Finding the remainder after division by $3$ is simple. What about the quotient? A well-known simple trick for obtaining the REMAINDER after division by 3 is just the digital root of the dividend after division by 3, the digital root itself further simply obtained by first casting out 9's, then every digit pair that add up to 9, etc.


  In the same vein, is there a short-cut method to rapidly obtain the QUOTIENT after division by 3?
Not sure if we can call it shortcut,

in base $10,$
Using Why is $a^n - b^n$ divisible by $a-b$? 
$$\dfrac{\sum_{r=0}^n10^ra_r}3=\sum_{r=0}^n\dfrac{10^r-1}3\cdot a_r+\dfrac{\sum_{r=0}^na_r}3$$

$$=\dfrac{\sum_{r=0}^na_r}3+\sum_{r=0}^n\underbrace{33\cdot33}_{r\text{ digits }}\cdot a_r$$
Not sure if we can call it shortcut,

in base $10,$
Using Why is $a^n - b^n$ divisible by $a-b$? 
$$\dfrac{\sum_{r=0}^n10^ra_r}3=\sum_{r=0}^n\dfrac{10^r-1}3\cdot a_r+\dfrac{\sum_{r=0}^na_r}3$$

$$=\dfrac{\sum_{r=0}^na_r}3+\sum_{r=0}^n\underbrace{33\cdot33}_{r\text{ digits }}\cdot a_r$$