A subset of the integers $1, 2, ..., 100$ has the property that none of its members is 3 times another. What is the largest number of members such a subset can have?

$ \textbf{(A)}\ 50 \qquad\textbf{(B)}\ 66 \qquad\textbf{(C)}\ 67 \qquad\textbf{(D)}\ 76 \qquad\textbf{(E)}\ 78 $
To solve this problem, we need to find the largest subset of the integers from 1 to 100 such that no member of the subset is three times another member. We will use a systematic approach to count the elements that can be included in the subset.

1. **Identify the multiples of 3:**
   - The multiples of 3 within the range 1 to 100 are: \(3, 6, 9, 12, \ldots, 99\).
   - There are \(\left\lfloor \frac{100}{3} \right\rfloor = 33\) multiples of 3.

2. **Remove the multiples of 3:**
   - Removing these 33 multiples from the set of 100 integers leaves us with \(100 - 33 = 67\) elements.

3. **Add back the multiples of 9:**
   - The multiples of 9 within the range 1 to 100 are: \(9, 18, 27, \ldots, 99\).
   - There are \(\left\lfloor \frac{100}{9} \right\rfloor = 11\) multiples of 9.
   - Adding these 11 elements back to the 67 elements gives us \(67 + 11 = 78\) elements.

4. **Remove the multiples of 27:**
   - The multiples of 27 within the range 1 to 100 are: \(27, 54, 81\).
   - There are \(\left\lfloor \frac{100}{27} \right\rfloor = 3\) multiples of 27.
   - Removing these 3 elements from the 78 elements gives us \(78 - 3 = 75\) elements.

5. **Add back the multiples of 81:**
   - The multiples of 81 within the range 1 to 100 are: \(81\).
   - There is \(\left\lfloor \frac{100}{81} \right\rfloor = 1\) multiple of 81.
   - Adding this 1 element back to the 75 elements gives us \(75 + 1 = 76\) elements.

Thus, the largest subset of the integers from 1 to 100 such that no member is three times another member contains 76 elements.

The final answer is \(\boxed{76}\).