If y is the average (arithmetic mean) of 15 consecutive positive integers, which of the following must be true? y...

1. Translate the problem requirements: We need to understand what "15 consecutive positive integers" means (like 3,4,5,6,7,8,9,10,11,12,13,14,15,16,17) and determine which properties their average must always have.
2. Establish the fundamental property of consecutive integer averages: Use the key insight that for any set of consecutive integers, the average equals the middle value.
3. Analyze each statement systematically: Test each Roman numeral statement (I, II, III) to determine if it must always be true for any possible set of 15 consecutive positive integers.
4. Verify with strategic examples: Use well-chosen examples to confirm our analysis, particularly testing boundary cases where statements might fail.

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what we're dealing with. When we say "15 consecutive positive integers," we mean 15 numbers that follow each other in order, like 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17. Another example could be 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24.

The key word here is "positive," which means all our integers must be greater than zero. So we can't have sequences that include negative numbers or zero.

The average (arithmetic mean) is what we get when we add all 15 numbers together and divide by 15. Our job is to figure out which properties this average must ALWAYS have, no matter which set of 15 consecutive positive integers we choose.

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

2. Establish the fundamental property of consecutive integer averages

Here's a beautiful property of consecutive integers: the average is always the middle number!

Let's see why this works with a simple example. Take 5 consecutive integers: 8, 9, 10, 11, 12.

• Average = \(\mathrm{(8 + 9 + 10 + 11 + 12) ÷ 5 = 50 ÷ 5 = 10}\)
• The middle number is 10

This works because consecutive integers are evenly spaced. The numbers below the middle balance out the numbers above the middle.

For 15 consecutive integers, the middle number is the 8th number in the sequence (since there are 7 numbers before it and 7 numbers after it).

So if our sequence starts at some positive integer n, then:

• 1st number: n
• 2nd number: n + 1
• 3rd number: n + 2
• ...
• 8th number (middle): n + 7
• ...
• 15th number: n + 14

Therefore, \(\mathrm{y = n + 7}\), where n is the first integer in our sequence.

3. Analyze each statement systematically

Now let's check each statement:

Statement I: y is an integer
Since \(\mathrm{y = n + 7}\), and n is a positive integer, y must also be an integer. Adding 7 to any integer gives us another integer.
Example: If our sequence starts at 5, then \(\mathrm{y = 5 + 7 = 12}\) (integer)
Example: If our sequence starts at 100, then \(\mathrm{y = 100 + 7 = 107}\) (integer)
Statement I is ALWAYS true.

Statement II: y > 7
Since our sequence must consist of positive integers, the smallest possible value for n is 1.
If n = 1, our sequence would be: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15
The average would be \(\mathrm{y = 1 + 7 = 8}\)
Since \(\mathrm{n ≥ 1}\), we have \(\mathrm{y = n + 7 ≥ 1 + 7 = 8}\)
Therefore, \(\mathrm{y > 7}\) is ALWAYS true.

Statement III: y < 100
There's no upper limit on how large our positive integers can be.
Example: If n = 100, then \(\mathrm{y = 100 + 7 = 107}\), which is greater than 100
Example: If n = 1000, then \(\mathrm{y = 1000 + 7 = 1007}\), which is much greater than 100
Statement III is NOT always true.

Process Skill: CONSIDER ALL CASES - Testing boundary conditions and extreme cases to verify our conclusions

4. Verify with strategic examples

Let's double-check our reasoning with concrete examples:

Example 1: Sequence starting at 1: {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}
Average = \(\mathrm{(1 + 15) × 15 ÷ 2 ÷ 15 = 16 ÷ 2 = 8}\)

• Statement I: 8 is an integer ✓
• Statement II: 8 > 7 ✓
• Statement III: 8 < 100 ✓

Example 2: Sequence starting at 95: {95, 96, 97, 98, 99, 100, 101, 102, 103, 104, 105, 106, 107, 108, 109}
Average = \(\mathrm{(95 + 109) × 15 ÷ 2 ÷ 15 = 204 ÷ 2 = 102}\)

• Statement I: 102 is an integer ✓
• Statement II: 102 > 7 ✓
• Statement III: 102 < 100 ✗

This confirms our analysis. Statements I and II are always true, but Statement III can be false.

5. Final Answer

Statements I and II must always be true for any set of 15 consecutive positive integers, while Statement III is not always true.

The answer is D. I and II

Common Faltering Points

Errors while devising the approach

1. Misunderstanding the "positive integers" constraint
Students may overlook that ALL 15 integers must be positive, thinking sequences like {-2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} are valid. This constraint is crucial because it establishes the minimum possible value for the average (which must be at least 8), making Statement II always true.

2. Attempting to calculate averages algebraically without recognizing the middle number property
Many students will set up the algebraic formula: \(\mathrm{average = (n + (n+1) + (n+2) + ... + (n+14))/15}\) and try to expand this fully rather than recognizing that for consecutive integers, the average always equals the middle number. This leads to unnecessary complexity and potential arithmetic errors.

3. Misinterpreting "must be true" vs. "could be true"
Students often confuse what MUST always be true with what COULD be true in some cases. For instance, they might think Statement III is correct because in their specific example (like 1-15), the average is less than 100, without considering that other sequences could yield averages greater than 100.

Errors while executing the approach

1. Incorrect identification of the middle number in a 15-number sequence
Students frequently miscalculate which number is in the middle position. They might think the 7th or 9th number is the middle instead of the 8th number. For a sequence starting at n, the middle (8th) number is n+7, but students often use n+6 or n+8.

2. Arithmetic errors when testing boundary cases
When checking Statement II, students need to find the minimum possible average. If the sequence starts at 1, the average should be 8, but students might miscalculate this as 7 or make errors in the addition, leading them to incorrectly conclude that Statement II is not always true.

Errors while selecting the answer

1. Choosing based on a single example rather than general analysis
After testing one specific sequence (often 1-15), students see that all three statements appear true and might select "E. II and III" or even an option that includes Statement III. They fail to test cases where Statement III fails, such as when the sequence starts at a large number like 95 or 100.

Alternate Solutions

Smart Numbers Approach

Step 1: Choose strategic concrete examples
Instead of working abstractly, let's test specific sets of 15 consecutive positive integers to verify each statement.

Example 1 (Small numbers): 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15
Average y = 8th number = 8

Example 2 (Larger numbers): 86, 87, 88, ..., 98, 99, 100
Average y = 8th number = 93

Example 3 (Very large numbers): 1000, 1001, 1002, ..., 1014
Average y = 8th number = 1007

Step 2: Test Statement I using our examples
• Example 1: y = 8 ✓ (integer)
• Example 2: y = 93 ✓ (integer)
• Example 3: y = 1007 ✓ (integer)
Pattern: For any 15 consecutive integers, the average is always the middle (8th) integer.

Step 3: Test Statement II using our examples
• Example 1: y = 8 > 7 ✓
• Example 2: y = 93 > 7 ✓
• Example 3: y = 1007 > 7 ✓
Testing the boundary: What's the smallest possible y?
If we start with 1: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15 → y = 8 > 7 ✓

Step 4: Test Statement III using our examples
• Example 1: y = 8 < 100 ✓
• Example 2: y = 93 < 100 ✓
• Example 3: y = 1007 < 100 ✗
Counterexample found! Statement III is not always true.

Step 5: Conclusion
Only statements I and II are always true for any set of 15 consecutive positive integers.

Answer: D (I and II)