$15 \cdot 12$ subsets of the set $\{1,2, \cdots, 100\}$, such that no number is exactly three times another, the maximum number of elements in such a subset is
(A) 50 .
(B) 66 .
(C) 67 .
(D) 76 .
(E) 78 .
(41st American High School Mathematics Examination, 1990)
［Solution］The subset with the most elements that meets the conditions is
$$
\{1\} \cup\{4,5,6,7,8,9,10,11\} \cup\{34,35,36, \cdots, 99,100\} \text {, }
$$

It contains a total of
$$
1+(11-4+1)+(100-34+1)=76 \text { elements. }
$$

Below is the proof that 76 is the maximum number of elements in such a subset.
Since $(2,6),(3,9),(12,36),(13,39), \cdots,(33,99)$, these 24 pairs of numbers, each pair can only have one number selected into the subset, so the subset can have at most
$$
100-24=76 \text { (elements). }
$$

Therefore, the answer is $(D)$.