1. ## divisibily

how many posotive integer divide 2^7 x 3^8 such that they are not prime or 1,
thanks for any help

2. Originally Posted by hmmmm
how many posotive integer divide 2^7 x 3^8 such that they are not prime or 1,
thanks for any help
How many numbers are there of the form $\displaystyle 2^{\alpha} \times 3^{\beta}$; $\displaystyle \alpha, \beta \in \mathbb{N}$

where $\displaystyle \alpha+\beta >1$, $\displaystyle 0\le \alpha\le 7$, $\displaystyle 0\le \beta \le 8$ ?

CB

3. hmmm really should have spotted that, 72 - 2 primes and 1 so 69, thanks alot for the help

4. Hello, hmmmm!

You got it!

How many positive integer divide $\displaystyle 2^7\cdot3^8$ such that they are not prime or 1?

There are 8 choices for the 2's: .$\displaystyle 2^0,\:2^1,\;2^2,\;\hdots \:2^7$

There are 9 choices for the 3's: .$\displaystyle 3^0,\:3^1,\:3^2,\hdots\:3^8$

Hence, there are: .$\displaystyle 8\times 9 \,=\,72$ choices for numbers of the form $\displaystyle 2^a3^b$

But this includes: .$\displaystyle 2^03^0 \,=\,1,\;2^13^0 = 2,\;2^03^1 = 3$
. .
The rest are all composite.

Therefore, the answer is: .$\displaystyle 72 - 3 \;=\;69$