If the smaller of 2 consecutive odd integers is a multiple of 5, which of the following could NOT be...

1. Translate the problem requirements: We need to understand what "consecutive odd integers" means (odd numbers that follow each other like 5, 7), identify that the smaller one must be divisible by 5, and determine which sum is impossible.
2. Establish the pattern for valid pairs: Since consecutive odd integers differ by 2, if the smaller is a multiple of 5, we can express any valid pair and find the pattern for their sums.
3. Determine the sum characteristics: Calculate what types of sums are possible when we add two consecutive odd integers where the smaller is a multiple of 5.
4. Test each answer choice against our pattern: Check which answer choice cannot be expressed as a sum following our established pattern.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what this problem is asking us to find.

First, what are consecutive odd integers? These are odd numbers that come right after each other. For example: \(3\) and \(5\), or \(7\) and \(9\), or \(15\) and \(17\). Notice that consecutive odd integers are always \(2\) apart.

Next, we're told the smaller of these two consecutive odd integers must be a multiple of 5. What odd multiples of 5 are there? Let's list some: \(\ldots, -15, -10, -5, 0, 5, 10, 15, 20, 25, \ldots\)

Wait! We need to be careful here. Multiples of 5 include both odd and even numbers. Since we need odd multiples of 5, we get: \(\ldots, -15, -5, 5, 15, 25, 35, \ldots\)

So our pairs of consecutive odd integers where the smaller is a multiple of 5 would be:

• \((-15, -13), (-5, -3), (5, 7), (15, 17), (25, 27), (35, 37)\), etc.

Process Skill: TRANSLATE - Converting the problem language into concrete mathematical examples

2. Establish the pattern for valid pairs

Now let's see the pattern more clearly. If the smaller odd integer is a multiple of 5, we can write it as \(5k\) where \(k\) is an odd integer.

But let's think about this in plain English first. Since we need the smaller number to be an odd multiple of 5, our smaller numbers look like: \(\ldots, -15, -5, 5, 15, 25, 35, \ldots\)

The corresponding larger numbers (adding 2 to each) are: \(\ldots, -13, -3, 7, 17, 27, 37, \ldots\)

So our valid pairs are:

• \((-15, -13), (-5, -3), (5, 7), (15, 17), (25, 27), (35, 37), \ldots\)

Let's express this pattern: If our smaller number is \(10n + 5\) where \(n\) is any integer, then our larger number is \(10n + 7\).

Why \(10n + 5\)? Because odd multiples of 5 follow this pattern:

• When \(n = -2\): \(10(-2) + 5 = -15\)
• When \(n = -1\): \(10(-1) + 5 = -5\)
• When \(n = 0\): \(10(0) + 5 = 5\)
• When \(n = 1\): \(10(1) + 5 = 15\)

3. Determine the sum characteristics

Now let's find what the sums look like. If our pair is \((10n + 5, 10n + 7)\), then their sum is:

\(\mathrm{Sum} = (10n + 5) + (10n + 7) = 20n + 12\)

This tells us something important: every valid sum must be of the form \(20n + 12\), where \(n\) is any integer.

Let's check this with our examples:

• For \((-15, -13)\): sum = \(-28\). Using our formula: \(20(-2) + 12 = -40 + 12 = -28\) ✓
• For \((-5, -3)\): sum = \(-8\). Using our formula: \(20(-1) + 12 = -20 + 12 = -8\) ✓
• For \((5, 7)\): sum = \(12\). Using our formula: \(20(0) + 12 = 12\) ✓
• For \((15, 17)\): sum = \(32\). Using our formula: \(20(1) + 12 = 32\) ✓

So any valid sum must be expressible as \(20n + 12\) for some integer \(n\).

Process Skill: INFER - Drawing the non-obvious conclusion about the mathematical pattern

4. Test each answer choice against our pattern

Now we test each answer choice to see if it can be written as \(20n + 12\) for some integer \(n\).

For each choice, we solve: \(20n + 12 = \text{[answer choice]}\)
This gives us: \(n = \frac{\text{[answer choice]} - 12}{20}\)

Choice A: \(-8\)
\(n = \frac{(-8 - 12)}{20} = \frac{-20}{20} = -1\) ✓ (\(n\) is an integer)

Choice B: \(12\)
\(n = \frac{(12 - 12)}{20} = \frac{0}{20} = 0\) ✓ (\(n\) is an integer)

Choice C: \(22\)
\(n = \frac{(22 - 12)}{20} = \frac{10}{20} = 0.5\) ✗ (\(n\) is not an integer)

Choice D: \(52\)
\(n = \frac{(52 - 12)}{20} = \frac{40}{20} = 2\) ✓ (\(n\) is an integer)

Choice E: \(252\)
\(n = \frac{(252 - 12)}{20} = \frac{240}{20} = 12\) ✓ (\(n\) is an integer)

Only Choice C gives us a non-integer value for \(n\), which means \(22\) cannot be expressed as \(20n + 12\) for any integer \(n\).

