goodgirlnotbad wrote:
We are supposed to make digits divisible by 5. The numbers are less than 1000
So we can form 1 digit, 2 digit and 3 digit numbers
Starting with 1 digit numbers: there is only one 1 digit number divisible by 5 and that is the number 5. Therefore number of such digits = 1
Now 2 digit numbers can have either 0 or 5 in their units place, so lets make two cases
case1: 0 in units place x 0 where x can be any of 1,2,3,4,5,6,7,8,9. So the number of ways of filling the tens digit = 9 and number of ways of filling the units digit = 1. That makes 9*1 such combinations
case2: 5 in units place x 5 where x can be any of 1,2,3,4,6,7,8,9. So the number of ways of filling the tens digit = 8 and number of ways of filling the units digit = 1. That makes 8*1 such combinations
Thus total 9 + 8 = 17 two digit numbers divisible by 5 and less than 1000
3 digit numbers can again have either 0 or 5 in their units place, making two cases again
case1: 0 in units place x y 0 where x can be any of 1,2,3,4,5,6,7,8,9. So the number of ways of filling the tens digit = 9, y will be any number of the list except for number x, thus number of ways of filling the tens digit = 8 and number of ways of filling the units digit = 1. That makes 9*8*1 = 72 such combinations
case1: 5 in units place x y 0 where x can be any of 1,2,3,4,6,7,8,9. So the number of ways of filling the tens digit = 8, y will be any number of the list except for number x but it could also be 0, thus number of ways of filling the tens digit = 8 and number of ways of filling the units digit = 1. That makes 8*8*1 = 64 such combinations
Thus total 72 + 64 = 136 three digit numbers divisible by 5 and less than 1000
So the total such numbers are 1+17+136 = 154
Starting with 1 digit numbers: there is only one 1 digit number divisible by 5 and that is the number 5.
Therefore number of such digits = 1but not only one digit but also 0 and 5(2 digit) both are also can be divide by 5.