Rotate the integer to the left...

Let the number be of the form (10x+y) where 'y' is the last digit and 'x' is the rest of the number. Also let the number of digits of the number be 'n' Therefore, according to the question: $\frac{2×(10x+y)}{3}=y×10^n+x$ $→20x+2y=3y×10^n+3x$ $→17x=3y×10^n-2y$ $→17=\frac{(y×(3×10^n - 2))}{x}$

Since y can only have values between 0 to 9, $(3×10^n - 2)$ has to be divisible by 17

The least value of n for which this is possible is 15

→ $\frac{x}{y}=\frac{(3×10^{15} - 2)}{17}$

→y=1

→x=176470588235294

Therefore, the number is 10x+y= 1764705882352941

$Say,\quad N=\sum _{ i=0 }^{ n } a_{ i }10^{ i }=\sum _{ i=1 }^{ n } a_{ i }10^{ i }+a_{ 0 }\\ Rot(N)=10^{ n }a_{ 0 }+\sum _{ i=1 }^{ n } a_{ i }10^{ i-1 }\\ 10\cdot Rot(N)=10^{ n+1 }a_{ 0 }+\sum _{ i=1 }^{ n } a_{ i }10^{ i }\\ 10\cdot Rot(N)-N=a_{ 0 }(10^{ n+1 }-1)\\ 10\cdot \frac { 2 }{ 3 } \cdot N-N=\frac { 17 }{ 3 } N=a_{ 0 }(10^{ n+1 }-1)\\ N=\frac { 3a_{ 0 }(10^{ n+1 }-1) }{ 17 } \\ Now,\quad 0\leq a_{ 0 }\leq 9\quad \Rightarrow 17\nmid 3a_{ 0 }\quad \Rightarrow 17|10^{ n+1 }-1$ The lowest number of the form $10^{ a }-1$ which is divisible by $17$ is $10^{ 16 }-1$ . Also, $a_{ 0 }$ is at least $1$ . (because, $a_{ 0 }=0 \Rightarrow N = 10\cdot Rot(N)$ ) $So\quad N=\frac { 3(10^{ 16 }-1) }{ 17 }=1764705882352941$

Since the main digit that plays a role in the rotation is the last digit, we can group the rest of $N$ as one number. Let this last digit of $n$ -digit number $N$ be $x$ , and the rest of $N$ be $Y$ . Then we have $N=Y+x$ and from the rotation information, we have $N=Y+x=\frac{3}{2}\left(10^{n-1}x+\frac{Y}{10}\right)$ $2Y+2x=30\cdot 10^{n-2}x+\frac{3Y}{10}$ $\frac{17}{10}Y=2\cdot \left(15\cdot 10^{n-2}-1\right)x$ Notice that the LHS is divisible by $17$ , so this must be the case on the RHS as well. Furthermore, $0<x<10$ , so we must have $17\mid 15\cdot 10^{n-2}-1$ . This is simple Modular Arithmetic: $15\cdot 10^{n-2}\equiv -2\cdot 10^{n-2}\equiv 1\pmod{17}$ $10^{n-2}\equiv 8\pmod{17} \implies n-2=14$ so the smallest possible $N$ is a $16$ digit integer. Now that we know this, let $Y=a\cdot 10^{15}+b\cdot 10^{14}+\cdots + 10\cdot o$ , then from the previous equation, we have $17\left(a\cdot 10^{14}+b\cdot 10^{13}+\cdots + o\right)=2\left(15\cdot 10^{14}-1\right)p\le 2\left(15\cdot 10^{14}-1\right)=176470588235294$ achieved when $p=1$ . This means that $N=\boxed{1764705882352941}$

It is easy to check that the "decimal parts" of $\frac{i}{7}\quad (0\leq i\leq 6)$ do not yield any solution. So, we try with "decimal parts" of $\frac{i}{17}\quad (0\leq i\leq 16)$ . Since the integer obtained after rotation is $\frac 2 3$ of the original number, $3\mid i$ . We see that the "decimal part" of $\frac{3}{17}$ is the required answer.

$\textbf{Bonus:}$ Can you tell why I used 7 and 17 in the denominator?