If n is a positive integer such that 2^(51) and \((2^n)(3^3)\) have the same number of positive factors, what is...

1. Translate the problem requirements: We need to find the value of n such that \(2^{51}\) and \((2^n)(3^3)\) have the same number of positive factors. This means we need to count factors for both expressions and set them equal.
2. Apply the factor counting formula: Use the rule that for a number with prime factorization \(p_1^{a_1} \times p_2^{a_2} \times \ldots \times p_k^{a_k}\), the number of positive factors is \((a_1 + 1)(a_2 + 1)\ldots(a_k + 1)\).
3. Calculate factors for the first expression: Determine the number of positive factors of \(2^{51}\).
4. Set up the equation: Express the number of factors of \((2^n)(3^3)\) and set it equal to the result from step 3.
5. Solve for n: Algebraically solve the equation to find the value of n.

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what we're looking for in everyday terms. We have two numbers: \(2^{51}\) and \((2^n)(3^3)\). The problem tells us these two numbers have the same number of positive factors, and we need to find what value of n makes this true.

Think of positive factors as all the whole numbers that divide evenly into our number. For example, the number 12 has factors 1, 2, 3, 4, 6, and 12 - that's 6 factors total.

So our job is to count how many factors each expression has and set them equal.

Process Skill: TRANSLATE

2. Apply the factor counting formula

There's a beautiful pattern for counting factors when we know the prime factorization of a number. Let me explain this with a simple example first.

Consider \(2^3 = 8\). The factors are: \(2^0 = 1\), \(2^1 = 2\), \(2^2 = 4\), \(2^3 = 8\). That's 4 factors total. Notice that we can choose the exponent of 2 to be 0, 1, 2, or 3 - that's 4 choices, which gives us 4 factors.

For \(2^3 \times 3^2 = 8 \times 9 = 72\), we can choose the exponent of 2 to be 0, 1, 2, or 3 (4 choices) AND independently choose the exponent of 3 to be 0, 1, or 2 (3 choices). This gives us \(4 \times 3 = 12\) total factors.

The general rule: If a number has prime factorization \(p_1^{a_1} \times p_2^{a_2} \times \ldots \times p_k^{a_k}\), then the number of positive factors is \((a_1 + 1)(a_2 + 1)\ldots(a_k + 1)\).

3. Calculate factors for the first expression

For \(2^{51}\), we have just one prime (2) raised to the power 51.

Using our formula: The number of factors = \(51 + 1 = 52\).

This means \(2^{51}\) has exactly 52 positive factors.

4. Set up the equation

Now let's look at \((2^n)(3^3)\). This has:

- The prime 2 raised to the power n

- The prime 3 raised to the power 3

Using our formula: The number of factors = \((n + 1)(3 + 1) = (n + 1)(4) = 4(n + 1)\).

Since both expressions must have the same number of factors:

\(4(n + 1) = 52\)

5. Solve for n

Now we solve the equation:

\(4(n + 1) = 52\)

Divide both sides by 4:

\(n + 1 = 13\)

Subtract 1 from both sides:

\(n = 12\)

Let's verify: If \(n = 12\), then \((2^{12})(3^3)\) has \((12 + 1)(3 + 1) = 13 \times 4 = 52\) factors, which matches the 52 factors of \(2^{51}\). ✓

4. Final Answer

The value of n is 12.

Looking at our answer choices, this corresponds to choice A.

Common Faltering Points

Errors while devising the approach

1. Confusing "number of factors" with the actual factors themselves

Students often get distracted by trying to list out all the actual factors (like 1, 2, 4, 8, 16...) instead of just counting them. This leads to unnecessary work and potential errors, especially with large numbers like \(2^{51}\).

2. Forgetting or misremembering the factor counting formula

Many students either don't recall that for \(p_1^{a_1} \times p_2^{a_2} \times \ldots \times p_k^{a_k}\), the number of factors is \((a_1 + 1)(a_2 + 1)\ldots(a_k + 1)\), or they remember it incorrectly as just \(a_1 \times a_2 \times \ldots \times a_k\) (forgetting to add 1 to each exponent).

Errors while executing the approach

1. Arithmetic errors when applying the factor formula

Students might correctly identify that \(2^{51}\) has \((51 + 1) = 52\) factors, but then make errors when calculating \((n + 1)(3 + 1) = 4(n + 1)\), perhaps writing it as \((n + 1)(4 + 1) = 5(n + 1)\) instead.

2. Algebraic mistakes when solving 4(n + 1) = 52

Common errors include dividing incorrectly (getting \(n + 1 = 52/4 = 13\) but then forgetting to subtract 1), or making basic arithmetic mistakes like \(52 \div 4 = 14\) instead of 13.

3. Errors while selecting the answer

No likely faltering points - once students correctly solve \(n + 1 = 13\) to get \(n = 12\), the answer choice A is straightforward to identify.