Quadruple Multiplication

Let $N=p_1^{a_1}p_2^{a_2}\cdots p_n^{a_n}$ For each distinct prime factor $p_i$ of $N$ , the $a_i$ $p_i$ 's must be distributed among $a,b,c,d$ . By stars and bars, there are $\dbinom{a_i+4-1}{4-1}=\dbinom{a_i+3}{3}$ Therefore, the number of ordered quadruples is $\dbinom{a_1+3}{3}\times\dbinom{a_2+3}{3}\times\cdots\times\dbinom{a_n+3}{3}$ Listing out small values, we can try to see how to combine numbers, $\dbinom{0+3}{3}=1,\dbinom{1+3}{3}=4,\dbinom{2+3}{3}=10,\dbinom{3+3}{3}=20,\\ \dbinom{4+3}{3}=35,\dbinom{5+3}{3}=56,\dbinom{6+3}{3}=84,\dbinom{7+3}{3}=120$ Then, for each power of a prime that divides $N$ , $a_i$ , $p_i|N\implies \dbinom{a_i+3}{3}|N$ If every prime appears only once, then $4|N$ , a contradiction, so at least one $a_i$ is at least 2. If an $a_i$ is 2, then $10|N$ . The smallest possible values of $N$ are 20, 40, 50, 60, 80, 90, and 100. The only one of these that works is $100$ .

Then, if there is a smaller $N$ it must be a multiple of 35, 56, or 84. It is easy to see that these do not work, so the answer is $\boxed{100}$ .

I would be interested to see a solution with less casework.

Let, N = $p_{1}^{k_{1}} \times p_{2}^{k_{2}} \times p_{3}^{k_{3}}$

then, the no. of ordered quadruples = $(k_{1} + 4 - 1) C (4 - 1) \times (k_{2} + 4 - 1) C (4 - 1) \times (k_{3} + 4 - 1) C (4 - 1)$

now, i tried a little bit of trial and error with few values of $k_{1}$

(1+4-1) C (4-1) = 4

(2+4-1) C (4-1) = 10

(3+4-1) C (4-1) = 20

with $k_{1}$ = 1 ... this is not possible ... as the number of quadruples will have a factor of 4 ... But the number cannot have a factor of 4, as the max power of 2 we can use is '1'

with $k_{1}$ = 2 ... we get 10 ... which is product of 2 primes 2*5. Now, we can use 2 powers of the primes. So, if we use that,

100 = $2^{2} \times 5^{2}$ ... which will have as many quadruples

so our answer must be $\boxed{100}$