All the 7-digit numbers containing each of the digits 1, 2, 3, 4, 5, 6, 7 exactly once, and not divisible by 5 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 2000-th number in this list.
My try:
The number of 7-digit numbers with 1 in the left most place and containing
each of the digits 1, 2, 3, 4, 5, 6, 7 exactly once is 6! = 720.The number of 7-digit numbers with 1 in the left most place and containing
each of the digits 1, 2, 3, 4, 5, 6, 7 exactly once is 6! = 720.

The number of 7-digit numbers with 1 in the left most place and containing
each of the digits 1, 2, 3, 4, 5, 6, 7 exactly once is 6! = 720. But 120 of these end in 5 and
hence are divisible by 5. Thus the number of 7-digit numbers with 1 in the left most place
and containing each of the digits 1, 2, 3, 4, 5, 6, 7 exactly once but not divisible by 5 is 600.
Similarly the number of 7-digit numbers with 2 and 3 in the left most place and containing
each of the digits 1, 2, 3, 4, 5, 6, 7 exactly once but not divisible by 5 is also 600 each. These
account for 1800 numbers. Hence 2000-th number must have 4 in the left most place.
Again the number of such 7-digit numbers beginning with 41,42 and not divisible by 5 is
120 − 24 = 96 each and these account for 192 numbers. This shows that 2000-th number
in the list must begin with 43.
The next 8 numbers in the list are: 4312567, 4312576, 4312657, 4312756, 4315267, 4315276,
4315627 and 4315672. Thus 2000-th number in the list is 4315672.
Thanks for the hint @almagest
