Firstly we add all the equations. $LHS = (n-11)(a_1 + a_2 + a_3 + ... + a_n)$ $RHS = 9 + 99 + 999 + ... + 10^n -1$

When $n \neq 11$ , we can get :

$a_1 + a_2 + a_3 + ... + a_n = \frac {9 + 99 + 999 + ... + 10^n -1}{n-11}$ .

Then, we use it to minus off each equation in the system. We can get $11a_i = \frac {9 + 99 + 999 + ... + 10^n -1}{n-11} - 10^i +1$ for $i = 1, 2, 3, ... , n$ . Then, we can easily get the solutions for the system.

When n = 11, $LHS = (11-11)(a_1 + a_2 + a_3 + ... + a_{11}) = 0$ $RHS = 9 + 99 + 999 + ... + 10^{11} -1$

Clearly $0 \neq 9 + 99 + 999 + ... + 10^{11} -1$ so $LHS \neq RHS$ and 11 is the only number which the system doesn't admit a solution.

Summing all equations together, we get

$(n - 11)(a_{1} + a_{2} + ... + a_{n}) = 9 + 99 + ... + 10^{n} - 1$

$= 111..10$ ( $n + 1$ digits) $- n$

Consider two cases,

a) $n \ne 11$ ,

Then dividing both sides of the above equation by $n - 11$ and subtracting $a_{1} + a_{2} + ... + a_{n}$ from each equation, we can get the value of $a_{i}$ where $i \in$ ${1,2,3,...,n}$ and thus we have a solution.

b) $n = 11$ ,

Again,substituting $n = 11$ in the above equation we get $0 = 111111111110 - 11$ , which is obviously false.

(Note : we can substitute $n = 11$ in the above equation because it is the result of summing up the $11$ equations.)

Thus, the answer is $11$ .    Q.E.D

Let us start solving the system by Gaussian elimination. In other words, add the first row times a constant to all other rows, so that the first term of them vanishes. Do this repeatedly for the square submatrices going down diagonally to the right. For a full solution of the system, we also would keep track of the right-hand sides, but we will take care of that at the very end.

Let's call the uppermost left element -k. In the first step, k = 10 and all other diagonal elements are also -10. (I consider the matrix form of the equation system, so the element is -10. In the original system, the coefficient of the diagonal terms is -10.)

In the first step, we add $\frac{1}{k}$ times the first row to the others. This makes 0 out of the first term of every row. The -k (diagonal) terms become $-k + \frac{1}{k} = \frac{1-k^2}{k}$ and the 1 terms become $1 + \frac{1}{k} = \frac{1+k}{k}$ . We multiply each row with $\frac{k}{1+k}$ and then get rows consisting of diagonal elements -k+1 and the other non-zero elements being 1.

In other words, after the first step, the second row looks like 0 -9 1 1 1 1 ..., the third row like 0 1 -9 1 1 1 ... and so on.

We can repeat this: the third row will become 0 0 -8 1 1 1 1..., the fourth row 0 0 0 -7 1 1 1 1 ... and at the eleventh step the diagonal element becomes 0. If there are further rows (and therefore columns), we can do a pivoting operation, and rename some nonzero element to be the new diagonal element in order to continue the elimination, but if this is the last element in the original diagonal, the process ends. Let us look at the step just before: we have a $2\times 2$ matrix

-1 1

1 -1

These two rows are linearly dependent (naturally, since the next step yields zero), and the question is now what has happened to the right-hand sides during the previous steps. During each step, the multiplicative factor has been multiplied with the right-hand side and been added to all the following right-hand sides. So in each step, the last two right-hand sides have received the same additions, and so, since they started out unequal, they are still unequal. Then there is no solution to the system we have arrived at.

So, if we start with a matrix of size $11 \times 11$ , we will arrive at a contradiction and the system has no solution.

(Sorry for not using the LaTeX array display technique in composing this text.)

Adding up all the equations we have $(n-11)(a_1+a_2+.....a_n) = 10+10^2+10^3+....+10^n -n$ $a_1+a_2+.....a_n = \frac{10(10^n-1)-9n}{9(n-11)}$ We can see that the r.h.s is only defined for $n \neq 11$ . So for $n=11$ , the system doesn't admits a solution.

