Given a number n with prime factorization n = p_1^(alpha_1)p_2^(alpha_2)...p_k^(alpha_k), each divisor of n is of the form p_1^(beta_1)p_2^(beta_2)...p_k^(beta_k) where beta_i in {0, 1, ..., alpha_i}. As there are alpha_i+1 choices for each beta_i, the number of divisors of n is given by
(alpha_1+1)(alpha_2+1)...(alpha_k+1)=prod_(i=1)^k(alpha_i+1)
As N=2^axx3^bxx5^cxx7^d, the number of divisors of N is given by (a+1)(b+1)(c+1)(d+1) = 378. Thus, our goal is to find (a, b, c, d) such that the above product holds and 2^axx3^bxx5^cxx7^d is minimal. As we are minimizing, we will assume from this point onward that a>=b>=c>=d (if this were not the case, we could swap exponents to get a lesser result with the same number of divisors).
Noting that 378 = 2xx3^3xx7, we can consider the possible cases in which 378 is written as a product of four integers k_1, k_2, k_3, k_4. We can inspect these to see which produces the least result for N.
Format: (k_1, k_2, k_3, k_4) => (a, b, c, d) => 2^axx3^bxx5^cxx7^d
(2, 3, 3^2, 7) => (8, 6, 2, 1) => ~3.3xx10^7
(2, 3, 3, 3*7) => (20, 2, 2, 1) => ~1.7xx10^9
color(red)((3, 3, 2*3, 7) => (6, 5, 2, 2) => ~1.9xx10^7)
(3, 3, 3, 2*7) => (13, 2, 2, 2) => ~9.0xx10^7
(1, 3, 2*3^2, 7) => (17, 6, 2, 0) => ~2.4xx10^9
We can stop here, as any further cases will have some k_i>=27, giving 2^a >= 2^26 ~~ 6.7xx10^7, which is already greater than our best case.
By the above work, then, the (a, b, c, d) which produces a minimal N with 378 divisors is (a, b, c, d) = (6, 5, 2, 2), giving N = 2^6xx3^5xx5^2xx7^2=19,051,200
