Does the integer k have a factor p such that 1 ? (k > 4! 13! + 2 leq k...

Understanding the Question

We need to determine whether integer k has a factor p such that \(1 < \mathrm{p} < \mathrm{k}\).

In simpler terms: Is k a composite number?

• If k is composite → It has factors other than 1 and itself → Answer is YES
• If k is prime or k = 1 → No such factors exist → Answer is NO

For this yes/no question, we need definitive information about whether k is composite.

Analyzing Statement 1

Statement 1: \(\mathrm{k} > 4! = 24\)

This means \(\mathrm{k} \geq 25\). Let's test whether all numbers ≥ 25 are composite:

• If \(\mathrm{k} = 25 = 5^2\), then 5 is a factor where \(1 < 5 < 25\) → Answer is YES
• If \(\mathrm{k} = 29\) (prime), then no factor exists where \(1 < \mathrm{p} < 29\) → Answer is NO

Since we can get both YES and NO answers, Statement 1 is NOT sufficient.

This eliminates choices A and D.

Analyzing Statement 2

Starting fresh—we must forget Statement 1 completely.

Statement 2: \(13! + 2 \leq \mathrm{k} \leq 13! + 13\)

This gives us a very specific range. Any k in this range can be written as \(\mathrm{k} = 13! + \mathrm{a}\), where \(2 \leq \mathrm{a} \leq 13\).

Key Insight: What factors does 13! have?

• \(13! = 1 \times 2 \times 3 \times 4 \times \ldots \times 13\)
• Therefore, 13! is divisible by every integer from 2 to 13

This creates a powerful pattern:

• When \(\mathrm{k} = 13! + 2\), since 13! is divisible by 2, we have \(\mathrm{k} \equiv 2 \pmod{2}\), so 2 divides k
• When \(\mathrm{k} = 13! + 3\), since 13! is divisible by 3, we have \(\mathrm{k} \equiv 3 \pmod{3}\), so 3 divides k
• When \(\mathrm{k} = 13! + 4\), since 13! is divisible by 4, we have \(\mathrm{k} \equiv 4 \pmod{4}\), so 4 divides k
• This pattern continues for every value in our range

Since \(13! = 6,227,020,800\) (huge!) and \(\mathrm{a} \leq 13\), we definitely have \(1 < \mathrm{a} < \mathrm{k}\).

Therefore, every integer in this range is composite.

The answer is definitively YES. Statement 2 is sufficient.

[STOP - Sufficient!]

The Answer: B

Statement 1 alone: Not sufficient (both composite and prime numbers exist above 24)
Statement 2 alone: Sufficient (all numbers in the range are composite)

Answer: B - Statement 2 alone is sufficient, but Statement 1 alone is not sufficient.