How Do I Remove These Floors?

Each integer n can be written as $n=6k+p$ where k and p are integers and $0\leq p \leq 5$ .

Now for each value of p the left hand side of the equation can be evaluated term by term to get

For p = 0, $n=6k$ and the left hand side becomes

$(3k)+(2k)+(k)=6k\neq n-1$ and so $n=6k$ is NOT a solution of the equation.

For p = 1, $n=6k+1$ and the left hand side becomes

$(3k)+(2k)+(k)=6k= n-1$ and so $n=6k+1$ is a solution of the equation.

For p = 2, $n=6k+2$ and the left hand side becomes

$(3k+1)+(2k)+(k)=6k+1= n-1$ and so $n=6k+2$ is a solution of the equation.

For p = 3, $n=6k+3$ and the left hand side becomes

$(3k+1)+(2k+1)+(k)=6k+2 = n-1$ and so $n=6k+3$ is a solution of the equation.

For p = 4, $n=6k+4$ and the left hand side becomes

$(3k+2)+(2k+1)+(k)=6k+3 = n-1$ and so $n=6k+4$ is a solution of the equation.

For p = 5, $n=6k+5$ and the left hand side becomes

$(3k+2)+(2k+1)+(k)=6k+3 \neq n-1$ and so $n=6k+5$ is NOT a solution of the equation.

By examining the above cases we see that the equation fails to have a solution whenever n is a multiple of 6 or one less than a multiple of six. So for n between 1 and 100 the number of failures is

$2 \times \lfloor \frac{100}{6} \rfloor = 32$

and so the number of successes is

$100-32=\boxed{68}$

Define, $f(n)=\left \lfloor\frac{n}{2}\right \rfloor+\left \lfloor\frac{n}{3}\right \rfloor+\left \lfloor\frac{n}{6}\right \rfloor\\g(n)=f(n)-(n-1)\\\implies f(n+6)=f(n)+6\\\implies g(n+6)=g(n)$

So, if $g(n)=0\implies g(n+6k)=0 \quad \forall k \in \mathbb{Z}$ .

Now, note that $g(1)=g(2)=g(3)=g(4)=0 \quad g(5)\neq0 \quad g(6)\neq0$ $g(n)=0\implies n \equiv k \pmod6\quad k=1,2,3,4$

Hence, the number of required solutions is $\boxed{68}$ .

We know that every integer can be written in form of $6\kappa + \psi$ . Where

$\kappa \in N$ and

$0\leq \psi \leq 5$ , $\psi$ is also a natural number.

Now

$\forall such \kappa , \psi = 0,5$ does not satisfies. Hence 32 solutions are to be emitted from $one$ hundred.

Hence answer is $\boxed{68}$ .

Via induction one can recognize that among the 100 possible values for n always 2 in equidistance must be removed resp.:

$n_{k}=5+6k$ ; $n_{k+1}=6(k+1)$ with 0<=k<=15)

As result there are 100 - 2*16 =68 solutions.