Let $$S = \{1, 2, 3, 4, 5, 6, 9\}$$. Then the number of elements in the set $$T = \{A \subseteq S : A \neq \phi$$ and the sum of all the elements of $$A$$ is not a multiple of 3$$\}$$ is _________.

We have the set $$S=\{1,2,3,4,5,6,9\}$$ containing seven elements. For every subset $$A\subseteq S$$ we are interested in the remainder obtained when the sum of its elements is divided by 3. Because divisibility by 3 depends only on the remainder, it is natural to classify each element of $$S$$ according to its remainder (or “residue”) modulo 3.

Computing the residues, we obtain

$$\begin{aligned} 1&\equiv1\pmod3,\\ 2&\equiv2\pmod3,\\ 3&\equiv0\pmod3,\\ 4&\equiv1\pmod3,\\ 5&\equiv2\pmod3,\\ 6&\equiv0\pmod3,\\ 9&\equiv0\pmod3. \end{aligned}$$

So the seven elements fall into three residue classes:

$$ \begin{array}{lcl} \text{Class }0:&\; \{3,6,9\} &\Longrightarrow n_0=3,\\[2pt] \text{Class }1:&\; \{1,4\} &\Longrightarrow n_1=2,\\[2pt] \text{Class }2:&\; \{2,5\} &\Longrightarrow n_2=2. \end{array} $$

Choosing elements from the “0‐class” never alters the remainder, because each of those numbers is itself a multiple of 3. Hence any subset chosen from $$\{3,6,9\}$$ contributes $$0$$ (mod 3) to the total sum. There are $$2^{n_0}=2^{3}=8$$ possible selections from this class, including the empty selection.

If we pick $$k_1$$ elements from the “1‐class”, each contributes a remainder $$1$$, so their combined remainder is

$$k_1\pmod3.$$

Because $$n_1=2$$, the value $$k_1$$ may be $$0,1,$$ or $$2$$, and the corresponding numbers of ways are

$$\binom{2}{0}=1,\quad\binom{2}{1}=2,\quad\binom{2}{2}=1.$$

Similarly, choosing $$k_2$$ elements from the “2‐class” gives a combined remainder

$$2k_2\pmod3,$$

and with $$n_2=2$$ the admissible values and counts are also

$$\binom{2}{0}=1,\quad\binom{2}{1}=2,\quad\binom{2}{2}=1.$$

For the total sum of a subset to be a multiple of 3 we require

$$k_1+2k_2\equiv0\pmod3.$$

We now list every pair $$(k_1,k_2)$$ satisfying this condition together with the number of ways it can occur.

1. $$k_1=0$$ gives residue $$0$$. We need $$2k_2\equiv0\pmod3$$, which is true only for $$k_2=0$$. Number of ways:

$$\binom{2}{0}\times\binom{2}{0}=1\times1=1.$$

2. $$k_1=1$$ gives residue $$1$$. We need $$1+2k_2\equiv0\pmod3\;\Longrightarrow\;2k_2\equiv2\pmod3\;\Longrightarrow\;k_2=1.$$ Number of ways:

$$\binom{2}{1}\times\binom{2}{1}=2\times2=4.$$

3. $$k_1=2$$ gives residue $$2$$. We need $$2+2k_2\equiv0\pmod3\;\Longrightarrow\;2k_2\equiv1\pmod3\;\Longrightarrow\;k_2=2.$$ Number of ways:

$$\binom{2}{2}\times\binom{2}{2}=1\times1=1.$$

Adding these counts, the total number of ways to choose elements from the “1‐class” and “2‐class” so that the running sum is divisible by 3 is

$$1+4+1=6.$$

For every such choice we may freely choose any subset of the three “0‐class” elements, and there are $$8$$ possibilities for that. Multiplying, the number of all subsets (including the empty one) whose element-sum is a multiple of 3 equals

$$6\times8=48.$$

Among these $$48$$ subsets, one is the empty set itself. Because the problem specifically excludes the empty set (it demands $$A\neq\varnothing$$), the count of non-empty subsets with sum divisible by 3 is

$$48-1=47.$$

The total number of non-empty subsets of a 7-element set is

$$2^{7}-1=128-1=127.$$

Therefore the number of non-empty subsets whose element-sum is not a multiple of 3 is obtained by simple subtraction:

$$127-47=80.$$

Hence, the correct answer is Option C ($$80$$).