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

thanks for any help

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- Feb 18th 2010, 02:34 AM #1

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- Feb 18th 2010, 03:04 AM #2

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- Feb 19th 2010, 04:33 AM #3

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- Feb 20th 2010, 07:56 AM #4

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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: .$\displaystyle 72 - 3 \;=\;69$