Let S={p_1, p_2, ..., p_{10}} be the set of first ten prime numbers. Let A = S ∪ P, where P is the s

Q: Wait, why is $A$ defined as $S \cup P$? Does that change anything?

That is a great observation! $S$ is the set of products of subsets of size 1, and $P$ is the set of products of subsets of size greater than or equal to 1. Since any element in $S$ is also in $P$, $S \cup P$ is simply $P$. It is just the set of all products of non-empty subsets of $S$.

Q: Why do we say $2^9$ and not $2^{10}$?

Excellent question. We are counting pairs $(x, y)$ for a fixed $x$. Since $x$ is already chosen, it is 'locked in' as a factor of $y$. We only have freedom to choose whether to include or exclude the remaining 9 primes in the product $y$. Thus, we have $2^9$ choices for $y$ for each $x$.

Q: What if $y$ is just $x$ itself? Does that count?

Yes, absolutely! If $y = x$, then $x$ divides $y$ is true. This corresponds to the case where we choose none of the other 9 primes to be part of the product. This is included in our $2^9$ count.

Question

Let $S=\{p_{1},p_{2}....,p_{10}\}$ be the set of first ten prime numbers. Let $A=S\cup P$, where P is the set of all possible products of distinct elements of S. Then the number of all ordered pairs $(x,y)$, $x\in S$, $y\in A$ such that x divides y is

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The foundation of our problem is the set $S = \{p_1, p_2, \dots, p_{10}\}$, which consists of the first ten prime numbers. We define the set $A$ as the collection of all products of non-empty subsets of $S$.

Every element $y \in A$ is a square-free integer formed by the product of a unique non-empty subset of $S$. This construction is a direct application of the **Fundamental Theorem of Arithmetic**.

### Decoding the Divisibility Constraint

We are tasked with finding the number of ordered pairs $(x, y)$ such that $x \in S$, $y \in A$, and $x \mid y$.

Because $x$ is a prime number and $y$ is a product of distinct primes, the condition $x \mid y$ is satisfied if and only if $x$ is one of the prime factors of $y$. If $x$ is not a prime factor of $y$, it cannot divide $y$ due to the unique factorization property of integers.

### The Combinatorial Counting

Let us fix $x$ to be a specific prime from the set $S$. Since there are 10 primes in $S$, there are 10 possible choices for $x$.

For a fixed $x$, we must construct $y$ such that $x$ is a factor. This means $y$ must be a product of a subset of $S$ that includes $x$.

Since $x$ is already included in the product, we only need to determine the inclusion of the remaining 9 primes in $S$. For each of these 9 primes, we have exactly two choices: either include it in the product $y$ or exclude it.

The number of ways to form such a product $y$ is given by:

$$2^9 = 512$$

### The Final Synthesis

For each of the 10 possible choices of $x$, there are $2^9$ valid values for $y$. The total number of ordered pairs $(x, y)$ is the product of these two values:

$$10 	imes 2^9 = 10 	imes 512$$

The total number of ordered pairs is **5120**.

Enter Numerical Value:

Visualized Solution

Defining Set $S$

Understanding Set $A$

The Condition $x ∣ y$

Fixing Element $x$

Implication of $x ∣ y$

Constructing $y$

Choices for Remaining Primes

Number of $y$ for a Fixed $x$

Total Ordered Pairs

The Sigma Insight: Combinations and Selection

Analyzing the Setup

Decoding the Divisibility Constraint

The Combinatorial Counting

The Final Synthesis

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Visualized Solution

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

Understanding Set .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A

The Condition .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x∣y

Fixing Element .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x

Implication of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x∣y

Constructing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y