Keeping it 360

$360 = 2^3×3^2×5$

We can get the smallest suitable value by taking the product of the smallest primes and putting them to the smallest possible powers (in a way that the smallest prime number (2) is put to the power of the ((biggest prime factor of 360) - 1), then the next smallest prime (3) is put to the power of the ((second biggest prime factor of 360) - 1) and so on).

$2^{5-1}×3^{3-1}×5^{3-1}×7^{2-1}×11^{2-1}×13^{2-1} = 2^4×3^2×5^2×7×11×13 = 3603600$

Indeed, this number has:

$(4+1)(2+1)(2+1)(1+1)(1+1)(1+1) = 5 × 3^2 × 2^3 = 360 \text { positive integer factors.}$

(See relevant wiki: Factors )

It is easy, but a bit longer to prove that 3603600 is the smallest positive integer with exactly 360 positive divisors. One way to do that, is by trying to leave out prime factors and combining its power component with that of one of the smallest primes. This would result in bigger numbers, because e.g.:

$2^{5-1}×13^{2-1} < 2^{5×2-1} \text { and } 2^{5-1}×3^{3-1} < 2^{5×3-1} \text { and } 3^{3-1}×13^{2-1} < 3^{3×2-1}$

$\text {Hence the first three digits are: } \boxed {360}$