, what we know is is an integer exactly.
Letting , we indeed see that but we have .
We clearly have for all .
Then, using the equation you gave with , we get
You can solve these inequalities by seperating into two parts.