Prove that any positive integer with exactly $1000$ divisors, which when arranged in increasing order have alternating parity (so the first divisor is odd, the second is even, the third is odd, etc. the 1000-th is even), have more than $150$ digits.
Little progress: We wish to show $n > 10^{150}$ and clearly the second divisor is $2$ and the second largest divisor is $\frac{n}{2}$, which must be odd, so $n$ is not divisible by $4$.
Any help appreciated!
Hint: if you divide by $2$ you get a number with $500$ divisors all of which are odd and all of which are divisors of your chosen integer. Now what is the condition on those to give alternating odd and even when you multiply by $2$?