求解一道組合數學題（用鴿籠原理即抽屜原理解答）？
西行寺·幽幽子2022-11-13 01:05:552。 Let S denote a set of 100 integers chosen from 1，2，。。。，200 such that i does not divide
j for all distinct i，j ∈ S。 We show that i 6∈ S for 1 ≤ i ≤ 15。 Certainly 1 6∈ S since
1 divides every integer。 By construction the odd parts of the elements in S are mutually
distinct and at most 199。 There are 100 numbers in the list 1，3，5，。。。，199。 Therefore each of
1，3，5，。。。，199 is the odd part of an element of S。 We have 3×5×13 = 195 ∈ S。 Therefore
none of 3，5，13，15 are in S。 We have 3 3 ×7 = 189 ∈ S。 Therefore neither of 7，9 is in S。 We
have 11×17 = 187 ∈ S。 Therefore 11 6∈ S。 We have shown that none of 1，3，5，7，9，11，13，15
is in S。 We show neither of 6，14 is in S。 Recall 3 3 ×7 = 189 ∈ S。 Therefore 3 2 ×7 = 63 6∈ S。
Therefore 2×3 2 ×7 = 126 ∈ S。 Therefore 2×3 = 6 6∈ S and 2×7 = 14 6∈ S。 We show 10 6∈ S。
Recall 3 × 5 × 13 = 195 ∈ S。 Therefore 5 × 13 = 65 6∈ S。 Therefore 2 × 5 × 13 = 130 ∈ S。
Therefore 2 × 5 = 10 6∈ S。 We now show that none of 2，4，8，12 are in S。 Below we list the
integers of the form 2 r 3 s that are at most 200：
1， 2， 4， 8， 16， 32， 64， 128，
3， 6， 12， 24， 48， 96， 192，
9， 18， 36， 72， 144，
27， 54， 108，
81， 162。
In the above array each element divides everything that lies to the southeast。 Also， each
row contains exactly one element of S。 For 1 ≤ i ≤ 5 let r i denote the element of row i that
is contained in S， and let c i denote the number of the column that contains r i 。 We must
have c i < c i−1 for 2 ≤ i ≤ 5。 Therefore c i ≥ 6 − i for 1 ≤ i ≤ 5。 In particular c 1 ≥ 5 so
r 1 ≥ 16， and c 2 ≥ 4 so r 2 ≥ 24。 We have shown that none of 2，4，8，12 is in S。 By the above
comments i 6∈ S for 1 ≤ i ≤ 15。
原題Richard 的組合數學 第三章第二題英文答案，我只是來求翻譯的233333
