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There are $73$ terms in the series and $\lfloor{\frac{542}{73}}\rfloor = 7$ and $\lceil{\frac{542}{73}}\rceil = 8$ Therefore $7 < x < 8$ . Now we can form an equation as -

Some $a$ terms give a value of $7$ and $(73 - a)$ terms give a value of $8$ .

$\therefore 7x + 8(73-x) = 542 \Rightarrow x = 42$ .

The first $42$ terms will give value $7$ and the $43$ rd term i .e. $\lfloor{(x + 0.61)}\rfloor = 8 \Rightarrow x = 7.39$

$\therefore \lfloor{100x}\rfloor = 739$

There are 73 terms. so 73 integers (suppose p ) will make 542 but 542 mod 73 = 31. so there will be 42 numbers of p and 31 numbers of (p + 1). as the terms are increasing so the last 31 terms will make (p+ 1).

assume that

x + 0.19 = k + d where k is an integer and 0 <= d < 1; the 31st term from the last is x + 0.61 = x+ 0.19 +0.42 = k +d +0.42

the value of d that can make the floor of this term (k+1) is 0.58<= d < 0. 59 (as the previous term can not be (k+1) )

542 = 7 x 73 +31. so, k = 7.

and 7.39 <= x < 7.40

and the ans is 100 x 7.39 = 739.

$\begin{aligned} S & = \left \lfloor x+\frac{19}{100} \right \rfloor + \left \lfloor x+\frac{20}{100} \right \rfloor +\ldots + \left \lfloor x+\frac{91}{100} \right \rfloor=542 \\ x & = \lfloor x\rfloor+ \{x\} \\ S & = \left \lfloor \lfloor x\rfloor+ \{x\} +\frac{19}{100} \right \rfloor + \left \lfloor \lfloor x\rfloor+ \{x\} +\frac{20}{100} \right \rfloor +\ldots + \left \lfloor \lfloor x\rfloor+ \{x\} +\frac{91}{100} \right \rfloor = 542 \\ S & = 73\lfloor x\rfloor+\displaystyle \sum^{73}_{n=1}\left \lfloor \{x\}+\frac{18+n}{100} \right \rfloor =542 \\ S & = 542, \quad \lfloor x\rfloor = \left \lfloor \frac{542}{73} \right \rfloor = 7 \\ T & = \displaystyle \sum^{73}_{n=1}\left \lfloor \{x\}+\frac{18+n}{100} \right \rfloor=542-73×7 = 31 \end{aligned}$

So, the first $73-31=42$ terms of $T$ must be $0$ and the next $31$ must be $1$ .

$\begin{aligned} \{x\} +\frac{18+43}{100} & \geq 1 \\ \{x\} +0.61 & \geq 1 \\ \{x\} & \geq 0.39 \\ x & = \lfloor x \rfloor + \{x\} =7+0.39=7.39 \\ \Rightarrow \lfloor 100x \rfloor & = \boxed{739} \end{aligned}$

Total number of terms = 73. Next 542 = 73 x 7 + 31, hence 31 terms are of value 8. Implies that x + 0.61 = 8 or x = 7.39