1000, 1005, 1110, 1115 and just keep adding five until you get to 9995.
-3
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.
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.
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
Multiples of 10 between 1000 and 9990 Any four-digit number ending in zero.
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.
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
5*5*4*4 = 400
There are an infinite number of multiples of 5
Multiples of 30 from 120 to 990
lcm(2, 3, 5) = 30 → 2 digit common multiples are 30, 60, 90.