What is the minimum value of a positive integer which has five three-digit divisors?
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Let N be this positive integer and N 1 , … , N 5 its five three-digit divisors. We can minimalise N if we minimalise the numbers N / N j ∈ N , therefore we set N j = N / j . This means that N must be divisible by 1 , 2 , 3 , 4 , 5 and hence by 6 0 . We require N / 5 ≥ 1 0 0 but also N / 6 < 1 0 0 (otherwise we will have too many three-digit divisors). This only leaves one possible multiple of 6 0 , which is N = 5 4 0 . The required divisors are 540, 270, 180, 135 and 108.