Move the last digit of $10k+9$ to the front, and obtain $70k+63$ "N" is a number and its last digit is 9. Delete the last digit (9) and write (9) to first digit, new number is 7 times "N". so whats that "N" ?
Could anyone show me a way, how to solve such tasks?
EDIT:
These are my thoughts so far:
N := 10 x + 9 ( 1 )
where 10 x is the ( integer ) value of anything but the last 9. Now put the 9 in front:
N ' := 9 * 10 ^ ( d - 1 ) + x = 7 N = 70 x + 63 ( 2 )
but i do not know how to continue from here
EDIT2:
d is the number of digits of N.
 A: You have made a good start.  Your equation 2 is $9\cdot 10^{d-1}+x=70x+63$ or $x=\frac 9{69}(10^{d-1}-7)$ so you need $10^{d-1} \equiv 7 \pmod {69}$  Now you can just start looking.  Increase $d$, multiply by $10$ and reduce the result $\pmod {69}$  It starts like this $$\begin {array}{c | c}d-1&10^{d-1}\\ \hline 1&10\\2&10\cdot 10 =100 \equiv 31 \pmod {69}\\ 3&10 \cdot 34 = 310 \equiv 34 \pmod {69}\end{array}$$  Keep going until you hit $7$
A: Let us call the original number $N$, and the modified number $M$.  Thus $N$ ends with a 9 and $M$ begins with a 9.  Since $N = M\div 7$, we can set up a short division problem that looks like this:
$$\require{enclose}
\begin{array}{rl}
& \ \,1\ldots\\
7&\enclose{longdiv}{9\ldots}
\end{array}$$
Here $N$ is the quotient and $M$ the dividend.  We see that $N$ begins with the digit 1, so this is the second digit of $M$:
$$\begin{array}{rl}
& \ \,1\ldots\\
7&\enclose{longdiv}{91\ldots} 
\end{array}$$
Now that we know the second digit of $M$, we can divide one more step, computing the next digit of $N$:
$$\begin{array}{rl}
& \ \,13\ldots\\
7&\enclose{longdiv}{913\ldots} \\ \\
& \ \,130\ldots\\
7& \enclose{longdiv}{9130\ldots} \\ \\
& \ \,1304\ldots\\
7& \enclose{longdiv}{91304\ldots} \\ \\
\end{array}$$
We must continue in this fashion until we reach a point where there the remainder is 0 and the quotient ends with 9. This first occurs at:
$$\begin{array}{rl}
& \ \,1304347826086956521739\\
7& \enclose{longdiv}{9130434782608695652173} \\ \\
\end{array}$$
This is the smallest answer. By continuing to divide until the next stopping point, we can reach larger solutions, such as 91304347826086956521739130434782608695652173.
A: We require 
$$y=7x$$
or in other words
$$9a_1a_2\dots a_n = 7(a_1a_2\dots a_n 9)$$
where the $a_i$ represent decimal digits.
Multiplying both sides by $10$, we have
$$9a_1a_2\dots a_n 0 = 70x$$
or
$$9\times 10^{n+1} + (x - 9) = 70x$$
or
$$69x = 9\times 10^{n+1} - 9\,.$$
The answer is not unique. For example, $$x = 1,304,347,826,086,956,521,739$$ and $$x = 13,043,478,260,869,565,217,391,304,347,826,086,956,521,739$$ both have the desired property (these correspond to $n=21$ and $n=42$, respectively.
