permutation (retrictions on arrangements) numbers

Question

a list of questions :D

1) how many four digit numbers are divisible by 5 if no two digits are the same?

My solution: 9*8*7*2=1008

2)From the digits 0,1,3,5,7,9 how many four digit numbers divisible by 5 can be written if no digit is repreated?

My solution=5*4*3*2

3) Discuss, without proceeding to a solution, how you would determine how many four digit numbers divisible by 3 could be made using any of the digits 0,2,4,6,8

my solution: Unknown but the knownledge that i have is that if the digits add up together is divisible 3 then it is divisible by 3 (e.g. 5211 ->5+2+1+1 is divisible by 3)

4) how many different arrangements are there of the keys of a calculator "1,2,3,4,5,6,X" taken three at a time. My solution: 7P3=210

Discuss how many different result are there when three of the keys are pressed followed by "=" without proceeding to a solution.

Answer 1

1. If the number ends in a $5$ : we have, $8 * 8 * 7 = 448$ such numbers
   If the number ends in a $0$ : we have, $9*8*7=504$ such numbers
   This gives us a cumulative of $952$ numbers which satisfy the given criteria

2. From the digits $0,1,3,5,7,9$ the four digit numbers (say ABCD), that end with $0$ are = $5*4*3=60$
   Number of required numbers that end with $5$ are $4*4*3 = 48$
   Hence total number of such numbers = $108$

3. We look at the $5$ different combinations of $4$ digits. Of the combinations, only $0,2,4,6$ and $0,4,6,8$ sum up to a number divisible by $3$.
   We now just need to find the sum of the number of numbers possible using the digits $0,2,4,6$ and $0,4,6,8$ and this should give us the result.