Process Skill: APPLY CONSTRAINTS - Systematically checking each option against our derived constraint

Final Answer

The answer is C. \(22\)

The sum \(22\) could NOT be the sum of two consecutive odd integers where the smaller is a multiple of 5, because \(22\) cannot be expressed in the form \(20n + 12\) for any integer \(n\).

Common Faltering Points

Errors while devising the approach

1. Misunderstanding what constitutes "odd multiples of 5"

Students often confuse "multiples of 5" with "odd multiples of 5". They might include even multiples like \(10, 20, -10\) as potential starting points, when the problem specifically requires the smaller integer to be both odd AND a multiple of 5. This leads to considering invalid pairs like \((10, 12)\) or \((20, 22)\).

2. Incorrectly setting up consecutive odd integers

Some students think consecutive odd integers differ by 1 (like consecutive integers) rather than by 2. For example, they might think \(5\) and \(6\) are consecutive odd integers instead of \(5\) and \(7\). This fundamental misunderstanding completely derails the solution from the start.

3. Missing the constraint that the SMALLER integer must be the multiple of 5

Students may overlook that specifically the "smaller" of the two consecutive odd integers must be a multiple of 5. They might consider pairs where the larger integer is the multiple of 5, such as \((3, 5)\) instead of \((5, 7)\), leading to a different pattern entirely.

Errors while executing the approach

1. Arithmetic errors when establishing the general pattern

When deriving that valid sums must be of the form \(20n + 12\), students often make calculation errors. For example, they might incorrectly add \((10n + 5) + (10n + 7)\) and get \(20n + 10\) instead of \(20n + 12\), which would lead them to test the wrong constraint.

2. Incorrectly solving for n when testing answer choices

When checking if each answer choice can be written as \(20n + 12\), students frequently make algebraic errors. For instance, when testing choice C (\(22\)), they might incorrectly calculate \(n = \frac{(22 - 12)}{20} = \frac{10}{20} = \frac{1}{2}\) as being equal to \(0.5\) but then mistakenly conclude this "works" instead of recognizing that \(n\) must be an integer.

Errors while selecting the answer

1. Selecting a choice that COULD be the sum instead of one that could NOT

This is a classic misreading error where students find all the valid sums and then select one of those instead of the invalid one. Since the question asks "which could NOT be the sum," students who identify that A, B, D, and E are all valid might accidentally select one of these instead of C.

2. Confusing the constraint verification results

Students might correctly identify that choice C gives \(n = 0.5\), but then incorrectly conclude that since they got "an answer" (even though it's not an integer), this choice is valid. They fail to recognize that \(n\) must be an integer for the sum to be possible.

Alternate Solutions

Smart Numbers Approach

Step 1: Choose smart numbers for consecutive odd multiples of 5
Let's work with concrete examples of consecutive odd integers where the smaller is a multiple of 5:

• Pair 1: \(5, 7\) (sum = \(12\))
• Pair 2: \(15, 17\) (sum = \(32\))
• Pair 3: \(25, 27\) (sum = \(52\))
• Pair 4: \(-5, -3\) (sum = \(-8\))
• Pair 5: \(125, 127\) (sum = \(252\))

Step 2: Identify the pattern from our concrete examples
From our smart number examples:

• \(5 + 7 = 12\) ✓
• \(15 + 17 = 32\)
• \(25 + 27 = 52\) ✓
• \(-5 + (-3) = -8\) ✓
• \(125 + 127 = 252\) ✓

Step 3: Test if 22 can be achieved
If the sum equals \(22\), then we need two consecutive odd integers that add to \(22\).
Let the smaller integer = \(n\), then the larger = \(n + 2\)
So: \(n + (n + 2) = 22\)
\(2n + 2 = 22\)
\(2n = 20\)
\(n = 10\)

But \(n = 10\) is even, not odd! This means we cannot have consecutive odd integers that sum to \(22\).

Step 4: Verify our conclusion
Using our smart number approach, we found that:

• We can construct pairs that sum to \(-8, 12, 52\), and \(252\)
• We cannot construct a valid pair that sums to \(22\) because it would require the first integer to be even

Answer: C (\(22\) could NOT be the sum)