gre problem %% combination42

Question

Find the number of ways in which 8064 can be resolved as the product of two factors? this answer is right or wrong ,can anyone explain this .......

8064=27*32*7 number of factors =(7+1)(2+1)(1+1)=48 no.of ways of writing 8064 as a product of two = 48 2 =24

Answer 1

$8064=2^7*3^2*7$ we want to find its divisors, by the rule of product there are $8$ options for the power of $2$ in the factor$(0,1,2,3,4,5,6,7)$, $3$ choices of the power of $3$ and 2 for the power of $7$ therefore it has $48$ divisors, however we need to divide this by two since every divisor pairs up with another divisor to give the product $8064$

Answer 2

$8064=2^7\times 3^2\times 7^1$ has $(7+1)\times (2+1)\times (1+1)=48$ positive divisors. Therefore there are $24$ ways of writing it as a product of $2$ factors, if we count $a\times b$ and $b\times a$ as $1$ way.

Answer 3

You need to redo this one. $N =d \frac{N}{d}$, where $d$ is some divisor of $N$. So the number of ways of writing $N$ as product of two factors is the number of UNordered pairs $(d,N/d)$.

Keep in mind that if $N$ is a perfect square then you need to be a bit more careful.

Answer 4

Choosing a factor of $8064$ involves choosing the number of times each prime factor of it will appear. Thus, for the prime factor $2$ you have $7+1$ choices (the $+1$ is for not choosing $2$), similarly, for $3$ you have $2+1$ choices and for $7$ you have $1+1$.