1000 divisors

If you write a number as its prime factorization, $2^m * 3^n * 5^p * 7^q ...$ , then you have m+1 choices for the power of 2 in a factor, n+1 choices for the power of 3, and so on, giving a total of $9m+1)(n+1)(p+1)(q+1)...$ distinct factors. We need to factor 1000 in the way which gives us the smallest product of primes possible. Given how rapidly exponentiation gives massive numbers, it seems probable that simply making all the factors as small as possible is the way to go. $1000=5*5*5*2*2*2$ , so a good guess is $2^4*3^4*5^4*7*11*13$ , or $30^4*7*11*13$ (Spoiler alert: this turns out to be the answer)

The next step I took was to find all the different factorizations of 1000, but this is a lot of work, and some of them clearly are too large. For example, $2^124*3^7$ is clearly much larger than most of the other possibilities. So what is the cutoff? Our current working guess of $30^4*7*11*13$ is a bit less than $32^4$ , or $2^20$ . This means any factorization with a factor greater than 21 will give too large of a power of 2, and not be the smallest. The factorizations which have all factors 20 or less are $5*5*5*2*2*2 , 10*5*5*2*2 , 5*5*5*4*2*2 , 20*5*5*2 , 10*10*5*2 , 20*10*5$ , and $10*10*10$ . All of the numbers these form are divisible by $30^4$ , which makes them easier to compare. We are left with $7*11*13 , 2^5*7*11 , 7^3*11 , 2^15*7 , 2^5*3^5*7 , 2^15*3^5$ , and $30^5$ . Of these, $7*11*13$ is clearly the smallest, so our answer is $30^4*7*11*13 = \boxed{810810000}$

Okay, so the first thing I did was to do the prime factorization of 1000. It gave me ${ 2 }^{ 3 }\times { 5 }^{ 3 }$ . Then, with this principle in mind: https://brilliant.org/practice/counting-divisors-2/?p=7 you want to give the first 3 primes (2,3,5) the 3 powers of 4. Knowing that ${ 7 }^{ 3 }>{ 11 }\times 13$ we will then take 7,11 and 13 as our next primes. ${ 2 }^{ 4 }\times { 3 }^{ 4 }\times { 5 }^{ 4 }\times 7\times 11\times 13=810810000\\ (4+1)\times (4+1)\times (4+1)\times (1+1)\times (1+1)\times (1+1)=1000$