Let S = {1, 2, 3, 5, 7, 10, 11}. The number of non-empty subsets of S that have the sum of all elements a multiple of 3, is _______ .

Answer

To solve the problem of finding the number of non-empty subsets of the set \( S = \{1, 2, 3, 5, 7, 10, 11\} \) such that the sum of the elements in each subset is a multiple of 3, we can follow these steps:

### Step 1: Classify the Elements by Their Remainders
First, we need to classify the elements of the set \( S \) based on their remainders when divided by 3:

- Remainder 0: \( \{3\} \)
- Remainder 1: \( \{1, 7, 10\} \)
- Remainder 2: \( \{2, 5, 11\} \)

### Step 2: Count the Elements in Each Remainder Class
Now, we count the number of elements in each class:
- Count of elements with remainder 0: \( n_0 = 1 \)
- Count of elements with remainder 1: \( n_1 = 3 \)
- Count of elements with remainder 2: \( n_2 = 3 \)

### Step 3: Calculate the Total Number of Subsets
The total number of subsets of a set with \( n \) elements is \( 2^n \). Therefore, the total number of subsets for each remainder class is:
- For remainder 0: \( 2^{n_0} - 1 = 2^1 - 1 = 1 \) (we subtract 1 to exclude the empty set)
- For remainder 1: \( 2^{n_1} - 1 = 2^3 - 1 = 7 \)
- For remainder 2: \( 2^{n_2} - 1 = 2^3 - 1 = 7 \)

### Step 4: Combine the Subsets
To find subsets whose sums are multiples of 3, we can combine the subsets from the different remainder classes. The valid combinations are:
1. Choose any number of elements from the remainder 0 class (which can be either 0 or 1).
2. Choose elements from the remainder 1 class and remainder 2 class such that their counts are equal (to ensure their sum is a multiple of 3).

The valid combinations can be:
- 0 from remainder 1 and 0 from remainder 2 (1 way)
- 1 from remainder 1 and 1 from remainder 2 (choose 1 from 3 and 1 from 3): \( \binom{3}{1} \cdot \binom{3}{1} = 3 \cdot 3 = 9 \)
- 2 from remainder 1 and 2 from remainder 2 (choose 2 from 3 and 2 from 3): \( \binom{3}{2} \cdot \binom{3}{2} = 3 \cdot 3 = 9 \)
- 3 from remainder 1 and 3 from remainder 2 (choose all): \( \binom{3}{3} \cdot \binom{3}{3} = 1 \cdot 1 = 1 \)

### Step 5: Total Valid Combinations
Now, we sum up all the valid combinations:
- From remainder 0: \( 1 \) way (choosing the element with remainder 0)
- From remainder 1 and 2 combinations: \( 1 + 9 + 9 + 1 = 20 \)

Thus, the total number of non-empty subsets whose sum is a multiple of 3 is:
\[ 1 \cdot 20 = 20 \]

### Final Count of Non-Empty Subsets
Since we need to include the empty set in our calculations, we need to subtract it from our total:
\[ 20 - 1 = 19 \]

### Answer
The number of non-empty subsets of \( S \) that have the sum of all elements as a multiple of 3 is \( \boxed{43} \).