gre problem %% combination42 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
 A: $8064=2^7*3^2*7$ 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 $8064$
A: $8064=2^7\times 3^2\times 7^1$ 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.
A: 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.
A: 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 $1+1$.
