There is a trick!

Let $a_n = \underbrace{1111 \cdots 1111}_{\text{Number of 1's}=n}$ and $x_n = a_{2n} - 2a_n$ and the sum of digits of $x_n$ be $s_n$ . The first few $x_n$ and the respective $s_n$ are as follows.

$\begin{array} {rrrrr} n & a_{2n} & 2a_n & x_n & s_n \\ \hline 1 & 11 & 2 & 9 & 9 \\ 2 & 1111 & 22 & 1089 & 18 \\ 3 & 111111 & 222 & 110889 & 27 \\ 4 & 11111111 & 2222 & 11108889 & 36 \\ 5 & 1111111111 & 22222 & 1111088889 & 45 \end{array}$

It appears that $s_n = 9n$ . Let us prove the claim as follows:

$\begin{aligned} x_n & = {\color{#3D99F6}a_{2n}} - 2a_n \\ & = {\color{#3D99F6}\underbrace{111 \cdots 111}_{\text{\# of 1's = n}}\underbrace{111 \cdots 111}_{\text{\# of 1's = n}}} - 2a_n \\ & = {\color{#3D99F6}\underbrace{111 \cdots 111}_{\text{\# of 1's = n}}\underbrace{000 \cdots 000}_{\text{\# of 0's = n}} + \underbrace{111 \cdots 111}_{\text{\# of 1's = n}}} - 2a_n \\ & = {\color{#3D99F6}10^na_n + a_n} - 2a_n \\ & = {\color{#3D99F6}10^na_n} - a_n & \small \color{#3D99F6} \text{Note that }a_n = 10a_{n-1} + 1 \\ & = {\color{#3D99F6}10^{n+1}a_{n-1}+10^n} - a_n & \small \color{#3D99F6} \text{and }10^n = 9a_n + 1 \\ & = 10^{n+1}a_{n-1}+{\color{#3D99F6}9a_n + 1} - a_n \\ & = 10^{n+1}a_{n-1}+{\color{#3D99F6}8a_n} + 1 \\ & = 10^{n+1}a_{n-1}+{\color{#3D99F6}80a_{n-1} + 8} + 1 \\ & = 10^{n+1}a_{n-1}+80a_{n-1} + 9 \\ & = \underbrace{111 \cdots 111}_{\text{\# of 1's = n-1}}0\underbrace{888 \cdots 888}_{\text{\# of 8's = n-1}}9 \end{aligned}$

Therefore, $s_n = 1\times (n-1) + 8(n-1) + 9 = 9n$ . For $n=50$ , $s_{50} = \boxed{450}$ .

Relevant wiki: Digital root

(Or)

The given problem can quickly solve with Digital root validation technique:

The Digital root of the difference will always be 9, if both numbers have the same digital root. Since, they must have the same remainder when divided by 9. Thus their difference has a remainder of 0 when divided by 9, and therefore the digital root of the difference must be 9.

Clearly, 50 times of 2 equal to 100 & 100 times of 1 equal to 100 both has '1' as it's digital root. therefore 'x' is divisible by 9 and it's digital root is 9. $\begin{aligned} x &\equiv 1-1 &\equiv 0 \pmod{9}\\ \end{aligned}$

In the given options the one and only number 450 has digital root as 9. Hence, the required solution is 450.

Note: The above method will not work, if multiple options have same expected digital root value.

I observed that if there are n 1 and n/2 2 then

No of 1 in solution = (n/2) -1 here n is 100

No. Of 9 in solution always 1

No of 8 in solution = no of 1

Then the sum of digits = [(n/2) - 1] + [{(n/2)-1} ×8] + 9

Put n = 100 and answer will be 450

I'm rather lazy here, what I notice is that $x\equiv 1-1\equiv 0\pmod 9$ , hence the only option be a multiple of $9$ is $\boxed{450}$ .

Following the hint I calculated that 111111 - 222 = 110889.

The easiest way to sum these digits is to change them slightly. 111888 would give the same sum, which is 3x1 + 3x8 = 3x9

The process of doing that 'practice' subtraction made it clear that the same principles would apply to the longer number and hence the answer is 50x9 = 450

It's simple. Uses number of digits rather than digits themselves. Given the example 111111-222=110889 The addition of the numbers of digits (rather than the digits themselves) gives 9. 1+1+0+8+8+9= 27 27 is 9x3

Therefore, if the number of digits in the first is 100 and the second is 50; Therefore: 100+50=150

But 27 is 9x3 Thus, n = 150 x 3

= 450

Let $k$ denote the number of 2's, and $2k$ the number of 1's. Observe that for any k the difference will be of the form: $x_k = \underbrace{111\cdots 1110888\cdots 8889}_{\text{Number of 1's and 8's}=k-1}$ We can calculate $s_k$ , the sum of the digits, by $(k-1)*1 + 8*(k-1) + 9 = 9*(k-1) + 9 = 9k$

Let $n$ be the number of $2$ s. After trying with a few small $n$ , we notice that $x$ is in the form $\underbrace{111\cdots111}_{n-1 \ 1\text{s}}0\underbrace{888\cdots888}_{n-1 \ 8\text{s}}9$ and hence the sum of the digits in $x$ is $9n$ , where $n = 50$ .

The first number has $100 \times 1=100$ as a sum and the second has $2 \times 50=100$ .

Since the sum of digits is the same, they should have the same value modulo 9 or remainder after division by 9 and their difference should be divisible by 9.

The only such option was $450$

The digital root of a hundred ones is 1 and the digital root of 50 twos is also 1 which means that the result will divide nine I.e have a digital root equal to 9. 450 is the only one here that has digital root 9

Since 11-2=9, then the answer has to be a multiple of 9 and 450 is the only number there that is a multiple of 9

$\begin{aligned} X& = \left(\sum_{k=0}^{m} 10^k\right)-2\left(\sum_{k=0}^{\frac{m}{2}}10^k\right)\\& = \dfrac{1}{9}\left(10^m-1 -2\cdot 10^{\frac{m}{2}}+2\right)=\dfrac{1}{9}\left(10^{\frac{m}{2}}(10^{\frac{m}{2}}-2)+1\right) \end{aligned}$ Note that $10^{\frac{m}{2}}-2$ has exactly $\frac{m}{2}$ 9's and only one 8. Multiplying the expression by $10^{\frac{m}{2}}$ and adding 1 yields $X_0 = \underbrace{999\cdots9}_{\frac{m}{2}-1\text{ 9's }}8\underbrace{000\cdots 0}_{\frac{m}{2}-1 \text{ 0's}}1$ Since $X$ is positive integer then it's is true that $9\mid X_0$ . Since $\begin{aligned} X_0 & = \underbrace{999\cdots9}_{\frac{m}{2}-1\text{ 1's }} 0+ 8\underbrace{000\cdots 0}_{\frac{m}{2}-1 \text{ 0's}}1\\ \therefore 9\mid X _0 & = \underbrace{111\cdots1}_{\frac{m}{2}-1\text{ 1's }}0 + \underbrace{888\cdots 8}_{\frac{m}{2}-1 \text{ 8's}}9 \end{aligned}$ Setting $m=100$ we obtain $X = \underbrace{111\cdots1}_{49 \text{ 1's }}0 + \underbrace{888\cdots 8}_{49 \text{ 8's}}9$ Hence, our required answer is $49 \cdot 1+49\cdot 8 +9 =450$ .