Question

All the $7$ digit numbers containing each of the digits $1,2,3,4,5,6,7$ exactly once and not divisible by $5$ are arranged in the increasing order. Find the ${2004}^{th}$ number in this list

Answer 1

The number of $7$-digit numbers with $1$ at the left most place$=61=720$
But $120$ of these numbers end in $5$, therefore, these numbers are divisible by $5$.
Thus, The number of $7$-digit numbers with $1$ at the left most place and not divisible by $5=600$
The number of $7$-digit numbers with $3$ at the left most place and not divisible by $5=600$
These account for $1800$ numbers. Therefore the ${2000}^{th}$ must have $4$ in its left most place.
Now, The number of such $7$-digit numbers beginning with $41,42$ and not divisible by $5$is$120-24=96$ each
These account for $192$ numbers. this shows that ${2000}^{th}$ number in he list must begin with $43$.
And ${2004}^{th}$ number is $4315627$.