Permutation problem!

[freeeeknik]

Determine how many numbers bigger than 30,000 and smaller than 9,999,999 and divisible by 5 can be formed using the digits 0, 1, 2, 3, 5, 8 and 9.

The answer is 156,407. Please help on the approach to get this answer!

Thank you.

 

[freeeeknik]

This is what I have tried:
total 5 digit numbers possible that will be divisible by 5 = 27776 (unit place on the left) total 6 digit numbers possible that will be divisible by 5 = 277776 total 7 digit numbers possible that will be divisible by 5 = 277777*6
Then I added all these answers & subtracted those 5 digit numbers lesser than 30,000 and divisible by 5
Then I subtracted 1 (9,999,999 is not to be included).
The answer i get does not match with the one on the book.

 

[Dr.Peterson]

This is what I have tried:
total 5 digit numbers possible that will be divisible by 5 = 27776 (unit place on the left) total 6 digit numbers possible that will be divisible by 5 = 277776 total 7 digit numbers possible that will be divisible by 5 = 277777*6
Then I added all these answers & subtracted those 5 digit numbers lesser than 30,000 and divisible by 5
Then I subtracted 1 (9,999,999 is not to be included).
The answer i get does not match with the one on the book.

Is that what you meant to say? Are the italic 7's supposed to be something else?

And what is your answer?

 

[pka]

Determine how many numbers bigger than 30,000 and smaller than 9,999,999 and divisible by 5 can be formed using the digits 0, 1, 2, 3, 5, 8 and 9.
The answer is 156,407. Please help on the approach to get this answer!

Here are some quick observations. Any number that is divisible by five must end in zero or five.
Therefore, the largest number we want is \(9,999,995\).
Using just those seven digits, there are \(7^6\cdot 2=235,298\) multiples of five less than \(9,999,999~.\)
From that collection we must remove all multiples of five less than \(30,005\).
I get \(3\cdot 7^3\cdot 2+1\) to be removed.