Limit inferior of difference of consecutive primes was 70 million

The factorization of $7 \times 10^7$ is simply $2^7 \times 5^7 \times 7$ .

Suppose for now we restrict the search for positive integer solutions only.

For non-negative integers $x_i, y_i, z_i$ , let $A = 2^{x_1} \cdot 5^{y_1} \cdot 7^{z_1}, B = 2^{x_2} \cdot 5^{y_2} \cdot 7^{z_2} , \ldots, F = 2^{x_6} \cdot 5^{y_6} \cdot 7^{z_6}$ .

Then their product is $2^7 \times 5^7 \times 7^1 = 2^{\sum_{j=1}^6 x_j} \times 5^{\sum_{j=1}^6 x_j} \times 7^{\sum_{j=1}^6 x_j}$ .

$\Rightarrow \displaystyle \sum_{j=1}^6 x_j = 7, \sum_{j=1}^6 y_j = 7, \sum_{j=1}^6 z_j = 1$ .

By Stars and Bars, we have ${7+6 - 1 \choose 6-1} = {12 \choose 5}$ ordered solution for $<x_i>$ and $<y_i>$ , and ${1+6 - 1 \choose 6-1} = 6$ ordered solution for $<z_i>$ .

Multiplying them together gives a total of ${12 \choose 5} ^2 \times 6 = 3763584$ ordered solution for positive integers $A$ to $F$ .

If we drop the constraint for positive integers, then we could have:

Case 1 : 6 Positive integers

Case 2 : 4 positive integers, 2 negative integers

Case 3 : 2 positive integers, 4 negative integers

Case 4 : 6 Negative integers

Combining all these cases together yields a total of

$3763584 \times \left [ {6 \choose 0} + {6 \choose 2} + {6 \choose 4} + {6 \choose 6} \right ] = \boxed{120434688}$ ordered solutions.