<s>[INST] Decimal pattern in division of two digit numbers by 9 Can some one explain how this is possible ?
    1)  13 / 9 = 1.(1 + 3) = 1.444 ...
    2)  23 / 9 = 2.(2 + 3) = 2.555 ...
    3)  35 / 9 = 3.(3 + 5) = 3.888 ...
    4)  47 / 9 = 4.(4 + 7) = 4.(11) → 4.(11 - 9) = 5.222 ...
    5)  63 / 9 = 6.(6 + 3) = 6.(9)  → 6.(9 - 9)  = 7.0 = 7

 [/INST] Let $a$ be the tens digit and $b$ be the ones digit so that our two digit number is $10a+b$. In general, the pattern seems to be:
$$
\dfrac{10a+b}{9} = (a+j).kkk...
$$
where:
$$(j,k)=
\begin{cases}
(0,a+b) & \text{if }a+b \in \{0,1,2,...,8\} \\
(1,a+b-9) & \text{if }a+b \in \{9,10,11,...,17\} \\
(2,a+b-18) & \text{if }a+b =18 \\
\end{cases}$$
In other words, $j$ and $k$ are the quotient and remainder (respectively) upon dividing $a+b$ by $9$. To see why this makes sense, recall that $\boxed{0.kkk... = k/9}$. This can be proven a number of ways (for example, it is a convergent geometric series). Hence, observe that:
$$
\dfrac{10a+b}{9} =\dfrac{9a+(a+b)}{9}=a+\dfrac{a+b}{9}=a+j+\dfrac{k}{9} = (a+j).kkk...
$$
</s>