What is the minimum value of a positive integer which has five three-digit divisors?

The answer is 540.

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Let $N$ be this positive integer and $N_1,\ldots,N_5$ its five three-digit divisors. We can minimalise $N$ if we minimalise the numbers $N/N_j\in\mathbb{N}$ , therefore we set $N_j=N/j$ . This means that $N$ must be divisible by $1,2,3,4,5$ and hence by $60$ . We require $N/5\geq 100$ but also $N/6<100$ (otherwise we will have too many three-digit divisors). This only leaves one possible multiple of $60$ , which is $N=\boxed{540}$ . The required divisors are 540, 270, 180, 135 and 108.