5 big divisors

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


The answer is 540.

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1 solution

Tijmen Veltman
Jan 18, 2015

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

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