Math Problem Statement

using this formula: N = (p1· p2· p3· . . . · pn) + 1 how can N be a non prime number but p1p2p3*... must be in sequence and N can only be divided by 2 or 3

Solution

To satisfy the conditions in the problem where NN must be a non-prime number, and N=(p1p2p3pn)+1N = (p_1 \cdot p_2 \cdot p_3 \cdot \ldots \cdot p_n) + 1, let's break down the steps:

  1. The sequence p1,p2,,pnp_1, p_2, \ldots, p_n must consist of consecutive prime numbers.
  2. NN must not be a prime number, meaning it should have divisors other than 1 and itself.
  3. The divisors of NN can only be 2 or 3, so NN must be divisible by either 2 or 3.

Example Process:

Step 1: Choose a sequence of consecutive primes

Start with small prime numbers for the sequence p1,p2,p3,p_1, p_2, p_3, \ldots.

For example, let’s take the first three primes: p1=2,p2=3,p3=5p_1 = 2, \quad p_2 = 3, \quad p_3 = 5

Step 2: Calculate NN

Using the formula: N=(p1p2p3)+1=(235)+1=30+1=31N = (p_1 \cdot p_2 \cdot p_3) + 1 = (2 \cdot 3 \cdot 5) + 1 = 30 + 1 = 31 In this case, N=31N = 31, which is a prime number, so this doesn't work.

Step 3: Try different sequences

Now try four consecutive primes: p1=2,p2=3,p3=5,p4=7p_1 = 2, \quad p_2 = 3, \quad p_3 = 5, \quad p_4 = 7 N=(2357)+1=210+1=211N = (2 \cdot 3 \cdot 5 \cdot 7) + 1 = 210 + 1 = 211 Again, 211 is a prime number, so it doesn't satisfy the condition.

Step 4: Continue expanding the sequence

Now, try five consecutive primes: p1=2,p2=3,p3=5,p4=7,p5=11p_1 = 2, \quad p_2 = 3, \quad p_3 = 5, \quad p_4 = 7, \quad p_5 = 11 N=(235711)+1=2310+1=2311N = (2 \cdot 3 \cdot 5 \cdot 7 \cdot 11) + 1 = 2310 + 1 = 2311 2311 is also a prime number.

Step 5: Adjusting for the divisibility rule

To find NN that satisfies the condition of being divisible by 2 or 3, we need to experiment further with sequences of primes and verify the conditions.

Final Answer:

The approach is to test sequences of primes and verify if NN satisfies the conditions. You can continue expanding the prime sequence until NN becomes divisible by 2 or 3 and is no longer prime. However, there's no straightforward pattern for such sequences; it requires checking larger products.

Would you like to see further calculations with larger primes, or do you have any questions about the process?

5 Related Questions:

  1. How do prime numbers behave in sequences when multiplied?
  2. Can a formula like this always generate a prime number?
  3. How can one prove that a number is divisible only by 2 or 3?
  4. What methods exist for checking the primality of a large number?
  5. Why does adding 1 to the product of primes often result in a prime number?

Tip:

When dealing with sequences of prime numbers and checking for divisibility, using a computer algorithm can help quickly test large sequences to see which satisfy the conditions.

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Math Problem Analysis

Mathematical Concepts

Prime Numbers
Multiplication of Primes
Divisibility

Formulas

N = (p1 * p2 * p3 * ... * pn) + 1

Theorems

Prime Number Theorem
Divisibility Rules for 2 and 3

Suitable Grade Level

Grades 10-12