The average (arithmetic mean) of the 5 positive integers, u, w, x, y, and z is 14, and u ....

1. Translate the problem requirements: We have 5 positive integers in ascending order where their average is 14, the largest value is 26, and we need to find the minimum possible median (middle value).
2. Establish the total constraint: Use the average to determine the sum of all five numbers, creating our primary mathematical constraint.
3. Set up the optimization framework: To minimize the median x, we need to minimize the two numbers before it (u and w) while respecting all given constraints.
4. Apply systematic minimization: Start with the smallest possible values for u and w, then work forward to find the resulting minimum value for the median x.

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what we have. We're told there are 5 positive integers arranged in order from smallest to largest: u, w, x, y, and z. Think of these as 5 students lined up by height, where u is the shortest and z is the tallest.

The average of these 5 numbers is 14. This means if we add all 5 numbers and divide by 5, we get 14.

We also know that z (the largest number) is 26, and we need to find the smallest possible value for x, which is the median (the middle number when arranged in order).

Since we have \(\mathrm{u < w < x < y < z}\), the median is x because it's the 3rd number out of 5.

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

2. Establish the total constraint

Since the average of the 5 numbers is 14, we can find their total sum:
\(\mathrm{Sum = Average \times Number\ of\ values = 14 \times 5 = 70}\)

So: \(\mathrm{u + w + x + y + z = 70}\)

We know \(\mathrm{z = 26}\), so:
\(\mathrm{u + w + x + y + 26 = 70}\)
\(\mathrm{u + w + x + y = 44}\)

This gives us our main constraint - the first four numbers must add up to exactly 44.

3. Set up the optimization framework

To minimize x (our median), we need to think strategically. Since \(\mathrm{u + w + x + y = 44}\), and we want x to be as small as possible, we should try to make the other numbers as large or as small as possible.

Looking at our ordering \(\mathrm{u < w < x < y < z}\), we have:

• u and w come before x, so they must be smaller than x
• y comes after x, so it must be larger than x

To minimize x, we should minimize u and w (the numbers before x) because this leaves more "room" for y to absorb the remaining sum.

Process Skill: INFER - Drawing the non-obvious conclusion about how to minimize the median

4. Apply systematic minimization

Since all numbers are positive integers and \(\mathrm{u < w < x}\), the smallest possible values are:

• \(\mathrm{u = 1}\) (smallest positive integer)
• \(\mathrm{w = 2}\) (next smallest, and \(\mathrm{w > u}\))

Now we have: \(\mathrm{1 + 2 + x + y = 44}\)
So: \(\mathrm{x + y = 41}\)

Since \(\mathrm{y > x}\) and both are integers, we need \(\mathrm{y \geq x + 1}\).
Substituting: \(\mathrm{x + (x + 1) \leq 41}\)
\(\mathrm{2x + 1 \leq 41}\)
\(\mathrm{2x \leq 40}\)
\(\mathrm{x \leq 20}\)

But we also need \(\mathrm{x > w = 2}\), and since we want the minimum value of x, let's try small values:

If \(\mathrm{x = 8}\), then \(\mathrm{y = 41 - 8 = 33}\)
Checking our constraints:
\(\mathrm{u = 1 < w = 2 < x = 8 < y = 33 < z = 26}\)

Wait! This violates \(\mathrm{y < z}\) since \(\mathrm{33 > 26}\). Let's reconsider.

Since \(\mathrm{y < z = 26}\), we need \(\mathrm{y \leq 25}\).
From \(\mathrm{x + y = 41}\) and \(\mathrm{y \leq 25}\):
\(\mathrm{x \geq 41 - 25 = 16}\)

But let's check if \(\mathrm{y}\) can equal 25:
If \(\mathrm{y = 25}\), then \(\mathrm{x = 41 - 25 = 16}\)
But we need \(\mathrm{x < y}\), and \(\mathrm{16 < 25}\) ✓

Actually, let's be more systematic. Since \(\mathrm{y < 26}\) and y is an integer, \(\mathrm{y \leq 25}\).
From \(\mathrm{x + y = 41}\) and \(\mathrm{y \leq 25}\):
\(\mathrm{x \geq 16}\)

Let's verify \(\mathrm{x = 8}\) works by being more careful:
If \(\mathrm{x = 8}\), then \(\mathrm{y = 41 - 8 = 33}\), but y must be \(\mathrm{< 26}\), so this doesn't work.

Let's try working backwards from the constraint \(\mathrm{y < 26}\):
Maximum value for y is 25
If \(\mathrm{y = 25}\), then \(\mathrm{x = 41 - 25 = 16}\)
But we want to minimize x, so let's see if we can make y smaller.

If we try \(\mathrm{x = 8}\): y would need to be 33, but \(\mathrm{y < 26}\), so impossible.
If we try \(\mathrm{x = 8}\): we need a different approach.

Let me reconsider by trying different values for u and w:
What if \(\mathrm{u = 1, w = 2}\), and we work with smaller y values?

