How Many Numbers Less Than 1000 Divisible by 5 Are Formed Using Unique Digits?

[mathdad]

The total number of numbers less than 1000 and divisible by 5 formed with 0,1,2,...9 such that each digit does not occur more than once in each number is what?

Solution:Divisible by 5 ==> number ending in 0 or 5.

Number of ways with no repeated digit:
[0, 9]---> ends in 5 = 1 way (only 5 here).

[10,99]
————> ends in 0 = 9 ways
————> ends in 5 = 8 ways (cannot use 55 )
————> 9 + 8 = 17 ways for two digit numbers in total

[100,999]
————> ends in 0 {once 0 is selected you are left with 9 digits)
{ 9 ways to select the 1st digit and 7 ways to select the 2nd digit}
9 x 8 = 72 ways
————> ends in 5 (cannot start with 0 but can use 0 for 2nd digit)
{8 ways to select 1st digit and 8 ways to select 2nd digit)
8 x 8 = 64 ways
————> 72 + 64 = 136 ways for three digit numbers in total

Number of numbers = 1 + 17 + 136 = 154 numbers

Is this right?

 

[Evgeny.Makarov]

Yes, this is right if you don't count 0.