PUMaC 2021 · 数论（B 组） · 第 3 题

3 . Note that we can’t have more than 2 other prime factors, the product of this is at most 5 / 3 · 9 / 5 · 13 / 7 = 39 / 7 > 3 . For 2 prime factors, the smallest possible value of the terms due 2 to the other factors is 5 / 3 · 9 / 3 = 3 , giving n = 2 · 3 · 5 = 60 . For 1 prime factor, we want e ( p − 1) 2 2 1+ = 3 , or that e ( p − 1) = 2 p , which requires p | e , or p − 1 | 2 , meaning that p = 3 , 2 2 2 2 2 2 2 p 2 2 3 and that e = 3 . This gives n = 2 · 3 = 108 . Our total sum is thus 108 + 1024 + 60 = 1192 . 2 3