There is a $n$ digit number $N$ with all distinct digits and none of them being $0$. If we multiply $N$ by $1,2,3,4 \cdots n$, we get a number which has a permutation of the digits of $N$. But if we multiply $N$ by $(n+1)$, we get all $9$'s. How to find this $N$?
The solution/hint given is :
we can find $n+1=7 \Rightarrow n=6$ so the number $N$ is $142857$.
Could anybody help me in understanding this?
This problem is actually very simmilar to this $PE-52$:
It can be seen that the number, $125874$, and its double, $251748$, contain exactly the same digits, but in a different order.
Find the smallest positive integer, $x$, such that $2x, 3x, 4x, 5x,$ and $6x,$ contain the same digits.
I don't remember solving this just by remembering the property of this particular number, I think I had devised some kind brute-force search which gave me the answer, anyways it seems like there is a perfectly mathematical approach to deduce this number?!