Complicated symbols make this problem easy

We note that the only part contributing to the fractional part of the RHS is the $\{a\}$ , $\{b\}$ , and $\{c\}$ terms.

Thus, if we let $a=a_i+a_f$ ( $a_i$ is integer part, and $a_f$ is fractional part) and ditto for the others, then $a_f=0.3$ , $b_f=0.7$ , and $c_f=0$ .

Now we simplify the system of equations into the following: $\begin{array}{l}b_i+a_i+1+0=16\\ c_i+b_i+1+0.3=11.3\\ a_i+c_i+0.7=9.7\end{array}$

Simplifying:

$\begin{array}{l}b_i+a_i=15\\ c_i+b_i=10\\ a_i+c_i=9\end{array}$

Adding the equations up and dividing by $2$ , we see that $a_i+b_i+c_i=17$ .

Thus, $a_i=17-10=7$ , $b_i=17-9=8$ , and $c_i=17-15=2$ .

Thus, $(a,b,c)=(7.3,8.7,2)$ and our answer is $7.3+8.7+2=\boxed{18}$ .

First, we note that $x=\lfloor{x}\rfloor+\{x\}$ .

Therefore, when we add the equations, we get:

$\begin{array}{c}\\ \lfloor{a}\rfloor+\lfloor{b}\rfloor+\lfloor{c}\rfloor+\{a\}+\{b\}+\{c\}+\lceil{a}\rceil+\lceil{b}\rceil+\lceil{c}\rceil=16+11.3+9.7 \\ a+b+c+\lceil{a}\rceil+\lceil{b}\rceil+\lceil{c}\rceil=37 \end{array}$

Since $\lfloor{x}\rfloor$ and $\lceil{x}\rceil$ are always integers and $0\le\{x\}<1$ , we conclude that $\{a\}=0.3$ , $\{b\}=0.7$ , and $\{c\}=0$ . Therefore,

$\begin{array}{c}\\ \lceil{a}\rceil=a-\{a\}+1=a+1-0.3=a+0.7 \\ \lceil{b}\rceil=b-\{b\}+1=b+1-0.7=b+0.3 \\ \lceil{c}\rceil=c \end{array}$

(Note, since $c$ has no fractional part (or $\{c\}=0$ ), $\lceil{c}\rceil=c$ .)

Plugging that into the above equation gives:

$\begin{array}{c}\\ a+b+c+(a+0.7)+(b+0.3)+c=37 \\ 2a+2b+2c+1=37 \\ 2a+2b+2c=36 \\ a+b+c=\boxed{18} \end{array}$

Since fractional sums of a b and c add up to 1, it leaves sum of ceiling and floor functions to be 19 + 17 respectively, but since they are counted twice, dividing by two gives the value of just the values of a+b+c to be 36/2 = 18

The equations reduced to the following, note......{a} =.3,.....{b} = .7,.....{c} = 0.
$\lfloor b \rfloor+\lfloor a \rfloor+1+0 = 16... \lfloor c \rfloor+\lfloor b \rfloor+1+ .3 = 11.3.. \lfloor a \rfloor+\lfloor c \rfloor +1+ .7 = 9.7$ Adding the three equations and noting that c is an integer,
$2( \lfloor a \rfloor+\lfloor b \rfloor + c )=34...... \lfloor a \rfloor + .3 + \lfloor b \rfloor + .7 + c + 0 = 17 + 1.... a+b+c=18$

take\quad \left\lfloor a \right\rfloor =a+\left{ a \right} \quad and\quad \left\lceil a \right\rceil =a+1-\left{ a \right} \ and\quad we\quad can\quad notice\quad that\quad { a} =0.3\quad and\quad { b} =0.7\quad and\quad { c} =0\ adding\quad all\quad three\quad equaltions\quad we\quad get\quad a+b+c=18

${\LaTeX}$ 'ed:

take $\left\lfloor a \right\rfloor =a+\left\{ a \right\}$ and $\left\lceil a \right\rceil =a+1-\left\{ a \right\}$

and we can notice that $\{ a\} =0.3$ and $\{ b\} =0.7$ and $\{ c\} =0$

adding all three equations we get $a+b+c=18$