Answer: D

Small Hint:

Group integers into chains $m,3m,9m,\ldots$ where $m$ is not a multiple of $3$

Big Hint:

Within each chain, alternating entries give a largest allowed selection

Solution:

Partition the integers into chains $m,3m,9m,\ldots,$ with $m$ not a multiple of $3.$ In each chain, no two adjacent terms may both be selected, so a maximum selection takes alternating terms starting with $m.$ Equivalently, select the integers whose exponent of $3$ is even. There are $$\begin{aligned} &\left(100-\left\lfloor\frac{100}{3}\right\rfloor\right)\\ &\quad+\left(\left\lfloor\frac{100}{9}\right\rfloor -\left\lfloor\frac{100}{27}\right\rfloor\right)\\ &\quad+\left\lfloor\frac{100}{81}\right\rfloor\\ &=67+8+1=76. \end{aligned}$$ The chain argument also proves no larger selection is possible.

Thus the correct answer is D.