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1 = d_1 < d_2 < ...< d_k = n=d_2d_3 + d_2d_5+d_3d_5

Source: Bosnia and Herzegovina EGMO TST 2019 p2

October 7, 2022
number theoryDivisors

Problem Statement

Let 1=d1<d2<...<dk=n1 = d_1 < d_2 < ...< d_k = n be all natural divisors of the natural number nn. Find all possible values ​​of the number kk if n=d2d3+d2d5+d3d5n=d_2d_3 + d_2d_5+d_3d_5.
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