Yes, that is correct.
All integers have an infinite amount of multiples.
No as 5, 15 and an infinite amount of other multiples of 5 are odd
Yes! Any multiple of 11 that is in the form of 11x10^n +209 will have an odd digit sum and will be divisible by 11. 209 is divisible by 11, and 11X10^n is too. Remember that n has to be over 1,000 for it too work. ex. 11x10^5+209=1100209. 1100209/11=100019
All of the odd numbers between 11 and 99 inclusive are two digit numbers that are not multiples of 2. There are 45 of them.
There is an infinite amount of prime numbers all of which are odd numbers
There are an infinite amount of numbers, so there can be no "greatest" odd number.
Since any odd number can be a factor, there is an infinite amount of them.
9 multiplied by any other odd number will still be odd. 9 multiplied by any even number will be even - so the multiples of 9 between 1 and 120 that are odd would be:1x9 = 93x9 = 275x9 = 457x9 = 639x9 = 8111x9 = 9913x9 =117The next odd number you could multiply 9 by is 15 but 15x9 = 135, which is greater than 120 so does not fall within the constraints of the question.
There are an infinite number of numbers that satisfy these criteria.
-3
The multiples of all odd numbers are odd and even. Odd x odd = odd. Odd x even = even. Since odd and even numbers alternate, the multiples will alternate as well.
They are members of the infinite set of numbers of the form 3*(2k-1) or 6*k-3 where k is an integer.