Let n be the least positive integer that has exactly 5 different prime factors. What is the greatest value of...

1. Translate the problem requirements: We need to find n (the smallest positive integer with exactly 5 different prime factors), then determine the highest power of 2 that divides \(\frac{12!}{n}\)
2. Identify the smallest number with exactly 5 prime factors: Find n by taking the product of the 5 smallest prime numbers
3. Count powers of 2 in the numerator and denominator: Calculate how many factors of 2 are in \(12!\) and how many are in n
4. Apply the division rule for powers: Subtract the power of 2 in the denominator from the power of 2 in the numerator to get the final answer

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what this problem is asking us to find step by step.

First, we need to understand what "exactly 5 different prime factors" means. A prime factor is a prime number that divides evenly into our number. For example, \(12 = 2^2 \times 3\), so 12 has 2 different prime factors (2 and 3).

We want the smallest positive integer that has exactly 5 different prime factors. This means we want a number that is divisible by exactly 5 different prime numbers.

Then we need to find the greatest value of k such that \(2^k\) divides \(\frac{12!}{n}\). This is asking: what's the highest power of 2 that goes into \(\frac{12!}{n}\)?

Process Skill: TRANSLATE - Converting the problem language into clear mathematical understanding

2. Identify the smallest number with exactly 5 prime factors

To get the smallest number with exactly 5 different prime factors, we should use the smallest possible prime numbers, each taken to the first power.

The first 5 prime numbers are: 2, 3, 5, 7, 11

Therefore: \(n = 2 \times 3 \times 5 \times 7 \times 11 = 2,310\)

This gives us the smallest positive integer with exactly 5 different prime factors.

3. Count powers of 2 in the numerator and denominator

Now we need to find how many factors of 2 are in \(12!\) and how many are in n.

First, let's count the powers of 2 in \(12!\):
In \(12!\), we count multiples of 2: 2, 4, 6, 8, 10, 12 → that's 6 numbers
Then multiples of 4 (\(2^2\)): 4, 8, 12 → that's 3 more factors of 2
Then multiples of 8 (\(2^3\)): 8 → that's 1 more factor of 2
Then multiples of 16: none in \(12!\)

Total powers of 2 in \(12!\) = 6 + 3 + 1 = 10

Next, let's count the powers of 2 in n:
Since \(n = 2 \times 3 \times 5 \times 7 \times 11\), there is exactly 1 factor of 2 in n.

Process Skill: MANIPULATE - Systematically counting prime factors using the standard technique

4. Apply the division rule for powers

When we divide \(\frac{12!}{n}\), the power of 2 in the result equals:
(Power of 2 in \(12!\)) - (Power of 2 in n)

Power of 2 in \(\frac{12!}{n}\) = 10 - 1 = 9

Therefore, \(2^9\) is the highest power of 2 that divides \(\frac{12!}{n}\).

4. Final Answer

The greatest value of k such that \(2^k\) divides \(\frac{12!}{n}\) is k = 9.

This matches answer choice C.

Common Faltering Points

Errors while devising the approach

Students often confuse "exactly 5 different prime factors" with "exactly 5 prime factors total." For example, they might think that \(2^3 \times 3 \times 5 \times 7 \times 11\) still has exactly 5 different prime factors, when the question asks for the least such number. The correct interpretation requires using each of the 5 smallest primes (2, 3, 5, 7, 11) exactly once to minimize the value.

Some students might randomly select 5 primes instead of systematically choosing the 5 smallest primes. Since we want the least positive integer, we must use 2, 3, 5, 7, and 11 (the first five primes) rather than any other combination like 3, 5, 7, 11, 13.

Students may get confused about what they're looking for in the final step. The question asks for "the greatest value of k such that \(2^k\) divides \(\frac{12!}{n}\)" but students might mistakenly try to find the total number of prime factors or work with other primes besides 2.

Errors while executing the approach

When computing \(n = 2 \times 3 \times 5 \times 7 \times 11\), students often make multiplication errors. While the exact value of n (2,310) isn't crucial for solving the problem, errors here can create confusion and lead to mistakes in subsequent steps.

The most critical calculation error occurs when counting powers of 2 in \(12!\). Students frequently forget to use the systematic approach: count multiples of 2 (contributing 6), then multiples of 4 (contributing 3 more), then multiples of 8 (contributing 1 more). Common mistakes include counting each number only once or forgetting higher powers of 2.

When finding the power of 2 in \(\frac{12!}{n}\), students must subtract (power of 2 in n) from (power of 2 in \(12!\)). Since n contains exactly one factor of 2, the calculation should be 10 - 1 = 9. Students sometimes forget this subtraction step or subtract incorrectly.

Errors while selecting the answer

No likely faltering points - the question directly asks for the value of k, and once students correctly calculate that \(2^9\) is the highest power of 2 that divides \(\frac{12!}{n}\), the answer k = 9 follows immediately without additional manipulation or interpretation.