Floor Ceiling Ceiling Floor

Assume: x=k+t, in which k is largest integer less than x. We have:
$\lceil{(k+t)\lfloor{k+t}}\rfloor\rceil+\lfloor{(k+t)\lceil{k+t}}\rceil\rfloor=111$ $=> 2k^2+k+\lceil{kt}\rceil+\lfloor{(k+1)t}\rfloor=111$ $=>2k^2+k+1<=111, 2k^2+3k>=111$ So only k=7 satisfies problem's criteria. $\Longrightarrow\lceil{7t}\rceil+\lfloor{8t}\rfloor=6$ $\Longrightarrow\frac{3}{8}<=t<=\frac{3}{7}=>7\frac{3}{8}<=x<=7\frac{3}{7}=>\alpha=7\frac{3}{8}, \beta=7\frac{3}{7}$ $8\alpha+7\beta=\boxed{111}$

Let $f(x)$ denote $\lceil x \lfloor x \rfloor \rceil + \lfloor x \lceil x \rceil \rfloor$

Notice that when $a \in \mathbb{Z}$ , then $f(a) = 2a^2$ . Thus $f(7) = 98 < 111 < 128 = f(8)$ .

It's not difficult to show that $f(x)$ is (not strictly) increasing, since the floor and ceiling functions are increasing. Therefore, $\alpha$ and $\beta$ are strictly between 7 and 8.

Therefore, we can write $x$ as $7 + t$ , where $0 < t < 1$ . Then we get: $f(x) = \lceil (7+t) * \lfloor 7+t \rfloor \rceil + \lfloor (7+t) * \lceil 7+t \rceil \rfloor = \lceil 7(7+t) \rceil + \lfloor 8(7+t) \rfloor = \lceil 49 + 7t \rceil + \lfloor 56 + 8t \rfloor$ $f(x) = 105 + \lceil 7t \rceil + \lfloor 8t \rfloor = 111$

Therefore, $\lceil 7t \rceil + \lfloor 8t \rfloor = 6$ .

Now $\lceil 7t \rceil$ and $\lfloor 8t \rfloor$ are both nonnegative integers which sum to 6. Thus, $0 \le \lfloor 8t \rfloor \le 6$ . Let $k = \lfloor 8t \rfloor$ . Then $k \le 8t < k+1$ , so $\frac{7k}{8} \le 7t < \frac{7(k+1)}{8}$ Since $0 \le k \le 6$ , this means that $k - 1 < 7t < k + 1$ .

Hence, $k \le \lceil 7t \rceil \le k+1$ . We can't have $\lceil 7t \rceil = k+1$ because that would imply that $2k + 1 = 6$ . Therefore, $\lceil 7t \rceil = k$ , and therefore $k = 3 = \lceil 7t \rceil = \lfloor 8t \rfloor$ .

Therefore, $3 \le 8t < 4$ and $2 < 7t \le 3$ . Hence, $t \in \left[ \frac{3}{8}, \frac{1}{2} \right)$ and $t \in \left( \frac{2}{7}, \frac{3}{7} \right]$ .

Thus $t \in \left[ \frac{3}{8}, \frac{3}{7} \right]$ .

Suppose that x = a + b with $a \in N$ and $0 < b < 1 , b \in R.$

We then have $\lfloor x \rfloor = a$ and $\lceil{x} \rceil= a + 1$ and so we get

$\lceil x \lfloor x \rfloor \rceil = \lceil (a + b) . a\rceil = \lceil a^2 + ab \rceil = a^2 + \lceil ab \rceil .$

Similarily will $\lfloor x \lceil x \rceil \rfloor = \lfloor (a +b). (a + 1) \rfloor = \lfloor a^2 + ab + a + b \rfloor = a^2 + a + \lfloor ab + b \rfloor = a^2 + a + \lfloor (a + 1) .b \rfloor .$

We get $\lceil x \lfloor x \rfloor \rceil + \lfloor x \lceil x \rceil \rfloor = 2 a^2 + a + \lceil ab \rceil + \lfloor (a + 1) .b \rfloor = 111$

Upon filling in some natural numbers into $2 a^2 + a$ , we find that for $a = 6$ we only get $78$ . The remaining terms clearly can't compensate for the missing part in $111$ when $a = 6$ and $b < 1$ .

