Yes. You should be able to multiply any number's multiple's into itself.
All numbers divisible by 9 are divisible by 3; since 9 = 3 x 3 all multiples of 9 are also multiples of 3. However, all numbers divisible by 3 are not divisible by 9, eg 6 = 2 x 3 but 6 is not divisible by 9 (since 6 is not a multiple of 9) - it only takes one counter example to disprove a theory.
yes, every multiple of 9 is a multiple of 3, but that doesn't cover all of them ...there are more multiples of 3 besides those.
Itself and any multiples of 9
Because they are multiples of 5. All multiples of 5 have 5 as a factor, which means they are divisible by 5.
All of its multiples are.
All multiples of 18 are divisible by 9. Not all multiples of 9 are divisible by 18.
yes since 9 is divisible by 3 all its multiples are as well
All numbers divisible by 9 are divisible by 3; since 9 = 3 x 3 all multiples of 9 are also multiples of 3. However, all numbers divisible by 3 are not divisible by 9, eg 6 = 2 x 3 but 6 is not divisible by 9 (since 6 is not a multiple of 9) - it only takes one counter example to disprove a theory.
Since 9 is not a prime, all [positive integral] multiples of 9 are exactly divisible.
They are all the multiples of 9, an infinite number of them.
The multiples of 9 are 0, 9, 18, etc. (add 9 at a time). These numbers are divisible by 9; all others are not.
yes, at least until the multiple of 12
yes, every multiple of 9 is a multiple of 3, but that doesn't cover all of them ...there are more multiples of 3 besides those.
The common multiples of 3 and 9 are 9, 18, 27 and so on.Any number that is divisible by 9 is also divisible by 3. This is because 9 is divisible by 3, so any multiples of 9 must also be divisible by three.78 117 156 195 234 273 312 351 390and so on ...
Itself and any multiples of 9
All multiples of 99 are divisible by 99
All multiples of 185 are divisible by it.