All of the whole numbers from 10 to 99.
Since a binary digit has only two possible values, each digit bears less information than in decimal, where each digit can have ten different values.
There are 21 two-digit prime numbers.
Three hundred sixty. 360. This was figured using math, not counting each possible combination.
16000
81
4500 of them.
99
60 numbers
There are 6,750 such numbers.
11 and 88
All of the whole numbers from 10 to 99.
I believe there are 81. There are 9 after every multiple of 10, including 90.
It is 2,500,300,005 in numbers. Not quite sure what nunbers are, though ;)
- 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.
15 and 30
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.