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
5*5*4*4 = 400
All multiples of 5 are numbers ending in either 5 or 0.
5! = 5 x 4 x 3 x 2 x 1 = 120
120
Any three numbers, a, b, and c, which satisfy the equation a2 + b2 = c2 will form the sides of a right triangle. Some common values are 3, 4, 5 (and all multiples), 5, 12, 13 (and all multiples), and 7, 24, 25 (and all multiples).
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.
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.
2250
There are 720 of them.
-4
There are 13 two-digit multiples of 7.
5
The first digit can have 5 possible numbers, the second digit can have 4, the third 3, the fourth 2. 5
there aren't any as otherwise they wouldn't be prime numbers
There are 9000 4-digit numbers. 8*9*9*9 = 5832 of them do not contain a 5 The remaining 3168 contain a 5.
There are 200 of them. Starting with 45*23 = 1035 and all the multiples of 45 up to 45*222 = 9990.
There are no numbers that fulfill that request.