The two digits

We note the position from the left of the $k$ th $a$ is given by $n_a(k) = T_k + k = \dfrac {k(k+1)}2+k = \dfrac {k(k+3)}2$ , where $T_n$ is the $n$ th triangular number. When $k=12$ , we have $n_a(12) = \dfrac {12\times 15}2 = 90$ . This confirms that the last digit is $a$ and that there are 12 $a$ 's and the number of 2's is $90-12= 78$ .

Let $N = \overline{2a22a222a\cdots a}$ . If $N$ is divisible by 9, its sum of digits must be divisible by 9. The digital sum $s_d = 12a+78\times 2 =12a + 156$ . Therefore, we have:

$\begin{aligned} N & \equiv 0 \text{ (mod 9)} \\ 12a + 156 & \equiv 0 \text{ (mod 9)} \\ 3a + 3 & \equiv 0 \text{ (mod 9)} \\ 3(a + 1) & \equiv 0 \text{ (mod 9)} \\ \implies a & = \boxed{2, 5, 8} \end{aligned}$

if we group the 2s and a's into pairs like

2a 22a 222a...

we will see that the terms per group progressively increase by 1.

(1) since we need to find divisibility of 9, we should follow the rule first: getting the digit sum.

we also get the idea of sum of first n positive integers which is n(n+1)/2. (since we know about the before mentioned statement (1) )

but in the line of numbers, the "1" is missing

so, plugging that in...

90+1=n(n+1)/2

91 = n(n+1)/2

n = 13

we can conclude that the last group of the string has 13 digits.

thus, there are 13-2+1=12 groups in our string.

in each group, we have 1 a and the rest consists of 2s.

we now have :

2 + a

2 + 2 + a

...

2 + 2 + 2... + a (12 2s)

thats 12(13)/2 2s and 12 a's.

the sum of the 12(13)/2 2s is 12(13)(2)/2=12(13).

the digit sum now is:

12(13) + 12a

=12(13+a)

this such number MUST be divisible by 9.

12(13+a)/9

=4(13+a)/3

since 4 is not divisible by 3, 13+a must be divisible be 3.

if we check and observe, a=2,5,8

As the number is divisible by 9 Therefore the sum of all the digits is divisible by 3

There are 12 a in the series

Therefore there are 78 "2"

The sum of all the digits is 12a+78*2

8(3a+39) must be a multiple of 9

3a+39 must be a multiple of 9

3(a+13) must be a multiple of 9

Therefore a+13 must be a multiple of 3

Therefore possible values of a are {2,5,8}

Hence the answer is {2,5,8}