Find four positive integers having more than $100$ divisors 
Find four positive integers each not exceeding $70000$ and each having more than $100$ divisors.

Since we are trying to maximize divisors and minimize value, we assume that $n = 2^{\alpha_1} 3^{\alpha_2} \cdots$ has $\alpha_1 \geq \alpha_2 \geq \cdots.$ Now we do casework.
How do we do the casework since there seem to be so many cases?
 A: It's just a hand heuristic which is possible because the limits of the puzzle ...
Having the 7 first primes power 1 is enough to get $2^7 > 100$ factors
Their product is too high : $2.3.5.7.11.13.17 = 510510$ which is $> 70000$
Let's replace 17 by 16 = $2^4$ : $2^4 .3.5.7.11.13 = 240240$ // too high
idem with 13 by 8 = $2^3$ : $2^7 .3.5.7.11 = 147840$ // too high
Let's replace 4 by 3 : $2^5 .3^2 .5.7.11 = 110880$ // too high
let's reduce the power of 2 by 1 :
$2^4 .3^2 .5.7.11 = 55440$
factors : $5 . 3 . 2^3 = 120$ // ok , first result !
let's replace 11 par 13 too see if it is in the bounds
$2^4 .3^2 .5.7.13 = 65520$
factors : $5 . 3 . 2^3 = 120$ // ok
play again between 2 and 5
$2^2 . 3^2 .5^2 .7.11 = 69300$ // ok
factors : $3 . 3 . 3 . 2 . 2 = 104$
Need a last one , let's try without 11 and 13
$2^6 . 3^3 . 5. 7 = 60480$ // ok
factors : $7. 4. 2. 2 = 112$
A: I'd love to just add it as a comment, but 50400 (108 divisors), 55440 (112 divisors), 60480 (120 divisors), 65520 (112 divisors), and 69300 (108 divisors) suit the numbers you are looking for.
A: Put $n=\prod_{i=1}^{n}p_i^{\alpha_i}$, the factorization of n in primes.
Every divisor have the same form, with different coefficents. Then, $n$ has $\prod_{i=1}^{n}(\alpha_i+1)$ divisors.
Now, you can play with this
