60 numbers
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45 multiples of 2 plus 30 multiples of 3 minus 15 multiples of 6 equals 60 numbers
The 3-digit counting numbers are 100 through 999 = 900 numbers.Half them are multiples of 2 (even numbers).The other half are not . . . 450 of them.
There are 720 of them.
8100
- There are an infinite number of "counting numbers" that are greater than 70. - There are also an infinite number of "counting number" that are multiples of 10. So.... since you used an "OR" statement, this mean how many number are true for both statements above. That would be: AN INFINITE NUMBER of counting numbers. In fact, if you had said "AND", it still would be an infinite number: 80, 90, 100, ... and so on FOREVER. * * * * * The above answer has interpreted the questions as "two-digit counting numbers greater than 70" OR "a multiple of ten". Apart from the fact that there are not an infinite number of two-digit counting numbers greater than 70, the answer would be correct. But the answer could be interpreted as "two digit counting numbers" that are "greater than 70" OR "a multiple of ten". In that case, the first set is {71, 72, ... 99} and the second is {10, 20, 30, ... 90} with an intersection set consisting of {80 and 90} So there are 29 + 9 - 2 = 36 such number.
45 multiples of 2 plus 30 multiples of 3 minus 15 multiples of 6 equals 60 numbers
The 3-digit counting numbers are 100 through 999 = 900 numbers.Half them are multiples of 2 (even numbers).The other half are not . . . 450 of them.
There are 450 of them.
There are 720 of them.
8100
81
4500 of them.
There are 898 three-digit even numbers. Nine of them are multiples of 55. That leaves 889 * * * * * There are 450 three-digit even numbers and 17 of them are multiples of 55. So that leaves 433.
There are 6,750 such numbers.
There are 9*10*9 = 810 such numbers.
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.
I believe there are 81. There are 9 after every multiple of 10, including 90.