If N is a positive odd integer, is N prime? N = 2^k + 1 for some positive integer k....

Understanding the Question

We need to determine whether N is prime, where N is a positive odd integer.

This is a yes/no question. For a statement to be sufficient, it must allow us to definitively answer either "Yes, N is prime" or "No, N is not prime." If we can find cases where N could be prime AND cases where N could be composite, then the statement is NOT sufficient.

Key Insight

A prime number has exactly two factors: 1 and itself. Any number with additional factors is composite. Since we're looking for a definitive answer about N's primality, we need information that rules out one possibility entirely.

Analyzing Statement 1

Statement 1 tells us that \(\mathrm{N = 2^k + 1}\) for some positive integer k.

Let's test different values of k to see what happens:

• \(\mathrm{k = 1: N = 2^1 + 1 = 3}\) (prime)
• \(\mathrm{k = 2: N = 2^2 + 1 = 5}\) (prime)
• \(\mathrm{k = 3: N = 2^3 + 1 = 9 = 3 \times 3}\) (NOT prime)

We've found that N can be prime (when k = 1 or k = 2) and N can be composite (when k = 3).

Since we get different answers to our question "Is N prime?" depending on the value of k, Statement 1 alone is NOT sufficient.

[STOP - Not Sufficient!] This eliminates choices A and D.

Analyzing Statement 2

Now let's forget Statement 1 completely and analyze Statement 2 independently.

Statement 2 tells us that \(\mathrm{N + 2}\) and \(\mathrm{N + 4}\) are both prime.

Since N is odd (given), both \(\mathrm{N + 2}\) and \(\mathrm{N + 4}\) are also odd. We have three consecutive odd numbers: N, \(\mathrm{N + 2}\), and \(\mathrm{N + 4}\), where the last two are prime.

Let's test some values:

• If \(\mathrm{N = 3}\): Then \(\mathrm{N + 2 = 5}\) (prime ✓) and \(\mathrm{N + 4 = 7}\) (prime ✓)
  And \(\mathrm{N = 3}\) is prime.
• If \(\mathrm{N = 9}\): Then \(\mathrm{N + 2 = 11}\) (prime ✓) and \(\mathrm{N + 4 = 13}\) (prime ✓)
  But \(\mathrm{N = 9 = 3 \times 3}\) is NOT prime.

We've found cases where N is prime (\(\mathrm{N = 3}\)) and cases where N is composite (\(\mathrm{N = 9}\)), even though \(\mathrm{N + 2}\) and \(\mathrm{N + 4}\) are prime in both cases.

Statement 2 alone is NOT sufficient.

[STOP - Not Sufficient!] This eliminates choice B.

Combining Both Statements

Now let's use BOTH statements together:

• \(\mathrm{N = 2^k + 1}\) for some positive integer k
• \(\mathrm{N + 2}\) and \(\mathrm{N + 4}\) are both prime

This means we need:

• \(\mathrm{2^k + 3}\) is prime
• \(\mathrm{2^k + 5}\) is prime

Let's check our previous values:

• \(\mathrm{k = 1: N = 3}\), so \(\mathrm{N + 2 = 5}\) (prime ✓) and \(\mathrm{N + 4 = 7}\) (prime ✓)
  Here \(\mathrm{N = 3}\) is prime.
• \(\mathrm{k = 2: N = 5}\), so \(\mathrm{N + 2 = 7}\) (prime ✓) and \(\mathrm{N + 4 = 9 = 3 \times 3}\) (NOT prime ✗)
  This case doesn't satisfy both conditions.
• \(\mathrm{k = 3: N = 9}\), so \(\mathrm{N + 2 = 11}\) (prime ✓) and \(\mathrm{N + 4 = 13}\) (prime ✓)
  Here \(\mathrm{N = 9 = 3 \times 3}\) is NOT prime.

We still have both possibilities:

• When \(\mathrm{k = 1: N = 3}\) (prime)
• When \(\mathrm{k = 3: N = 9}\) (composite)

Even with both statements combined, we cannot definitively determine whether N is prime.

Both statements together are NOT sufficient.

[STOP - Not Sufficient!] This eliminates choice C.

The Answer: E

The statements together are not sufficient because we can find values of N that satisfy both conditions where N is prime (\(\mathrm{N = 3}\)) and other values where N is composite (\(\mathrm{N = 9}\)).

Answer Choice E: Both statements together are still not sufficient.