Consider a string of $$n$$ $$1$$'s. We wish to place $$+$$ signs in between so that the sum is $$1000$$. For instance, if $$n=190$$, one may put $$+$$ signs so as to get $$11$$ ninety times and $$1$$ ten times, and get the sum $$1000$$. If $$a$$ is the number of positive integers $$n$$ for which it is possible to place $$+$$ signs so as to get the sum $$1000$$, then find the sum of the digits of $$a$$.

Let the string contain $$n$$ digits ‘1’. We partition it into blocks of equal‐length ‘1’s; after inserting $$+$$ signs every block represents a repunit

$$R_m=\underbrace{\,11\ldots1\,}_{m\text{ digits}}=\dfrac{10^{m}-1}{9},\qquad m\in\{1,2,3,\ldots\}$$

Because the target sum is only $$1000$$, any block longer than $$3$$ digits would already exceed $$1000$$ (since $$R_4=1111\gt1000$$).
Hence every block is $$1,\;11$$ or $$111$$, i.e. blocks of length $$1,\;2$$ or $$3$$.

Let
$$x_1=\text{number of 1-digit blocks},\qquad x_2=\text{number of 2-digit blocks},\qquad x_3=\text{number of 3-digit blocks}$$

Then

$$x_1+11x_2+111x_3=1000\quad -(1)$$

and the total number of digits is

$$n=x_1+2x_2+3x_3\quad -(2)$$

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From (1) write $$x_1=1000-11x_2-111x_3$$ and substitute into (2):

$$n=(1000-11x_2-111x_3)+2x_2+3x_3 =1000-9x_2-108x_3\quad -(3)$$

All variables are non-negative. Put $$x_3=k\;(0\le k\le 9)$$. Then (1) becomes

$$x_1+11x_2=1000-111k=:r_k$$

The remainder $$r_k$$ must be non-negative, so indeed $$k\le 9$$. Now $$x_2$$ can vary from $$0$$ up to

$$m_k=\Bigl\lfloor\dfrac{r_k}{11}\Bigr\rfloor$$

For each fixed $$k$$ relation (3) gives

$$n=1000-108k-9x_2,\qquad 0\le x_2\le m_k$$

Thus each admissible $$k$$ generates the set

$$S_k=\{\,1000-108k-9t\mid t=0,1,\dots ,m_k\}$$

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Every element of every $$S_k$$ is $$\equiv1\pmod9$$, because

$$1000-108k-9t\equiv 1-0-0\equiv1\pmod9$$

Write any candidate length as $$n=9q+1\;(q\ge0)$$. With $$t=\dfrac{1000-n}{9}=111-q$$, equation (3) becomes

$$t=12k+x_2,\qquad 0\le k\le9,\;0\le x_2\le m_k$$

So for each $$k$$ the attainable values of $$t$$ form the interval

$$I_k=[\,12k,\;12k+m_k\,]$$

The table of $$m_k$$ is

$$m_0=90,\;m_1=80,\dots ,m_8=10,\;m_9=0$$

Hence

$$\begin{aligned} I_0&=[0,90]\\ I_1&=[12,92]\\ I_2&=[24,94]\\ &\;\;\vdots\\ I_8&=[96,106]\\ I_9&=[108,108] \end{aligned}$$

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The union $$\bigcup_{k=0}^{9}I_k$$ covers every integer $$t$$ from $$0$$ up to $$108$$ except $$t=107$$, because
  • $$I_8$$ ends at $$106$$,
  • $$I_9$$ is the single point $$108$$.

Thus the number of attainable $$t$$ is

$$108-0+1-1=108$$

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Every admissible $$t$$ gives a unique $$n=1000-9t\;(=9(111-t)+1)$$, so exactly $$108$$ different values of $$n$$ are possible.

Therefore $$a=108$$ and the required digit-sum is $$1+0+8=9$$.

Answer: 09