Math Problem Statement

State a set of integers of the form 3q rs14 with 3 < q < r < s < 14 for which the PDL contains no repeated integers.u

Solution

We need to find a set of integers of the form 3qrs143qrs14 where:

  • 3<q<r<s<143 < q < r < s < 14 (i.e., q,r,sq, r, s are distinct integers between 4 and 13)
  • The PDL (Prime Divisor List) contains no repeated integers

Step 1: Understanding the Prime Divisor List (PDL)

The Prime Divisor List (PDL) of a number is the list of all distinct prime factors of that number.

For 3qrs143qrs14, we first determine its prime factorization:

3qrs14=3qrs×14=3qrs×(2×7)3qrs14 = 3qrs \times 14 = 3qrs \times (2 \times 7)

Thus, the PDL of 3qrs143qrs14 is:

PDL={2,3,7}PDL of (qrs)\text{PDL} = \{2, 3, 7\} \cup \text{PDL of } (qrs)

To ensure no repeated integers in the PDL, q,r,sq, r, s must be chosen such that their prime factorizations introduce no duplicates from {2,3,7}\{2,3,7\} or among themselves.

Step 2: Choosing q,r,sq, r, s

We select values from 4q<r<s134 \leq q < r < s \leq 13, ensuring they introduce only new prime factors.

Example Set:

Let’s choose q=5q = 5, r=9r = 9, and s=11s = 11.

  1. Prime factorizations:
    • 55 (prime)
    • 9=329 = 3^2 (introduces 33, which is already in PDL)
    • 1111 (prime)

Since 33 is already in the PDL due to 3 itself, including 99 would repeat 33, which we must avoid.

Let’s try another set: q=5,r=8,s=11q = 5, r = 8, s = 11.

  1. Prime factorizations:
    • 55 (prime)
    • 8=238 = 2^3 (introduces 22, already in PDL)
    • 1111 (prime)

Again, 22 is repeated.

A better set is:
q=5,r=9,s=13q = 5, r = 9, s = 13

  1. Prime factorizations:
    • 55 (prime)
    • 9=329 = 3^2 (introduces 33, already in PDL)
    • 1313 (prime)

Since 9 introduces a repeated 33, we need to remove it.

A valid choice is q=5,r=10,s=11q = 5, r = 10, s = 11.

  1. Prime factorizations:
    • 55 (prime)
    • 10=2×510 = 2 \times 5 (introduces 2,52, 5)
    • 1111 (prime)

Thus, the PDL is:

{2,3,5,7,11}\{2, 3, 5, 7, 11\}

which contains no repeated numbers.

Answer:

A valid set is {3,5,10,11,14}\{3, 5, 10, 11, 14\}.


Do you need more examples or a deeper breakdown? Here are five related questions to think about:

  1. How can we systematically find all valid sets satisfying the conditions?
  2. What happens if we allow repeated integers in the PDL?
  3. How does the PDL change if we replace 14 with another number?
  4. What is the effect of choosing composite numbers for q,r,sq, r, s?
  5. Can we construct a general formula for sets that satisfy this condition?

Tip: When constructing sets with distinct prime factor lists, start by ensuring new primes are introduced at each step!

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Math Problem Analysis

Mathematical Concepts

Prime Factorization
Number Theory
Set Theory

Formulas

Prime Divisor List (PDL) = Union of distinct prime factors

Theorems

Prime Factorization Theorem

Suitable Grade Level

Grades 10-12