If n is a positive integer and n^2 is divisible by 72, then the largest positive integer that must divide...

1. Translate the problem requirements: We need to find the largest positive integer that MUST divide n, given that n is positive and \(\mathrm{n}^2\) is divisible by 72. This means we're looking for the greatest common factor that all possible values of n must share.
2. Find the prime factorization structure: Break down 72 into its prime factors to understand what divisibility by 72 requires.
3. Determine minimum requirements for n: Since \(\mathrm{n}^2\) must be divisible by 72, figure out what prime factors n must contain at minimum to satisfy this condition.
4. Identify the largest guaranteed divisor: From the minimum requirements, determine the largest number that must divide every valid n.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what this question is really asking. We know that n is a positive integer, and \(\mathrm{n}^2\) (n squared) is divisible by 72. The question wants the largest positive integer that MUST divide n.

The key word here is "MUST" - this means we're looking for the biggest number that will divide ALL possible values of n that satisfy our condition. In other words, we need to find the greatest common factor shared by every valid n.

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

2. Find the prime factorization structure

To understand what "divisible by 72" means, let's break 72 down into its basic building blocks (prime factors):

\(72 = 8 \times 9 = 2^3 \times 3^2\)

So \(72 = 2^3 \times 3^2\)

This tells us that for \(\mathrm{n}^2\) to be divisible by 72, \(\mathrm{n}^2\) must contain at least three factors of 2 and at least two factors of 3.

3. Determine minimum requirements for n

Here's the crucial insight: if \(\mathrm{n}^2\) must be divisible by \(2^3 \times 3^2\), what does this tell us about n itself?

Think about it this way: when we square a number, we double all the exponents in its prime factorization. For example, if \(\mathrm{n} = 2^\mathrm{a} \times 3^\mathrm{b} \times \text{(other primes)}\), then \(\mathrm{n}^2 = 2^{(2\mathrm{a})} \times 3^{(2\mathrm{b})} \times \text{(other primes)}^2\).

Since \(\mathrm{n}^2\) must be divisible by \(2^3 \times 3^2\), we need:

• \(\mathrm{n}^2\) must have at least 3 factors of 2, so \(2\mathrm{a} \geq 3\), which means \(\mathrm{a} \geq 1.5\). Since a must be a whole number, \(\mathrm{a} \geq 2\)
• \(\mathrm{n}^2\) must have at least 2 factors of 3, so \(2\mathrm{b} \geq 2\), which means \(\mathrm{b} \geq 1\)

Therefore, n must contain at least \(2^2 = 4\) factors of 2 and at least \(3^1 = 3\) factors of 3.

This means every valid n must be divisible by \(2^2 \times 3^1 = 4 \times 3 = 12\).

Process Skill: INFER - Drawing the non-obvious conclusion about minimum prime factor requirements

4. Identify the largest guaranteed divisor

From our analysis, every valid value of n must be divisible by at least \(2^2 \times 3^1 = 12\).

Let's verify this with an example: if \(\mathrm{n} = 12\), then \(\mathrm{n}^2 = 144\). Let's check: \(144 \div 72 = 2\) ✓

Could the answer be larger than 12? Let's check if every valid n must be divisible by 24:

If \(\mathrm{n} = 12\), then \(12 \div 24 = 0.5\), which is not a whole number. So not every valid n is divisible by 24.

Therefore, the largest positive integer that MUST divide n is 12.

5. Final Answer

The largest positive integer that must divide n is 12.

This corresponds to answer choice B.

Common Faltering Points

Errors while devising the approach

1. Misinterpreting "must divide" vs "can divide"

Students often confuse what the question is asking. They might look for the largest possible divisor of n rather than the largest divisor that is guaranteed to divide ALL valid values of n. This leads them to find maximum possible values instead of the greatest common factor across all cases.

2. Working directly with n instead of using the \(\mathrm{n}^2\) constraint

Students may try to find divisors of n directly without properly using the given information that \(\mathrm{n}^2\) is divisible by 72. They miss the key insight that they need to work backwards from the \(\mathrm{n}^2\) condition to determine what this tells them about n itself.

3. Forgetting that prime factor exponents double when squaring

Students often fail to recognize the relationship between the prime factorization of n and \(\mathrm{n}^2\). They may not realize that if \(\mathrm{n} = 2^\mathrm{a} \times 3^\mathrm{b}\), then \(\mathrm{n}^2 = 2^{(2\mathrm{a})} \times 3^{(2\mathrm{b})}\), which is crucial for determining the minimum requirements for n.

Errors while executing the approach

1. Incorrect handling of minimum exponent requirements

When determining that \(\mathrm{n}^2\) must have at least 3 factors of 2, students may incorrectly conclude that n needs at least 3 factors of 2 (instead of the correct answer: at least 2 factors). They forget to divide the required exponents by 2 when working backwards from \(\mathrm{n}^2\) to n.

2. Arithmetic errors in prime factorization

Students may make basic computational mistakes when finding \(72 = 2^3 \times 3^2\) or when calculating the minimum value \(2^2 \times 3^1 = 12\). These arithmetic slips can lead to wrong intermediate results.

Errors while selecting the answer

1. Choosing a larger answer without proper verification

After finding that 12 must divide n, students may assume that larger values like 24 or 36 must also divide n without testing counterexamples. They fail to verify that \(\mathrm{n} = 12\) (which satisfies all conditions) is not divisible by these larger values, proving these cannot be universal divisors.

Alternate Solutions

Smart Numbers Approach

Step 1: Choose a strategic smart number for n

Since we need \(\mathrm{n}^2\) to be divisible by 72, let's work backwards from the prime factorization of 72.

\(72 = 2^3 \times 3^2 = 8 \times 9\)

For \(\mathrm{n}^2\) to be divisible by \(2^3 \times 3^2\), we need \(\mathrm{n}^2\) to contain at least three factors of 2 and two factors of 3. Since \(\mathrm{n}^2\) means each prime factor in n appears twice, n must contain enough prime factors so that when squared, we get at least \(2^3 \times 3^2\).

Step 2: Determine the minimum prime factorization of n

For \(\mathrm{n}^2\) to contain \(2^3\), n must contain at least \(2^2\) (since we need an odd power of 2 in \(\mathrm{n}^2\) and \(2^2 \times 2^2 = 2^4 > 2^3\)).
For \(\mathrm{n}^2\) to contain \(3^2\), n must contain at least \(3^1\) (since \(3^1 \times 3^1 = 3^2\)).

Therefore, n must be divisible by \(2^2 \times 3^1 = 4 \times 3 = 12\).

Step 3: Test with concrete smart numbers

Let's verify with \(\mathrm{n} = 12\):
\(\mathrm{n}^2 = 144\)
\(144 \div 72 = 2\) ✓ (divisible)

Let's try \(\mathrm{n} = 24\) (which is \(12 \times 2\)):
\(\mathrm{n}^2 = 576\)
\(576 \div 72 = 8\) ✓ (divisible)

Let's try \(\mathrm{n} = 36\) (which is \(12 \times 3\)):
\(\mathrm{n}^2 = 1296\)
\(1296 \div 72 = 18\) ✓ (divisible)

Step 4: Verify 12 is the largest number that MUST divide n

All valid values of n (12, 24, 36, 60, 72, etc.) are multiples of 12, but not all are multiples of any number larger than 12. For example, \(\mathrm{n} = 12\) is not divisible by 24.

Therefore, 12 is the largest positive integer that must divide n.

Answer: B