How many positive integer divisors does 8! have?

1. Translate the problem requirements: We need to find how many positive integers divide evenly into \(8!\) (\(8\) factorial = \(8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1\)). A divisor is any positive integer that divides \(8!\) with no remainder.
2. Express \(8!\) in terms of prime factors: Break down \(8!\) into its prime factorization by counting how many times each prime appears in the product \(8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1\).
3. Apply the divisor counting principle: Any divisor of \(8!\) must be formed by taking some (or none) of each prime factor. If a prime p appears n times in the factorization, we can choose it 0, 1, 2, ..., or n times in our divisor.
4. Calculate the total combinations: Multiply the number of choices for each prime factor to get the total number of positive divisors.

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what we're looking for. We want to find how many positive integer divisors \(8!\) has.

First, let's calculate what \(8!\) actually equals:
\(8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 40,320\)

A divisor of \(8!\) is any positive integer that divides evenly into \(40,320\) with no remainder. For example, 1, 2, 3, 4, 5, 6, 7, 8, 10, 12, etc. are all divisors of \(8!\).

Rather than trying to list all divisors (which would be tedious with \(40,320\)), we'll use a systematic approach based on prime factorization.

Process Skill: TRANSLATE - Converting the factorial notation into a concrete multiplication and understanding what "divisor" means in this context.

2. Express \(8!\) in terms of prime factors

To count divisors systematically, we need to break down \(8!\) into its prime factorization. Let's think about this step by step.

When we multiply \(8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1\), we're including all the prime factors that appear in each of these numbers:

• \(8 = 2^3\) (contributes three 2's)
• \(7 = 7^1\) (contributes one 7)
• \(6 = 2 \times 3\) (contributes one 2 and one 3)
• \(5 = 5^1\) (contributes one 5)
• \(4 = 2^2\) (contributes two 2's)
• \(3 = 3^1\) (contributes one 3)
• \(2 = 2^1\) (contributes one 2)
• \(1 = 1\) (contributes nothing)

Now let's count how many times each prime appears:
• Prime 2 appears: 3 times (from 8) + 1 time (from 6) + 2 times (from 4) + 1 time (from 2) = 7 times total
• Prime 3 appears: 1 time (from 6) + 1 time (from 3) = 2 times total
• Prime 5 appears: 1 time (from 5) = 1 time total
• Prime 7 appears: 1 time (from 7) = 1 time total

Therefore: \(8! = 2^7 \times 3^2 \times 5^1 \times 7^1\)

3. Apply the divisor counting principle

Here's the key insight: any divisor of \(8!\) must be formed by taking some combination of these prime factors, but we can't use more of any prime than what's available.

Think of it like choosing ingredients from a pantry:
• For the prime 2: we can choose to use 0, 1, 2, 3, 4, 5, 6, or 7 copies (that's 8 choices)
• For the prime 3: we can choose to use 0, 1, or 2 copies (that's 3 choices)
• For the prime 5: we can choose to use 0 or 1 copy (that's 2 choices)
• For the prime 7: we can choose to use 0 or 1 copy (that's 2 choices)

For example:
• If we choose \(2^2\), \(3^1\), \(5^0\), \(7^0\), we get the divisor \(4 \times 3 \times 1 \times 1 = 12\)
• If we choose \(2^1\), \(3^0\), \(5^1\), \(7^1\), we get the divisor \(2 \times 1 \times 5 \times 7 = 70\)
• If we choose \(2^0\), \(3^0\), \(5^0\), \(7^0\), we get the divisor 1

Each valid combination gives us exactly one divisor of \(8!\).

4. Calculate the total combinations

Since we make an independent choice for each prime factor, we multiply the number of choices:

Total number of divisors = (number of ways to choose powers of 2) × (number of ways to choose powers of 3) × (number of ways to choose powers of 5) × (number of ways to choose powers of 7)

Total number of divisors = \(8 \times 3 \times 2 \times 2 = 96\)

Let's verify this calculation:
\(8 \times 3 = 24\)
\(24 \times 2 = 48\)
\(48 \times 2 = 96\)

Final Answer

The number of positive integer divisors of \(8!\) is 96.

Looking at our answer choices:
A. 14 ❌
B. 15 ❌
C. 24 ❌
D. 56 ❌
E. 96 ✓

The answer is E.

Common Faltering Points

Errors while devising the approach

• Attempting to calculate \(8! = 40,320\) and then trying to find all divisors by trial division: Students often think they need to compute the actual value of \(8!\) and then test each number from 1 to \(40,320\) to see if it divides evenly. This approach is extremely time-consuming and impractical for the GMAT.
• Not recognizing that prime factorization is the key strategy: Students may understand what divisors are but fail to see that the systematic way to count them is through prime factorization, leading them to attempt unsystematic counting methods.
• Confusing divisors with factors in the multiplication: Some students think the question is asking for how many numbers are being multiplied (which would be 8), rather than how many positive integers divide the final product evenly.

Errors while executing the approach

• Miscounting prime factors when breaking down each number: Students often make errors when determining how many times each prime appears. For example, they might forget that \(8 = 2^3\) contributes 3 copies of the prime 2, or they might miscount the total occurrences when adding up contributions from different numbers (like getting 6 instead of 7 for the power of 2).
• Forgetting to include the option of using 0 copies of each prime: When applying the divisor formula, students frequently forget that they can choose to use 0, 1, 2, ..., up to the maximum power of each prime. For instance, for \(2^7\), there are 8 choices (powers 0 through 7), not 7 choices.
• Arithmetic errors in the final multiplication: Students may correctly identify that they need to multiply \((7+1) \times (2+1) \times (1+1) \times (1+1) = 8 \times 3 \times 2 \times 2\), but then make calculation mistakes, such as getting \(8 \times 3 \times 2 \times 2 = 48\) instead of 96.

Errors while selecting the answer

• Selecting a partial calculation as the final answer: Students might stop at an intermediate step in their calculation. For example, they might calculate \(8 \times 3 = 24\) and select choice C (24) without completing the full multiplication \(8 \times 3 \times 2 \times 2 = 96\).