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 $\boxed{76}$.