1000, 1005, 1010, 1015, 1020, 1025 and just keep adding 5 until you get to 9995.
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
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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.
The main generalization you can make about all multiples of 5 would be that they will either end in the digit 0 or 5. So 5, 487975, 100, and so on all are multiples of 5. I don't believe there are any other generalizations that can be made about it.
There are 720 of them. The three digit counting numbers are 100-999. All multiples of 5 have their last digit as 0 or 5. There are 9 possible numbers {1-9} for the first digit, There are 10 possible numbers {0-9} for each of the first digits, There are 8 possible numbers {1-4, 6-9} for each of the first two digits, Making 9 x 10 x 8 = 720 possible 3 digit counting numbers not multiples of 5.
To determine the number of 3-digit numbers that are multiples of 5, we need to find the first and last 3-digit multiples of 5. The first 3-digit multiple of 5 is 100, and the last 3-digit multiple of 5 is 995. To find the total number of such multiples, we can use the formula (Last - First) / 5 + 1 = (995 - 100) / 5 + 1 = 180. Therefore, there are 180 3-digit numbers that are multiples of 5.
70 and 75
Because 20 is a multiple of 4. All multiples of 20 are also multipples of 2, 5 and 10
Any and all multiples of 60.
Multiples of 30 from 120 to 990
lcm(2, 3, 5) = 30 → 2 digit common multiples are 30, 60, 90.