For simplicity, replace the expression

$$\displaystyle a \times (a+1) \times \dots \times b \mid c \times (c+1) \times \dots \times d$$

with the equivalent

$$\displaystyle (c-1)! \times b! \mid (a-1)! \times d!$$

For each prime ~p~, define the map ~v_p(x)~ which returns the largest power of ~p~ which divides ~x~.

The left-hand side divides the right-hand side if an only if: for each ~p~, we have ~v_p~ of the left-hand side is at most the ~v_p~ of the right-hand side.

It's easy to see ~v_p(xy) = v_p(x)+v_p(y)~. We now want only ~v_p(x!)~.

Proposition: We have

$$\displaystyle v_p(x!) = \left\lfloor \frac{x}{p} \right\rfloor + \left\lfloor \frac{x}{p^2} \right\rfloor + \left\lfloor \frac{x}{p^3} \right\rfloor + \dots$$

Proof: The first summand counts multiples of ~p~, the second counts multiples of ~p^2~, and so on.

The solution is thus: find all primes up to ~10^7~, implement ~v_p(x!)~, and check the relevant inequality of ~v_p~ for each prime.