$z/27=x$, Digital root of the first three digits of the decimal places of x = digital root of z. Why? I noticed when playing around with numbers that the digital root of the first 3 digits in the decimal portion of the answer to $z/27$ is equal to the digital root of z itself (z is an integer)
For example: 
$329833/27 = 12216.037037$
The first three digits of the decimal portion of this answer are 0, 3, and 7. Their digital sum is "1" because 0+3+7=10-->1+0=1
The number we are dividing by, 329833, also has a digital sum of "1". 
329833-->3+2+9+8+3+3=28-->2+8=10-->1+0=1
This is the case for all* integers z.

Why is this? Can it be proven algebraically? I discovered this back in 2011 and never thought to ask here until just now. 

*Exceptions sort of include numbers evenly divisible by 27, see comments in this answer for more: https://math.stackexchange.com/a/1656503/56935
 A: What you are aiming to prove is that if the remainder of $n$ divided by $9$ is $r$, then the sum of the first three digits divided by $9$ is also $r$.
To use your example, $329833$ divided by $9$ has a remainder of $1$, and in $0.037 \dots$ , note that $3+7$ divided by $9$ is also $1$. 
In your other example, $27$ divided by  $9$ is $0$, or $9$ if you prefer. Note that the digit sum of the first three digits of $1.0000...$ is $0$. 
Note the fact that $\frac{9k+r}{27}$ is $(0.333\dots)k+\frac{r}{27}$, where  $1 \le r \le 8$
Note that the sum of the first three digits of the decimal representation of $(0.333\dots)k$ are always divisible by $9$.
For $1 \le r \le 8$, Note that the first three digits of $\frac{r}{27}$ are of the form $111m+37k$, where $r$ is $3m+k$ and $1 \le k \le 2$.  
This implies the sum of the first three digits of the decimal representation of $\frac{n}{27}$ divided by $9$ is $3m+10k \equiv 3m+k=r \pmod 9$. Our proof is done. 
A: This is not true for $z=27$:
$$\frac{27}{27} = 1.00000000,$$ so the digital sum of the decimal portion is $0$. On the other hand, the number we are dividing, $27$, has a digital sum of $9$.
