If 3 , for how many values of x is (x/3 the square of a prime number?

1. Translate the problem requirements: We need \(\mathrm{x/3}\) to equal the square of a prime number, where x is between 3 and 100 (exclusive). This means \(\mathrm{x/3 = p}^2\) where p is prime, so \(\mathrm{x = 3p}^2\).
2. Identify relevant prime numbers: Find all primes whose squares, when multiplied by 3, give values in our range (\(\mathrm{3 < x < 100}\)).
3. Test each prime systematically: Calculate \(\mathrm{x = 3p}^2\) for each prime and verify it falls within our bounds.
4. Count valid solutions: Determine how many values of x satisfy all conditions.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what the problem is asking in plain English. We need to find values of x where:
• x is between 3 and 100 (but not equal to 3 or 100)
• When we divide x by 3, we get the square of a prime number

So if \(\mathrm{x/3}\) equals the square of a prime, that means \(\mathrm{x/3 = p}^2\) where p is some prime number.

We can rearrange this to get: \(\mathrm{x = 3 × p}^2\)

This tells us that x must be 3 times the square of a prime number.

Process Skill: TRANSLATE - Converting the problem statement into a clear mathematical relationship

2. Identify relevant prime numbers

Now we need to figure out which prime numbers could work. Let's think about this step by step.

Since x must be between 3 and 100, we have:
\(\mathrm{3 < x < 100}\)

Substituting \(\mathrm{x = 3p}^2\):
\(\mathrm{3 < 3p}^2 \mathrm{< 100}\)

Dividing everything by 3:
\(\mathrm{1 < p}^2 \mathrm{< 33.33...}\)

Taking square roots:
\(\mathrm{1 < p < 5.77...}\)

Since p must be a prime number, let's list the primes in this range:
• \(\mathrm{p = 2}\) (since 2 is prime)
• \(\mathrm{p = 3}\) (since 3 is prime)
• \(\mathrm{p = 5}\) (since 5 is prime)

We need to check if \(\mathrm{p = 5}\) actually works since \(\mathrm{5}^2 \mathrm{= 25}\), and we need \(\mathrm{p}^2 \mathrm{< 33.33...}\)
Since \(\mathrm{25 < 33.33}\), \(\mathrm{p = 5}\) is valid to test.

3. Test each prime systematically

Let's calculate \(\mathrm{x = 3p}^2\) for each prime and see if it falls in our range:

For \(\mathrm{p = 2}\):
\(\mathrm{x = 3 × (2}^2\mathrm{) = 3 × 4 = 12}\)
Check: Is \(\mathrm{3 < 12 < 100}\)? Yes! ✓

For \(\mathrm{p = 3}\):
\(\mathrm{x = 3 × (3}^2\mathrm{) = 3 × 9 = 27}\)
Check: Is \(\mathrm{3 < 27 < 100}\)? Yes! ✓

For \(\mathrm{p = 5}\):
\(\mathrm{x = 3 × (5}^2\mathrm{) = 3 × 25 = 75}\)
Check: Is \(\mathrm{3 < 75 < 100}\)? Yes! ✓

For the next prime \(\mathrm{p = 7}\):
\(\mathrm{x = 3 × (7}^2\mathrm{) = 3 × 49 = 147}\)
Check: Is \(\mathrm{3 < 147 < 100}\)? No! \(\mathrm{147 > 100}\), so this doesn't work.

Any larger prime will give an even bigger value of x, so we can stop here.

Process Skill: APPLY CONSTRAINTS - Systematically checking that each solution satisfies all given conditions

4. Count valid solutions

Let's verify our solutions:
• \(\mathrm{x = 12}\) (when \(\mathrm{p = 2}\)): \(\mathrm{12/3 = 4 = 2}^2\), and 2 is prime ✓
• \(\mathrm{x = 27}\) (when \(\mathrm{p = 3}\)): \(\mathrm{27/3 = 9 = 3}^2\), and 3 is prime ✓
• \(\mathrm{x = 75}\) (when \(\mathrm{p = 5}\)): \(\mathrm{75/3 = 25 = 5}^2\), and 5 is prime ✓

We found exactly 3 values of x that satisfy all conditions: \(\mathrm{x = 12, 27}\), and \(\mathrm{75}\).

Final Answer

There are three values of x for which \(\mathrm{x/3}\) is the square of a prime number.

The answer is (B) Three.

Common Faltering Points

Errors while devising the approach

1. Misunderstanding what "square of a prime" means
Students might confuse "square of a prime" with "prime squared" or think it means finding prime numbers that happen to be perfect squares. They need to clearly understand that if p is prime, then \(\mathrm{p}^2\) is the square of that prime (like \(\mathrm{2}^2 \mathrm{= 4, 3}^2 \mathrm{= 9, 5}^2 \mathrm{= 25}\)).

2. Incorrectly handling the inequality constraints
Students often forget that \(\mathrm{3 < x < 100}\) means x cannot equal 3 or 100. They might include boundary values or set up the constraint incorrectly when substituting \(\mathrm{x = 3p}^2\), leading to wrong ranges for the prime p.

3. Setting up the wrong equation
Instead of correctly identifying that \(\mathrm{x/3 = p}^2\) leads to \(\mathrm{x = 3p}^2\), students might write \(\mathrm{x = p}^2\mathrm{/3}\) or some other incorrect relationship, fundamentally misunderstanding the algebraic manipulation required.

Errors while executing the approach

1. Arithmetic errors in calculating \(\mathrm{x = 3p}^2\)
Students might make simple computational mistakes like calculating \(\mathrm{3 × 5}^2 \mathrm{= 3 × 25 = 65}\) instead of 75, or \(\mathrm{3 × 3}^2 \mathrm{= 3 × 9 = 24}\) instead of 27, leading to incorrect values of x.

2. Forgetting to check if calculated values satisfy the original constraint
After finding \(\mathrm{x = 147}\) for \(\mathrm{p = 7}\), students might forget to verify that \(\mathrm{147 < 100}\), and incorrectly include this as a valid solution. They need to systematically check each calculated x against \(\mathrm{3 < x < 100}\).

3. Missing prime numbers in the valid range
Students might skip checking \(\mathrm{p = 2}\) (thinking 2 isn't prime or is too small) or \(\mathrm{p = 5}\), or they might incorrectly include composite numbers like 4 or 6 as primes when testing values.

Errors while selecting the answer

1. Counting the wrong quantity
Students might count the number of prime numbers found (which is 3) instead of verifying they found exactly 3 values of x, or they might accidentally count some incorrect values they calculated during their work.

2. Double-checking verification errors
When verifying their final answer, students might make errors like checking whether \(\mathrm{75/3 = 25}\) is actually \(\mathrm{5}^2\) correctly, potentially miscounting their valid solutions due to verification mistakes rather than calculation mistakes.