The value of \((\sqrt{8!} + \sqrt{9!})^2\) is an integer. What is the greatest integer n such that 2^n is a...

1. Translate the problem requirements: We need to find the highest power of 2 that divides \((\sqrt{8!} + \sqrt{9!})^2\). The problem states this expression equals an integer, which gives us a crucial constraint.
2. Factor out common terms from the square roots: Simplify \(\sqrt{8!}\) and \(\sqrt{9!}\) by identifying their relationship and factoring out common elements.
3. Apply algebraic expansion: Use the \((a + b)^2\) formula to expand the squared expression and identify all terms.
4. Count powers of 2 systematically: Determine the power of 2 in each term of the expansion, then find the minimum power that appears in all terms to get the final answer.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what we're being asked to find. We have the expression \((\sqrt{8!} + \sqrt{9!})^2\) and we're told this equals an integer. We need to find the highest power of 2 that divides this integer.

First, let's understand what factorials give us:

- \(8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 40,320\)
- \(9! = 9 \times 8! = 9 \times 40,320 = 362,880\)

The key insight is that we're looking for powers of 2, so we need to think about how many factors of 2 are hidden inside these factorials.
Process Skill: TRANSLATE - Converting the problem from finding factors to understanding factorial structure

2. Factor out common terms from the square roots

Since \(9! = 9 \times 8!\), we can rewrite our square roots:
- \(\sqrt{9!} = \sqrt{9 \times 8!} = \sqrt{9} \times \sqrt{8!} = 3\sqrt{8!}\)

So our original expression becomes:
\((\sqrt{8!} + \sqrt{9!})^2 = (\sqrt{8!} + 3\sqrt{8!})^2 = (\sqrt{8!}(1 + 3))^2 = (4\sqrt{8!})^2\)

This simplifies to: \(16 \times 8!\)

Now we can see clearly what we're working with - we need to find the highest power of 2 that divides \(16 \times 8!\).
Process Skill: SIMPLIFY - Recognizing the relationship between 8! and 9! to factor out common terms

3. Apply algebraic expansion

We've already done the key algebraic work in the previous step by factoring out \(\sqrt{8!}\) and using the fact that \((4\sqrt{8!})^2 = 16 \times 8!\).

The expansion gave us:
\((\sqrt{8!} + 3\sqrt{8!})^2 = (4\sqrt{8!})^2 = 16 \times 8!\)

Since \(16 = 2^4\), we have: \(2^4 \times 8!\)

So the total power of 2 in our expression will be 4 plus however many powers of 2 are in \(8!\).

4. Count powers of 2 systematically

Now we need to count how many factors of 2 are in \(8!\). We can do this by looking at each number from 1 to 8 and counting their factors of 2:

- 1: 0 factors of 2
- 2: 1 factor of 2 (\(2^1\))
- 3: 0 factors of 2
- 4: 2 factors of 2 (\(2^2\))
- 5: 0 factors of 2
- 6: 1 factor of 2 (\(6 = 2 \times 3\))
- 7: 0 factors of 2
- 8: 3 factors of 2 (\(2^3\))

Total factors of 2 in \(8!\): \(1 + 2 + 1 + 3 = 7\)

Therefore, \(8!\) contains \(2^7\) as a factor.

Our complete expression is: \(2^4 \times 8! = 2^4 \times 2^7 \times \text{(other factors)} = 2^{11} \times \text{(other factors)}\)

The highest power of 2 that divides \((\sqrt{8!} + \sqrt{9!})^2\) is \(2^{11}\), so \(n = 11\).
Process Skill: APPLY CONSTRAINTS - Using the systematic counting method to ensure we don't miss any factors of 2

4. Final Answer

The greatest integer n such that \(2^n\) is a factor of \((\sqrt{8!} + \sqrt{9!})^2\) is \(n = 11\).

This matches answer choice D.

Common Faltering Points

Errors while devising the approach

1. Misunderstanding what the question is asking for
Students may confuse finding the value of \((\sqrt{8!} + \sqrt{9!})^2\) with finding the highest power of 2 that divides it. They might calculate the actual numerical value instead of focusing on the prime factorization approach needed to find powers of 2.

2. Not recognizing the factorial relationship
Many students fail to immediately see that \(9! = 9 \times 8!\), which is crucial for simplifying the square root expressions. Without this insight, they may attempt to calculate \(\sqrt{8!}\) and \(\sqrt{9!}\) separately, leading to very complex calculations.

3. Overlooking the constraint that the expression must be an integer
The problem states that \((\sqrt{8!} + \sqrt{9!})^2\) is an integer, but students might not use this information strategically. This constraint actually helps confirm that our approach is correct, but students may ignore it entirely.

Errors while executing the approach

1. Arithmetic errors when counting factors of 2 in 8!
When systematically counting powers of 2 in each number from 1 to 8, students commonly miscount. For example, they might say \(8 = 2^3\) gives 2 factors instead of 3, or forget that \(6 = 2 \times 3\) contributes 1 factor of 2.

2. Incorrect algebraic manipulation of square roots
Students may make errors when factoring out \(\sqrt{8!}\) from the expression \((\sqrt{8!} + 3\sqrt{8!})\). They might incorrectly expand \((4\sqrt{8!})^2\) as \(4^2 \times 8! = 16 \times 8!\) without properly handling the square root, leading to wrong power calculations.

3. Forgetting to add the powers of 2
After finding that \(16 = 2^4\) and \(8!\) contains \(2^7\), students may forget to add these exponents together. They might report either 4 or 7 as their final answer instead of 11.

Errors while selecting the answer

1. Confusing the final calculation
Even after correctly finding that the expression equals \(2^{11} \times \text{(other factors)}\), students might select \(2^{11} = 2048\) as their answer instead of recognizing that \(n = 11\) is what the question asks for.

2. Selecting an intermediate result
Students might select 7 (the power of 2 in \(8!\) alone) or 4 (the power of 2 in 16 alone) instead of the total power of 11, especially if they lose track of what the question is ultimately asking for.