Imagine, given some $n$ , how to actually solve the system of equations. One of the easiest ways would be to add up all the equations and divide by $n-11$ . This will result in an equation with a coefficient of one for each variable. From there, we could take this new equation and subtract the first equation to obtain an equation with only $a_1$ , which can be easily solved to find the value of $a_1$ . Similarly, we can subtract the second equation to get the value of $a_2$ , and so on. Using this method, we can solve this system for any $n$ , with one important exception. When we divided the sum of the equations by $n-11$ , we had to assume that $n-11\neq0$ , so when $n=11$ , the equations are unsolvable.

Notice that we have a very symmetric system of equations here. In each "column", we have $n-1$ terms of the $a_i$ variable, as well as the $-10$ term, for a total of $(n-11)a_i$ . Since each variable has this coefficient, we can factor it out. In order for the system to have no solutions, this coefficient must be 0, which would yield a sum of 0 on the LHS and a positive sum on the RHS. The only value for which this is possible is $n=\boxed{11}$ .

We express a 1 + a 2 + a 3 + ... - 10 \cdot a n as a 1 + a 2 + a 3 + ... + a n - 11 \cdot a_n.

When n is assigned, the sum of all linear equations is as given: (n \cdot a 1 + n \cdot a 2 + n \cdot a 3 + ... + n \cdot a n) + (- 11 \cdot a 1 - 11 \cdot a 2 - 11 \cdot a 3 - ... - 11 \cdot a n) = 9 + 99 + 999 + 10^n-1 > 0 \rightarrow (n-11) \cdot a 1 + (n-11) \cdot a 2 + (n-11) \cdot a 3 + ... + (n-11) \cdot a n > 0

When we assign n = 11, the following property should also be true: (11-11) \cdot a 1 + (11-11) \cdot a 2 + (11-11) \cdot a 3 + ... + (11-11) \cdot a n > 0 \rightarrow 0 > 0 thus, we have a contradiction.

Therefore, when n=11, the system doesn't admit a solution.

here we construct a augmented matrix [-10 1 1 1........1 ! 9 ] [ 1 -10 1 1........1 ! 99 ] [ 1 1 -10 1.........1 ! 999] [ .. .. .. .. ......... ! .......] [1 1 1 1 -10 ! 10^n-1] now rank of augmented matrix and rank of coefficient are different if n = 11. therefore the system of equation is inconsistent if n=11 .

Adding all the equations together, we get:

$\sum_{i=1}^n (a_1(n-1)-10a_i)=9+99+999+9999+...+10^n-1$

If the left side is 0, equality cannot occur, since the right side is clearly positive. The case where the left side is 0 is when $n=11$ .

Thus, the system of equations is inconsistent and does not have a solution.

Consider the system of equations in matrix form $Aa = b$

\[ \begin{pmatrix} -10 & 1 & 1 & \ldots & 1 \\ 1 & -10 & 1 & \ldots & 1\\ 1 & 1 & -10 & \ldots & 1\\ \vdots & & & \ldots & \vdots\\ 1 & 1 & 1 & \ldots & -10 \\ \end{pmatrix} \begin{pmatrix} a_1\\ a_2\\ a_3\\ \vdots\\ a_n\\

\end{pmatrix}

\begin{pmatrix} 9\\ 99\\ 999\\ \vdots\\ 10^n-1 \\ \end{pmatrix} \]

Let $x_j$ with $j \geq 2$ be an $n$ -dimensional vector with 1 in the 1st coordinate, -1 in the jth coordinate, and 0 in the other coordinates. Then $x_j$ is an eigenvector of $A$ with eigenvalue -11. Also, $x_1= (1, 1, \ldots, 1)$ is an eigenvector with eigenvalue $-10 + n-1$ . Since these eigenvectors are linearly independent, $\det(A)= (n -11 ) (-11)^{n-1}$ . Thus $A$ is singular (determinant is 0) if and only if $n=11$ . For $n = 11$ , the range space would be a subspace of vectors whose coordinate sum is 0, hence there are no solutions to the given equation.

Note: The determinant of matrix A can be calculated from elementary row operations (add each row to the first, pull out common factor of $n - 11$ , then subtract each column by the first and expand along the first row). However, to show that no solution exists, we would still need to determine the vector that gets mapped to 0, which is $x_1$ .