The Missing Exponent

Let $a$ and $b$ be the tenth and unit digits of $x$ respectively so that $x=10a+b$ .

We note that the difference is of the form:

\begin {equation} \begin{aligned} 10^x-x&=\overline{\underbrace {999...999}_{(×-2) \text{ of } 9}cd} \end{aligned} \end {equation}

There are $x$ digits of which $x-2$ are $9$ , $c=9-a$ and $d=10 - b$ .

Therefore, the sum of digits is given by:

\begin {equation} \begin{aligned} 9(x-2) + 9-a + 10 -b & = 300 \\ 9(10a+b-2) + 19 - a-b & = 300 \\ 89a+8b & = 299 \quad \quad \small \color{#3D99F6}{\text{We note that } a = \left \lfloor \frac{299}{89} \right \rfloor = 3} \\ 89(\color{#3D99F6}{3}) + 8b & = 299 \\ 8b & = 299 - 267 = 32 \\ \implies b & = 4 \\ \implies x & = \boxed{34} \end{aligned} \end {equation}

$x$ is a two digit number, and the expression difference is equal to 300.

Now, keeping these things in mind, we will approach the solution of the question.

The expression $10^{x}-x$ will contain $x-2$ number of 9's. For an explicit example, $10^{13}-13$ will have $eleven$ 9's in it.

Now, $9 \times 33=297$ , lets check for the number $x-2=33$ i.e. $x=35$ .

$10^{35}-35=$ a result in which there are $thirty-three$ $9's$ and last two digits are $100-35=65$ , so total sum of the digits= $33\times9+6+5=306$ , which is more than the required result.

Now check for $x=34$ .

$10^{34}-34=$ a result in which there are $thirty-two$ $9's$ and last two digits are $100-34=66$ , so total sum of the digits $=32 \times 9+(6+6)=\boxed{300}$ $\boxed{tada}$ , .