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.
This is what I have tried:
- total 5 digit numbers possible that will be divisible by 5 $= 2 \cdot 7 \cdot 7 \cdot 7 \cdot 6$ (unit place on the left)
- total 6 digit numbers possible that will be divisible by $5 = 2\cdot7\cdot7\cdot7\cdot7\cdot6$
- total 7 digit numbers possible that will be divisible by 5 $= 2\cdot7\cdot7\cdot7\cdot7\cdot7\cdot6$
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.
You were on the right track.
5 digits numbers
We will keep only numbers bigher than $30\ 000$. There is $2$ possibilities for the unit, $7$ for the middle digits and $4$ for the leftmost digit. And we will substract one for $30\ 000$. $$2\times7\times7\times7\times4-1=2743$$ 6 digits numbers
You were right there is $2$ possibilities for the unit, $7$ for the middle digits and $6$ for the leftmost digit. $$2\times7\times7\times7\times7\times6=28812$$ 7 digits numbers
You were right there is $2$ possibilities for the unit, $7$ for the middle digits and $6$ for the leftmost digit. $$2\times7\times7\times7\times7\times7\times6=201684$$
Adding those numbers, we find $233\ 239$ possibilities.
There might be a mistake in the book, or there might be something missing in the problem.