If $N=2^26^45^6$, how many ways can $N$ be written as a product of an even number and an odd number? Given N= $2^2*6^4*5^6=A*B$
In how many ways N can be written as a product of two numbers such that one of them is even and the other is odd?
My answer
$N= 2^6*3^4*5^6$
Number of odd factors= $5*7=35$ factors multiplied by total number of even factor that is $6$.
$35*6=210$
But the answer is $35$. How?
 A: Your logic simply does not follow.  The odd factors are $3^i*5^j$ and there are $35$ of them.  But the even factors are not the $6$ of the form $2^k; k \ge 1$.  There are of the factors of the form $2^k3^i5^j; k > 1$ and there are $210$ of them.
But that doesn't matter.  Because you don't multiply the number of even factors by the number of odd factors to get the answer.  Why would you?  Just because one term is even and another odd doesn't mean they will multiply out to $N$.
You need $a*b$ $a$ odd and be even and $a*b = 2^6*3^4*5^6$.  So $a = 2^j3^k5^l; b = 2^{6-j}3^{4-k}5^{6-k}$.  But as $a$ is odd, $j=0$ and $6-j= 6$.  
So $a = 3^k5^l$ and $b=3^{4-k}5^{6-l}$.  $k$ can be anything from $0$ to $4$ so there $5$ choices, and $l$ can be anything from $0$ to $6$ so there and $7$ choices for that.
So there are $35$ ways of doing this.
Another, and far easier way to do it is to note that for any value of $a$ then $b$ must equal $N/a$.  So the for each odd factor, the other even factor must be fixed.  SO there are as many ways to do this as there are odd factors.
===old answer ===
There are 35 odd factors.  True.
There are 6 even factors.  False.  
There are $6$ even powers of $2$; $2^1, 2^2,.... , 2^6$.  And each of those can be multiplied by an odd factor to get another even factor.  Examep: $2^i*3$ or $2^i*3^3 *5^2$ etc. So there are $6*35 = 210$ even factors.
There are 210 even factors. True.
There are $210*35$ ways to combine these factors.  True.
Evey way to combine these factors will equal $N$.  !!!!!FALSE!!!!
Example.  $3$ is an odd factor.  $2$ is an even factor.  $3*2 \ne N$ so we can not count this example.  Taking all $210*35$ combinations does take that example.
If $a$ is an odd factor.  And $a*b=N$ then $b = \frac Na$ and that is the only possible even factor that can be multiplied by $a$ to get $N$.
So for each of the $35$ odd factors, we do NOT have $210$ options for an even factor.  We have ONE option.  One only!.  So we do not have $35*210$ ways of doing this.
We had $35*1$ ways of doing this.
A: Once you choose the odd factor (say, $A$), there's only one possibility for the other (it has to be $\frac{N}{A}$).
A: $N=2^6*3^4*5^6$
The 2 factors must be even or 1 factor odd another factor even. There is no possibility that both factors are odd as you have a 2 as one of the factors of $N$
No. of odd factors $=35$
For each odd factor there must an even factor to get product as $N$
So no. of ways $=35$
