olymmath

RU-ALLRUSSIAN 1989 Problem 4

    A positive integer n has exactly 12 positive divisors 1=d1<d2<d3<…<d12=n1 = \mathrm{d}_1 < \mathrm{d}_2 < \mathrm{d}_3 < \ldots < \mathrm{d}_{12} = \mathrm{n} . Let m=d4\mathrm{m} = \mathrm{d}_4 - 1. We have dm=(d1+d2+d4)d8\mathrm{d}_{\mathrm{m}} = (\mathrm{d}_1 + \mathrm{d}_2 + \mathrm{d}_4) \mathrm{d}_8 . Find n.

    Dataset text NuminaMath olympiads. Quoted with source.

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