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

Q: Wait, why are we using complex numbers like $\omega$ for a simple counting problem?

It feels like magic, doesn't it? The Roots of Unity Filter is a powerful algebraic sieve. Because $1 + \omega^k + \omega^{2k}$ equals 3 if $k$ is a multiple of 3, and 0 otherwise, it acts as a perfect switch to 'turn on' only the terms we care about. It is a standard, elegant technique in combinatorics for problems involving modular arithmetic.

Q: Why do we subtract 1 at the end? Is the empty set really a multiple of 3?

Yes, mathematically, the sum of elements in an empty set is defined as 0. Since $0 = 3 \times 0$, it is indeed a multiple of 3. However, the question specifically asks for 'non-empty' subsets, so we must remove that one case from our total count.

Q: Can't I just use brute force since the set $S$ is small?

With 7 elements, there are $2^7 = 128$ subsets. While you *could* list them, it is prone to human error and takes too long. Learning the Generating Function approach is vital because it scales to much larger sets where brute force becomes impossible. Master the method, not just the answer!

Question

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 ____.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

To find the number of non-empty subsets of $S = \{1, 2, 3, 5, 7, 10, 11\}$ whose sum is a multiple of 3, we first classify the elements based on their remainders modulo 3.

The set $S$ is partitioned as follows:
*   Type $R_0$ (remainder 0): $\{3\}$ (1 element)
*   Type $R_1$ (remainder 1): $\{1, 7, 10\}$ (3 elements)
*   Type $R_2$ (remainder 2): $\{2, 5, 11\}$ (3 elements)

### The Generating Function

We define the generating function $P(x)$ to track the subset sums:
$$P(x) = \prod_{a \in S} (1 + x^a)$$

When expanded, the coefficient of $x^n$ in $P(x)$ represents the number of subsets of $S$ that sum to $n$. Our goal is to find the sum of coefficients of $x^n$ where $n$ is a multiple of 3.

### The Roots of Unity Filter

We employ the Roots of Unity Filter to extract the sum of coefficients for powers that are multiples of 3. The number of such subsets $N$ is given by:
$$N = \frac{1}{3} [P(1) + P(\omega) + P(\omega^2)]$$

Here, $\omega = e^{i \frac{2\pi}{3}}$ is a primitive cube root of unity. This filter works because the sum $1 + \omega^k + \omega^{2k}$ equals 3 if $k$ is a multiple of 3, and 0 otherwise.

### The Calculation

First, we calculate $P(1)$:
$$P(1) = (1+1)^7 = 2^7 = 128$$

Next, we evaluate $P(\omega)$ by substituting $\omega$ into the product:
$$P(\omega) = (1+\omega^3) \cdot (1+\omega)^3 \cdot (1+\omega^2)^3$$

Since $\omega^3 = 1$, we have $1+\omega^3 = 2$. Using the identity $1 + \omega + \omega^2 = 0$, we substitute $1+\omega = -\omega^2$ and $1+\omega^2 = -\omega$:
$$P(\omega) = 2 \cdot (-\omega^2)^3 \cdot (-\omega)^3 = 2 \cdot (-1) \cdot (-1) = 2$$

Because $P(x)$ has real coefficients, $P(\omega^2)$ is the complex conjugate of $P(\omega)$, yielding $P(\omega^2) = 2$.

### Final Calculation

Plugging these values into our filter formula:
$$N = \frac{1}{3} [128 + 2 + 2] = \frac{132}{3} = 44$$

This count of 44 includes the empty set, which has a sum of 0 (a multiple of 3). Since the problem asks for non-empty subsets, we subtract 1 from the total.

The final answer is **43**.

Animated Solution for Mathematics - Permutations and Combinations: 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 ____.

Enter Numerical Value:

Visualized Solution

Categorizing Elements of $S$

The Generating Function Approach

The Roots of Unity Filter

Evaluating $P (1)$

Evaluating $P (ω)$ Setup

Simplifying $P (ω)$

Evaluating $P (ω^{2})$

Final Calculation

The Non-empty Condition

The Sigma Insight: Combinations and Selection

Analyzing the Setup

The Generating Function

The Roots of Unity Filter

The Calculation

Final Calculation

Similar Questions

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Enter Numerical Value:

Visualized Solution

Categorizing Elements of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S

The Generating Function Approach

The Roots of Unity Filter

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(1)

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(ω) Setup

Simplifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(ω)

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(ω2)

Final Calculation

The Non-empty Condition

The Sigma Insight: Combinations and Selection

Analyzing the Setup

The Generating Function

The Roots of Unity Filter

The Calculation

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Similar Questions

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