Since \(\mathrm{x < y < 26}\), let's try \(\mathrm{y = 25}\):
\(\mathrm{x + 25 = 41}\), so \(\mathrm{x = 16}\)
Check: \(\mathrm{1 < 2 < 16 < 25 < 26}\) ✓

Can we do better? Let's try \(\mathrm{y = 24}\):
\(\mathrm{x + 24 = 41}\), so \(\mathrm{x = 17}\)
This makes x larger, not smaller.

Let's try adjusting u and w. What if \(\mathrm{u = 1, w = 3}\)?
Then: \(\mathrm{1 + 3 + x + y = 44}\)
\(\mathrm{x + y = 40}\)

With \(\mathrm{y \leq 25}\): \(\mathrm{x \geq 40 - 25 = 15}\)
If \(\mathrm{y = 25}\), then \(\mathrm{x = 15}\)
Check: \(\mathrm{1 < 3 < 15 < 25 < 26}\) ✓

Can we make \(\mathrm{w = 4}\)?
\(\mathrm{1 + 4 + x + y = 44}\)
\(\mathrm{x + y = 39}\)
With \(\mathrm{y \leq 25}\): \(\mathrm{x \geq 39 - 25 = 14}\)
But we need \(\mathrm{x > w = 4}\), and \(\mathrm{14 > 4}\) ✓

If \(\mathrm{y = 25}\), then \(\mathrm{x = 14}\)
Check: \(\mathrm{1 < 4 < 14 < 25 < 26}\) ✓

Continuing this pattern:
If \(\mathrm{w = 5}\): \(\mathrm{x + y = 38}\), so \(\mathrm{x \geq 13}\) when \(\mathrm{y = 25}\)
If \(\mathrm{w = 6}\): \(\mathrm{x + y = 37}\), so \(\mathrm{x \geq 12}\) when \(\mathrm{y = 25}\)
If \(\mathrm{w = 7}\): \(\mathrm{x + y = 36}\), so \(\mathrm{x \geq 11}\) when \(\mathrm{y = 25}\)
If \(\mathrm{w = 8}\): \(\mathrm{x + y = 35}\), so \(\mathrm{x \geq 10}\) when \(\mathrm{y = 25}\)
If \(\mathrm{w = 9}\): \(\mathrm{x + y = 34}\), so \(\mathrm{x \geq 9}\) when \(\mathrm{y = 25}\), but we need \(\mathrm{x > w = 9}\), so \(\mathrm{x \geq 10}\)

Wait, let me check \(\mathrm{w = 7}\) more carefully:
\(\mathrm{u = 1, w = 7, y = 25}\)
\(\mathrm{x + 25 = 36}\), so \(\mathrm{x = 11}\)
Check: \(\mathrm{1 < 7 < 11 < 25 < 26}\) ✓

Trying \(\mathrm{w = 8}\):
\(\mathrm{u = 1, w = 8, y = 25}\)
\(\mathrm{x + 25 = 35}\), so \(\mathrm{x = 10}\)
Check: \(\mathrm{1 < 8 < 10 < 25 < 26}\) ✓

Trying \(\mathrm{w = 9}\):
\(\mathrm{u = 1, w = 9, y = 25}\)
\(\mathrm{x + 25 = 34}\), so \(\mathrm{x = 9}\)
But this violates \(\mathrm{x > w}\) since \(\mathrm{9 = 9}\)

So we need \(\mathrm{x = 10}\), achieved with \(\mathrm{u = 1, w = 8, x = 10, y = 25, z = 26}\).

But wait, let's check if we can reduce further by making y smaller:
With \(\mathrm{u = 1, w = 8}\): \(\mathrm{x + y = 35}\)
If \(\mathrm{y = 24}\): \(\mathrm{x = 11}\) (worse)
If \(\mathrm{y = 23}\): \(\mathrm{x = 12}\) (worse)

So the minimum value of x is 10... but that's not among our answer choices. Let me double-check.

Actually, let me try \(\mathrm{u = 2}\):
If \(\mathrm{u = 2, w = 8}\), then: \(\mathrm{2 + 8 + x + y = 44}\), so \(\mathrm{x + y = 34}\)
With \(\mathrm{y = 25}\): \(\mathrm{x = 9}\)
Check: \(\mathrm{2 < 8 < 9 < 25 < 26}\) ✓

Can we push further?
If \(\mathrm{u = 3, w = 8}\): \(\mathrm{x + y = 33}\)
With \(\mathrm{y = 25}\): \(\mathrm{x = 8}\)
Check: \(\mathrm{3 < 8 < 8}\)... this violates \(\mathrm{x > w}\)

Therefore, the minimum possible value for the median x is 8, achieved with \(\mathrm{u = 2, w = 7, x = 8, y = 27}\).

Wait, let me recalculate:
If \(\mathrm{u = 2, w = 7}\): \(\mathrm{2 + 7 + x + y = 44}\), so \(\mathrm{x + y = 35}\)
To minimize x, maximize y: \(\mathrm{y = 25}\), so \(\mathrm{x = 10}\)
But we need \(\mathrm{x > 7}\), and we want smaller x.

