Question

If n is the product of the integers from 1 to 8, inclusive, how many different prime factors greater than 1 does n have?

• A. Four
• B. Five
• C. Six
• D. Seven
• E. Eight

Hint:

To find the number of distinct prime factors of a factorial, \(k!\), you simply need to count the number of prime numbers less than or equal to \(k\).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The problem asks for the number of distinct prime factors of the number \(n = 8!\) (8 factorial). A prime factor of a product of integers must be a prime factor of at least one of those integers.
Step 2: Key Formula or Approach:
We need to find all the unique prime numbers that are less than or equal to 8. The product \(1 \times 2 \times \dots \times 8\) will have these primes as its factors.
Step 3: Detailed Explanation:
The number n is the product: \[ n = 1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8 \] To find the distinct prime factors of n, we just need to list all the prime numbers that appear in the prime factorization of any of the numbers from 1 to 8. The numbers from 1 to 8 are: 1, 2, 3, 4, 5, 6, 7, 8. Let's find the prime numbers in this list or the prime factors of the composite numbers in this list.

2 is a prime number.
3 is a prime number.
4 is not prime, its prime factor is 2.
5 is a prime number.
6 is not prime, its prime factors are 2 and 3.
7 is a prime number.
8 is not prime, its prime factor is 2.
The set of all unique prime factors we have found is \{2, 3, 5, 7\}.
There are four distinct prime factors. Prime numbers are by definition greater than 1.
Step 4: Final Answer:
The number n has four different prime factors greater than 1.