MathDB

490

Part of 1989 All Soviet Union Mathematical Olympiad

Problems(1)

ASU 490 All Soviet Union MO 1989 12 positive divisors, d_m=(d_1+d_2+d_4)d_8

Source:

8/13/2019
A positive integer nn has exactly 1212 positive divisors 1=d1<d2<d3<...<d12=n1 = d_1 < d_2 < d_3 < ... < d_{12} = n. Let m=d4−1m = d_4 - 1. We have dm=(d1+d2+d4)d8d_m = (d_1 + d_2 + d_4) d_8. Find nn.
number theorydivisor