Let me try: \(\mathrm{u = 1, w = 7, x = 8}\), then \(\mathrm{y = 36 - 8 = 28}\)
But \(\mathrm{y = 28 > z = 26}\), which violates our constraint.

Let me be very systematic:
\(\mathrm{u = 1, w = 7}\): \(\mathrm{x + y = 36}\)
Since \(\mathrm{y < 26}\), the maximum y can be is 25.
So \(\mathrm{x = 36 - 25 = 11}\), and we need \(\mathrm{x > 7}\) ✓

Trying \(\mathrm{u = 2, w = 7}\): \(\mathrm{x + y = 35}\)
With \(\mathrm{y \leq 25}\): \(\mathrm{x \geq 10}\), and we need \(\mathrm{x > 7}\) ✓
So \(\mathrm{x = 10}\) when \(\mathrm{y = 25}\).

Trying \(\mathrm{u = 3, w = 7}\): \(\mathrm{x + y = 34}\)
With \(\mathrm{y \leq 25}\): \(\mathrm{x \geq 9}\), and we need \(\mathrm{x > 7}\) ✓
So \(\mathrm{x = 9}\) when \(\mathrm{y = 25}\).

Trying \(\mathrm{u = 4, w = 7}\): \(\mathrm{x + y = 33}\)
With \(\mathrm{y \leq 25}\): \(\mathrm{x \geq 8}\), and we need \(\mathrm{x > 7}\) ✓
So \(\mathrm{x = 8}\) when \(\mathrm{y = 25}\).

Trying \(\mathrm{u = 5, w = 7}\): \(\mathrm{x + y = 32}\)
With \(\mathrm{y \leq 25}\): \(\mathrm{x \geq 7}\), but we need \(\mathrm{x > 7}\), so \(\mathrm{x \geq 8}\)
If \(\mathrm{x = 8}\), then \(\mathrm{y = 24}\), and \(\mathrm{5 < 7 < 8 < 24 < 26}\) ✓

Trying \(\mathrm{u = 6, w = 7}\): \(\mathrm{x + y = 31}\)
With \(\mathrm{y \leq 25}\): \(\mathrm{x \geq 6}\), but we need \(\mathrm{x > 7}\), so \(\mathrm{x \geq 8}\)
If \(\mathrm{x = 8}\), then \(\mathrm{y = 23}\) ✓

So the minimum value of x is 8.

4. Final Answer

The least possible value of the median is 8.

This can be achieved with values like \(\mathrm{u = 4, w = 7, x = 8, y = 25, z = 26}\).
Verification: \(\mathrm{4 + 7 + 8 + 25 + 26 = 70}\), and \(\mathrm{70 ÷ 5 = 14}\) ✓
Ordering: \(\mathrm{4 < 7 < 8 < 25 < 26}\) ✓

The answer is (C) 8.

Common Faltering Points

Errors while devising the approach

1. Misidentifying the median in an ordered set

Students often confuse which position represents the median. In a set of 5 numbers arranged as \(\mathrm{u < w < x < y < z}\), some students might think the median is the average of w and y, or incorrectly identify w or y as the median, when x (the middle/3rd position) is actually the median.

2. Misunderstanding the optimization objective

When asked to find the "least possible value of the median," students may incorrectly think they should minimize all variables equally. They fail to recognize that to minimize x (the median), they need to strategically choose values for the other variables - specifically making u and w as large as possible while respecting constraints, rather than making them as small as possible.

3. Overlooking the strict inequality constraints

Students often treat the ordering constraint \(\mathrm{u < w < x < y < z}\) as \(\mathrm{u \leq w \leq x \leq y \leq z}\), forgetting that all inequalities are strict. This leads them to allow equal values (like setting \(\mathrm{u = w}\) or \(\mathrm{x = y}\)), which violates the given conditions.

Errors while executing the approach

1. Violating the upper bound constraint y < z

When trying to minimize x, students correctly identify that they should maximize y, but they forget that y must be less than \(\mathrm{z = 26}\). They might set \(\mathrm{y = 26}\) or even larger values, which violates the ordering constraint \(\mathrm{y < z}\).

2. Arithmetic errors in constraint management

With multiple variables and constraints to track simultaneously (sum equals 70, strict ordering, positive integers), students often make calculation errors. For example, when \(\mathrm{u + w + x + y = 44}\) and they're testing different values, they might incorrectly calculate the remaining sum or lose track of which constraint they're currently checking.

3. Systematic testing errors

When trying different combinations of u and w values to minimize x, students may not test systematically or may stop too early. They might find one valid combination (like \(\mathrm{u=1, w=8, x=10}\)) and assume it's optimal without testing other possibilities (like \(\mathrm{u=4, w=7, x=8}\)).

Errors while selecting the answer

1. Selecting a non-optimal value found during testing

Students often find a valid solution early in their testing process (such as \(\mathrm{x = 10}\) or \(\mathrm{x = 11}\)) and select that answer choice without continuing to search for the true minimum value. They fail to exhaust all possibilities to ensure they've found the smallest possible value of x.