$a = 7$ looks promising with $105$ .

$a = 8$ is too big with $136$ (the other terms can't be negative).

So I conclude that $a = 7$ . For those who would find this too much 'guessing', I don't. When you inspect the first equation it already is clear that a value greater than 8 for x would be too much. This can give you the idea here.

Remains to find the value for b. It is clear that $\lceil 7b \rceil + \lfloor 8 b \rfloor = 6 .$ Clearly both terms could be $3$ . To guarantee this, I checked whether the first term could be 4 and the second one 2. It would mean that $7b > 3$ and $8 . b \leq 3$ or $\frac{3}{7} < b \leq \frac{3}{8}$ which is impossible.

If you wonder why it cant' be the other way around (2 and 4), I found that trivial because of the floor and ceiling. A similar argument can however be used.

Now this means $\lceil 7b \rceil \wedge \lfloor 8 .b \rfloor = 3$ which reduces to the following conditions :

$2< 7b \leq 3 \Rightarrow \frac{2}{7} < b \leq \frac{3}{7}$

$3 \leq 8b < 4 \Rightarrow \frac{3}{8} \leq b < \frac{4}{8}$ .

Clearly $\frac{3}{8} \leq b \leq \frac{3}{7}$ so $\alpha = 7 + \frac{3}{8} \wedge \beta = 7 + \frac{3}{7}$ which gives 111 for $8 \alpha + 7 \beta$

Since 2x^2 = 98 for 7 and 128 for 8, obviously 7<x<8 say x = 7 +k where k is between 0 and 1 , the equation then reduces to floor(7k)+ceil(8k) = 6

The only possibility is both these terms equal to 3 (if you try 4,2 or 2,4 u get infeasible solutions) which means 2<7k <3 , and 3<8k<4 These reduce to k lying between 3/8 and 3/7 and hence the answer is 111.

PS : wondering if there is a more elegant soln

The given function $\begin{aligned} f:&&\mathbb{R}^+&\rightarrow\mathbb{R}^+,& f(x)&:=\lceil x\lfloor x\rfloor\rceil+\lfloor x\lceil x\rceil\rfloor \end{aligned}$ is a composition of increasing functions floor, ceil and multiplication with $x > 0$ . Therefore $f(x)$ is increasing and the solution of $f(x)=111$ is an interval (if it exists). Let's find upper and lower bounds for $f(x)$ :: $\begin{aligned} f(x)&\leq (x\lfloor x\rfloor+1)+x\lceil x\rceil\leq (x^2+1)+x(x+1)=2x^2+x+1=:u(x)\\[1em] f(x)&\geq x\lfloor x\rfloor+(x\lceil x\rceil-1)\geq x(x-1)+(x^2-1)=2x^2-x-1=:l(x) \end{aligned}$ A necessary condition for a solution $x>0$ is $\begin{aligned} l(x)\overset{!}{\leq}&111\overset{!}{\leq} u(x)&\Rightarrow&& 7.17\leq\frac{\sqrt{881}-1}{4}\leq x \leq \frac{\sqrt{897}+1}{4}\leq 7.74 \end{aligned}$ Now take an approximate solution $x=7+\frac{2}{7}+r\overset{!}{\in} (7.17;\:7.74)\subset (7;\:8)$ : $\begin{aligned} f(x)=\lceil x\cdot 7\rceil+\lfloor x\cdot 8\rfloor=49+2+\green{\lceil 7r\rceil}+56+2+\green{\left\lfloor8r+\frac{2}{7}\right\rfloor}=:109+\green{R(r)}\overset{!}{=}111 \end{aligned}$ We need to find $r$ such that $R(r)\overset{!}{=}2$ :

$\begin{aligned} R(r)&\overset{!}{=}2&\Rightarrow&&0&<7r\leq 1&\wedge&&1&\leq 8r+\frac{2}{7}<2&\Rightarrow&&\frac{5}{56}\leq r\leq \frac{1}{7} \end{aligned}$ Insert it all into our approximate solution: $x=7+\frac{2}{7}+r \in\left[\frac{59}{8};\:\frac{52}{7}\right]=:[\alpha;\:\beta]\quad\Rightarrow\quad 8\alpha+7\beta=\boxed{111}$