If m is an even integer, v is an odd integer, and m > v > 0, which of the...

1. Translate the problem requirements: We need to count how many even integers exist in the range between v (odd) and m (even), excluding both endpoints since we want integers "less than m" and "greater than v".
2. Identify the boundary even integers: Determine the first even integer greater than v and the last even integer less than m to establish our counting range.
3. Apply the counting pattern for evens: Use the fact that even integers are spaced 2 units apart to count how many even integers exist between our identified boundaries.
4. Verify with concrete examples: Test our formula with specific values to ensure it produces the correct count.

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what we're looking for in everyday language. We have two numbers: m (which is even) and v (which is odd), and we know m is bigger than v, and both are positive.

We need to count how many even integers sit between these two numbers, but we're not including m or v themselves in our count. Think of it like counting the even house numbers on a street between two specific addresses, but not counting those two addresses.

For example, if \(\mathrm{v = 3}\) and \(\mathrm{m = 10}\), we want to count the even integers that are greater than 3 but less than 10. Those would be: 4, 6, and 8. So our answer should be 3.

Process Skill: TRANSLATE - Converting the problem language "less than m and greater than v" into a clear counting task

2. Identify the boundary even integers

Now let's figure out the first and last even integers in our range.

Since v is odd and we want even integers greater than v, the first even integer greater than v is always \(\mathrm{v + 1}\). Why? Because if v is odd (like 3, 5, 7, etc.), then \(\mathrm{v + 1}\) is always even (4, 6, 8, etc.).

Since m is even and we want even integers less than m, the last even integer less than m is always \(\mathrm{m - 2}\). Why? Because even integers are spaced 2 apart, so the even integer right before m is \(\mathrm{m - 2}\).

Using our example where \(\mathrm{v = 3}\) and \(\mathrm{m = 10}\):

• First even integer greater than 3: \(\mathrm{3 + 1 = 4}\)
• Last even integer less than 10: \(\mathrm{10 - 2 = 8}\)

So we're counting even integers from 4 to 8, inclusive.

3. Apply the counting pattern for evens

Now we need to count how many even integers exist from \(\mathrm{(v + 1)}\) to \(\mathrm{(m - 2)}\), inclusive.

Since even integers are spaced exactly 2 units apart, we can think of this as counting how many "jumps of 2" we make from our first even integer to our last even integer, then adding 1 to include both endpoints.

The distance from \(\mathrm{(v + 1)}\) to \(\mathrm{(m - 2)}\) is: \(\mathrm{(m - 2) - (v + 1) = m - 2 - v - 1 = m - v - 3}\)

Since we jump by 2's, the number of jumps is: \(\mathrm{(m - v - 3) ÷ 2}\)

But we need to add 1 to count both the starting and ending positions: \(\mathrm{(m - v - 3) ÷ 2 + 1}\)

Simplifying: \(\mathrm{(m - v - 3 + 2) ÷ 2 = (m - v - 1) ÷ 2}\)

Using our example: \(\mathrm{v = 3, m = 10}\)
Count = \(\mathrm{(10 - 3 - 1) ÷ 2 = 6 ÷ 2 = 3}\)
This matches our manual count of 4, 6, 8!

Process Skill: INFER - Recognizing that counting evenly spaced integers requires the standard counting formula

4. Verify with concrete examples

Let's test our formula \(\frac{\mathrm{m - v - 1}}{2}\) with another example to make sure it works.

Try \(\mathrm{v = 5, m = 12}\):

• Even integers greater than 5 and less than 12: 6, 8, 10
• Manual count: 3
• Formula: \(\frac{\mathrm{12 - 5 - 1}}{2} = \frac{6}{2} = 3\) ✓

Try \(\mathrm{v = 1, m = 8}\):

• Even integers greater than 1 and less than 8: 2, 4, 6
• Manual count: 3
• Formula: \(\frac{\mathrm{8 - 1 - 1}}{2} = \frac{6}{2} = 3\) ✓

Our formula works!

Final Answer

The number of even integers less than m and greater than v is \(\frac{\mathrm{m - v - 1}}{2}\).

Looking at our answer choices, this matches choice B: \(\frac{\mathrm{m-v-1}}{2}\)

The answer is B.

Common Faltering Points

Errors while devising the approach

1. Misinterpreting the boundary conditions

Students often confuse whether to include or exclude the endpoint values m and v. The question asks for integers "less than m and greater than v," which means we exclude both m and v from our count. However, students might mistakenly include one or both endpoints, leading them to count integers from v to m or from \(\mathrm{(v+1)}\) to m, resulting in an incorrect approach.

2. Overlooking the parity constraints

The problem specifically states that m is even and v is odd, but students might ignore these constraints and try to develop a general formula that works for any integers. This oversight can lead to incorrect boundary identification - for example, not recognizing that when v is odd, the first even integer greater than v is simply \(\mathrm{v+1}\).

