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Let p and q each represent a digit from 0 through 9 and let pq represent a 2-digit positive prime number such that \(20 < \mathrm{pq} < 99\). For example, if \(\mathrm{p} = 2\) and \(\mathrm{q} = 3\), then the 2-digit prime number \(\mathrm{pq}\) is 23. If the 2-digit integer \(\mathrm{qp}\) is 27 greater than \(\mathrm{pq}\), what is the sum of the digits p and q?
Let's break down what we're looking for in everyday language:
We have two single digits, \(\mathrm{p}\) and \(\mathrm{q}\). When we put them together as \(\mathrm{pq}\), we get a 2-digit number that must be a prime number between 20 and 99. For example, if \(\mathrm{p = 2}\) and \(\mathrm{q = 3}\), then \(\mathrm{pq = 23}\).
Now, if we flip these digits around to get \(\mathrm{qp}\), this new number is exactly 27 more than our original number \(\mathrm{pq}\). We need to find what \(\mathrm{p + q}\) equals.
So we're looking for a situation where: (reversed number) = (original number) + 27
Process Skill: TRANSLATE - Converting the word problem into clear mathematical relationships
Let's express our two-digit numbers in terms of the individual digits:
The 2-digit number \(\mathrm{pq}\) means: \(\mathrm{p}\) is in the tens place, \(\mathrm{q}\) is in the units place
So \(\mathrm{pq = 10p + q}\)
The 2-digit number \(\mathrm{qp}\) means: \(\mathrm{q}\) is in the tens place, \(\mathrm{p}\) is in the units place
So \(\mathrm{qp = 10q + p}\)
The problem tells us that \(\mathrm{qp}\) is 27 greater than \(\mathrm{pq}\), so:
\(\mathrm{qp = pq + 27}\)
Substituting our expressions:
\(\mathrm{10q + p = (10p + q) + 27}\)
Let's solve this equation step by step:
\(\mathrm{10q + p = 10p + q + 27}\)
Moving all terms with \(\mathrm{p}\) and \(\mathrm{q}\) to one side:
\(\mathrm{10q + p - 10p - q = 27}\)
\(\mathrm{9q - 9p = 27}\)
Factoring out 9:
\(\mathrm{9(q - p) = 27}\)
Dividing both sides by 9:
\(\mathrm{q - p = 3}\)
This tells us that \(\mathrm{q}\) must be exactly 3 more than \(\mathrm{p}\)!
Now we know that \(\mathrm{q = p + 3}\). Since both \(\mathrm{p}\) and \(\mathrm{q}\) must be single digits (0-9), and we need \(\mathrm{pq}\) to be between 20 and 99, let's check our possibilities:
If \(\mathrm{p = 1}\), then \(\mathrm{q = 4}\), so \(\mathrm{pq = 14}\) (too small, less than 20)
If \(\mathrm{p = 2}\), then \(\mathrm{q = 5}\), so \(\mathrm{pq = 25}\) (checking: \(\mathrm{25 = 5 \times 5}\), so it's not prime)
If \(\mathrm{p = 3}\), then \(\mathrm{q = 6}\), so \(\mathrm{pq = 36}\) (checking: \(\mathrm{36 = 6 \times 6}\), so it's not prime)
If \(\mathrm{p = 4}\), then \(\mathrm{q = 7}\), so \(\mathrm{pq = 47}\) (checking: 47 is prime!)
If \(\mathrm{p = 5}\), then \(\mathrm{q = 8}\), so \(\mathrm{pq = 58}\) (checking: \(\mathrm{58 = 2 \times 29}\), so it's not prime)
If \(\mathrm{p = 6}\), then \(\mathrm{q = 9}\), so \(\mathrm{pq = 69}\) (checking: \(\mathrm{69 = 3 \times 23}\), so it's not prime)
If \(\mathrm{p \geq 7}\), then \(\mathrm{q \geq 10}\), which violates our single-digit requirement
Let's verify our answer: \(\mathrm{p = 4, q = 7}\)
\(\mathrm{pq = 47, qp = 74}\)
Difference: \(\mathrm{74 - 47 = 27}\) ✓
Is 47 prime? Yes ✓
Process Skill: APPLY CONSTRAINTS - Systematically checking which values satisfy both the algebraic relationship and the prime requirement
We found that \(\mathrm{p = 4}\) and \(\mathrm{q = 7}\) satisfy all our conditions.
Therefore: \(\mathrm{p + q = 4 + 7 = 11}\)
Looking at our answer choices, 11 corresponds to choice D.
Final Answer: 11
1. Misinterpreting the digit reversal relationship
Students often confuse which number should be larger. The problem states "\(\mathrm{qp}\) is 27 greater than \(\mathrm{pq}\)" but students might set up the equation as \(\mathrm{pq = qp + 27}\) instead of \(\mathrm{qp = pq + 27}\), leading to \(\mathrm{q - p = -3}\) instead of \(\mathrm{q - p = 3}\).
2. Overlooking the prime number constraint
Students may focus only on the algebraic relationship (\(\mathrm{qp = pq + 27}\)) and forget that \(\mathrm{pq}\) must be a prime number between 20 and 99. This leads them to accept any solution that satisfies \(\mathrm{q - p = 3}\) without checking if the resulting two-digit number is actually prime.
3. Misunderstanding two-digit number representation
Students sometimes struggle with expressing \(\mathrm{pq}\) as \(\mathrm{10p + q}\) and \(\mathrm{qp}\) as \(\mathrm{10q + p}\). They might incorrectly think \(\mathrm{pq = p \times q}\) (multiplication) rather than understanding it represents a two-digit number where \(\mathrm{p}\) is the tens digit and \(\mathrm{q}\) is the units digit.
1. Arithmetic errors in algebraic manipulation
When solving \(\mathrm{10q + p = 10p + q + 27}\), students often make sign errors or coefficient errors. Common mistakes include getting \(\mathrm{9p - 9q = 27}\) instead of \(\mathrm{9q - 9p = 27}\), or incorrectly simplifying to get \(\mathrm{q - p = -3}\).
2. Incorrect primality testing
Students may incorrectly identify whether numbers are prime. For example, they might think 25 is prime (forgetting \(\mathrm{25 = 5^2}\)), or 58 is prime (missing that \(\mathrm{58 = 2 \times 29}\)), or doubt that 47 is prime when it actually is.
3. Incomplete systematic checking
When testing values where \(\mathrm{q = p + 3}\), students might not systematically check all possible values of \(\mathrm{p}\) from the valid range, missing the correct solution or accepting an incorrect one without verifying all constraints.
1. Confusing \(\mathrm{p + q}\) with individual digit values
After finding \(\mathrm{p = 4}\) and \(\mathrm{q = 7}\), students might mistakenly select 4 or 7 as their final answer instead of calculating \(\mathrm{p + q = 11}\). They may think the question is asking for one of the individual digits rather than their sum.
2. Not verifying the final answer
Students may calculate \(\mathrm{p + q = 11}\) but fail to double-check that this value satisfies all original conditions (\(\mathrm{pq = 47}\) is prime, \(\mathrm{qp = 74}\), and \(\mathrm{74 - 47 = 27}\)), leading them to second-guess their correct answer and select a different option.