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.
Oh, dude, let me break it down for you. So, to find the number of 3-digit multiples of 5, you just gotta figure out the first and last numbers that are multiples of 5 in the 3-digit range, which are 100 and 995. Then you just do some simple math, like subtracting and dividing, and voila, you'll get your answer. It's like counting candy, but with numbers.
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.
There are 200 of them. Starting with 45*23 = 1035 and all the multiples of 45 up to 45*222 = 9990.
Multiples of 10 between 1000 and 9990 Any four-digit number ending in zero.