That would be all the numbers that end with 0 or 5. So:* For the first digit, you have 9 options (digits 0-9).
* For the second digit, you have 10 options.
* For the third digit, you have 10 options.
* For the fourth digit, you have 2 options (the digits 0 or 5).
You can multiply all those together.
There are an infinite number of multiples of 5
lcm(2, 3, 5) = 30 → 2 digit common multiples are 30, 60, 90.
The integers divisible by those 5 numbers are exactly the multiples of 1008. The largest multiple of 1008 which is still a four-digit number is 9*1008 = 9072.
There are 7200 such numbers.
If the digits can be repeated, then there are (5 x 5 x 5 x 5) = 625 different four-digit numbers.If the digits can't be repeated, then there are (5 x 4 x 3 x 2) = 120 different four-digit numbers.
99
-4
2250
There are 720 of them.
Multiples of 10 between 1000 and 9990 Any four-digit number ending in zero.
1000, 1005, 1110, 1115 and just keep adding five until you get to 9995.
There are an infinite number of multiples of 5
5*5*4*4 = 400
Multiples of 30 from 120 to 990
lcm(2, 3, 5) = 30 → 2 digit common multiples are 30, 60, 90.
There are 2000 4-digit numbers that are multiples of 5, so, instead of listing them all, it is equally valid to say: Any4-digit number whose final digit is either a 5 or a 0 is a multiple of 5. Get Right? :P
NO. All multiples of 5 have a final digit of 0 or 5. Therefore 1001 with its final digit of 1 is NOT a multiple of 5.