Count The Divisors

Relevant wiki: Number of Factors

Key fact: if the prime factorization of $N$ is $N = 2^{e_2} \cdot 3^{e_3} \cdot 5^{e_5} \cdots p^{e_p} \cdots,$ then the number of divisors of $N$ is $\delta(N) = (1+e_2)(1+e_3)(1+e_5)\cdots (1+e_p)\cdots.$

From the fact that $\delta(N)$ is odd, we know that all exponents $e_i$ must be even, so that $N$ is a perfect square. In particular, this means that $N$ contains no prime factors greater than 50.

Using $128 = 2^7$ , we have $\frac{\delta(128N)}{\delta(N)} = \frac{(1+e_2+7)(1+e_3)\cdots}{(1+e_2)(1+e_3)\cdots} = \frac{8+e_2}{1+e_2} = \frac{18}{11},$ which tells us immediately that $e_2 = 10$ .

Now $213 = 3\cdot 71$ , so $\frac{\delta(213N)}{\delta(N)} = \frac{(2+e_3)(2+e_{71})}{(1+e_3)(1+e_{71})} = \frac{8}{3}.$ We already know that $e_{71} = 0$ , so that we find $\frac{2+e_3}{1+e_3} = \frac{4}{3},$ making $e_3 = 2$ .

Next, $175 = 7\cdot 5^2$ , giving $\frac{\delta(175N)}{\delta(N)} = \frac{(3+e_5)(2+e_7)}{(1+e_5)(1+e_7)} = \frac 8 5.$ To solve this, we cross-multiply, work out brackets, and subtract, and refactor (I find it convenient to multiply everything by 3): $5(3+e_5)(2+e_7) = 8(1+e_5)(1+e_7);$ $30 + 10 e_5 + 15 e_7 + 5 e_5 e_7 = 8 + 8 e_5 + 8 e_7 + 8 e_5 e_7;$ $(3 e_5 - 7)(3 e_7 - 2) = 80;$ there are ten positive integer factorizations of 80, and only four of them result in integer values for $e_5$ and $e_7$ . Of these four, only one has even values: $e_5 = 4$ and $e_7 = 6$ .

Thus we know that $N = 2^{10}\cdot 3^2\cdot 5^4\cdot 7^6\cdots$ Using $2016 = 2^5\cdot 3^2\cdot 7$ , we get $2016N = 2^{15}\cdot 3^4\cdot 5^4\cdot 7^7\cdots$ and $\frac A B = \frac{\delta(2016N)}{\delta(N)} = \frac{16\cdot 5\cdot 5\cdot 8}{11\cdot 3\cdot 5\cdot 7} = \frac{640}{231},$ making the answer $\boxed{871}$ .