3. Attempting to count all integers instead of just even integers

Students might misread the question and think they need to count all integers between m and v, rather than specifically counting only the even integers. This fundamental misunderstanding would lead them toward a formula like \(\mathrm{(m-v-1)}\) instead of the correct approach that accounts for the spacing of even integers.

Errors while executing the approach

1. Incorrectly identifying the first and last even integers in range

Even when students understand they need to find boundary even integers, they might incorrectly determine that the first even integer greater than odd v is \(\mathrm{v+2}\) instead of \(\mathrm{v+1}\), or that the last even integer less than even m is \(\mathrm{m-1}\) instead of \(\mathrm{m-2}\). These boundary errors lead to counting the wrong set of integers.

2. Misapplying the counting formula for evenly spaced sequences

Students might remember that for counting evenly spaced terms, they need to use \(\frac{\mathrm{last - first}}{\mathrm{spacing}} + 1\), but they could make errors in the application. For example, they might forget to add 1, or incorrectly calculate the spacing as 1 instead of 2, or make arithmetic errors when substituting \(\mathrm{(v+1)}\) and \(\mathrm{(m-2)}\) into the formula.

3. Algebraic manipulation errors

When simplifying the expression \(\frac{\mathrm{(m-2)-(v+1)}}{2} + 1\), students might make algebraic errors such as: incorrectly distributing the negative sign to get \(\frac{\mathrm{(m-2-v+1)}}{2}\) instead of \(\frac{\mathrm{(m-2-v-1)}}{2}\), or making mistakes when combining the +1 with the fraction to get the final form \(\frac{\mathrm{(m-v-1)}}{2}\).

Errors while selecting the answer

1. Choosing a superficially similar but incorrect expression

After arriving at the correct formula \(\frac{\mathrm{(m-v-1)}}{2}\), students might select choice A: \(\frac{\mathrm{(m-v)}}{2} - 1\), thinking it's equivalent to their answer. While these expressions look similar, they're actually different: \(\frac{\mathrm{(m-v)}}{2} - 1 = \frac{\mathrm{(m-v-2)}}{2}\), not \(\frac{\mathrm{(m-v-1)}}{2}\). Students might not carefully verify the algebraic equivalence before selecting.

no likely faltering points

Alternate Solutions

Smart Numbers Approach

This problem is well-suited for the smart numbers approach because we can choose specific values for m and v that satisfy the constraints and verify our answer formula.

Step 1: Choose Smart Numbers

We need m to be even, v to be odd, and \(\mathrm{m > v > 0}\). Let's choose values that create a clear pattern:

• Let \(\mathrm{v = 3}\) (odd integer)
• Let \(\mathrm{m = 10}\) (even integer, and \(\mathrm{10 > 3 > 0}\) ✓)

Step 2: Count Even Integers Between v and m

We need even integers that are:

• Greater than \(\mathrm{v = 3}\)
• Less than \(\mathrm{m = 10}\)

The even integers in this range are: 4, 6, 8

So we have 3 even integers.

Step 3: Test Each Answer Choice

With \(\mathrm{v = 3}\) and \(\mathrm{m = 10}\):

• A. \(\frac{\mathrm{(m-v)}}{2} - 1 = \frac{\mathrm{(10-3)}}{2} - 1 = \frac{7}{2} - 1 = 3.5 - 1 = 2.5\) ❌
• B. \(\frac{\mathrm{(m-v-1)}}{2} = \frac{\mathrm{(10-3-1)}}{2} = \frac{6}{2} = 3\) ✓
• C. \(\frac{\mathrm{(m-v)}}{2} = \frac{\mathrm{(10-3)}}{2} = \frac{7}{2} = 3.5\) ❌
• D. \(\mathrm{m-v-1 = 10-3-1 = 6}\) ❌
• E. \(\mathrm{m-v = 10-3 = 7}\) ❌

Step 4: Verify with Another Set of Smart Numbers

Let's confirm with different values:

• Let \(\mathrm{v = 1}\) (odd), \(\mathrm{m = 8}\) (even)
• Even integers between 1 and 8: 2, 4, 6 → Count = 3
• Using formula B: \(\frac{\mathrm{(8-1-1)}}{2} = \frac{6}{2} = 3\) ✓

Let's try one more:

• Let \(\mathrm{v = 5}\) (odd), \(\mathrm{m = 12}\) (even)
• Even integers between 5 and 12: 6, 8, 10 → Count = 3
• Using formula B: \(\frac{\mathrm{(12-5-1)}}{2} = \frac{6}{2} = 3\) ✓

Conclusion

The smart numbers approach clearly shows that answer choice B, \(\frac{\mathrm{(m-v-1)}}{2}\), consistently gives us the correct count of even integers between any odd integer v and even integer m.