how many four digits numbers are there that is divisible by 5, repetition not allowed

Question

use 0 to 9 digits

Answer 1

"Four digit numbers" does not allow leading zeros, so consider 1000 - 9999.

 

If repetition is allowed, then every number ending with either 0 or 5 is divisible by five.  There are 9000 values and 2 out of 10 (or 1 out of 5) of them are divisible by 5, so 9000/5 = 1800 are divisible by 5 -- if repetition is allowed.

 

However, if repetition of digits is not allowed, we have:

   4th digit must be either 0 or 5  --  2 choices

   1st digit may be (1,2,3,4,5,6,7,8,9) -- 9 choices

   2nd digit may be (0,1,2,3,4,5,6,7,8,9) -- 10 choices, but not same as 1st digit or 4th digit

   3rd digit may be (0,1,2,3,4,5,6,7,8,9) -- 10 choices, but not same as 1st or 2nd digit or 4th digit

 

So, since the 1st digit cannot be 0 but can be 5, there are:

  If the last digit is 0:

   9 * 8 * 7 * 1 = 504 if the 4th digit is 0, the 1st digit can be (1,2,3,4,5,6,7,8,9), the 2nd digit can be (1,2,3,4,5,6,7,8,9 but not same as 1st digit) and the 3rd digit can be (1,2,3,4,5,6,7,8,9 but not same as 1st digit or 2nd digit).

  If the last digit is 0:

   8 * 8 * 7 * 1 = 448 if the 4th digit is 5, the 1st digit can be (1,2,3,4,6,7,8,9), the 2nd digit can be (1,2,3,4,6,7,8,9 but not same as 1st digit) and the 3rd digit can be (1,2,3,4,6,7,8,9 but not same as 1st digit or 2nd digit).

 

That gives a total of 504+448 = 952 numbers that are divisible by 5 with no repeating digits.

Answer 2

The numbers under consideration would be 1000 through 9999, inclusive.

 

The first of these (1000) is evenly divisible by 5, and every fifth one is, also. So we can consider 1000/5 = 200, 1005/5 = 201, etc. up through 9995/5 = 1,999...all the integer quotients.

 

if we look at these integer quotients (w/o remainder when ÷ by 5) we have the means of counting, to give your answer.

200 ,201, ...through 1995 is 1999-199 different results, a total of 1800.