A problem for math lovers to count the digits Today a classmate of mine asked a question which is based on counting.
Question.
Find a positive integer which when multiplied up to $6$ times will give numbers having the same digits but rearranged and after that will give a number with all nines. 
I have checked up to $100000$ but no such number is found. 
Please help if you could suggest any ideas. Is there any general rule?
 A: There's no need to check values up to 100000 --- there's a simple way to find the answer. You want 7 times the number to be all nines. So you want to look at the numbers 9, 99, 999, and so on, until you find one that is a multiple of 7. And the first one that works is 999999, which is 7 times 142857. Then you just have to check that the other multiples of 142857 behave the way they are asked to behave. 
EDIT: Bonus question --- what if you want the 16th addition (instead of the 6th) to give all nines?
A: Multiples of $142857$:
\begin{align}
1\times 142857=142857\\
2\times 142857=285714\\ 
3\times 142857=428571\\ 
4\times 142857=571428\\
5\times 142857=714285\\ 
6\times 142857=857142\\ 
7\times 142857=999999
\end{align}
Note. If $N=0588235294117647$, then $2N,3N,\ldots,18N$ have the same digits as $N$ cyclically rearranged (that's why the 0 in front of $N$), and $18N=999999999999999999$, and this phenomenon is related with the fact that $19$ divides $10^{18}-1$, but it does not divide $10^{k}-1$, for $k<18$, as in the case of $142857$, where $7\mid 10^6-1$, but $7\not\mid 10^k-1$, for $k<6$. 
Also, this phenomenon is present in every number system - See  Cyclic rearrangements of periods of the decimal expansions of certain rationals.
