Consider six distinct natural numbers such that the average of the two smallest numbers is $14$, and the average of the two largest numbers is $28$. Then, the maximum possible value of the average of these six numbers is

We have six distinct natural numbers. We call them $a < b < c < d < e < f$.

Given information:

Average of two smallest numbers = 14, so $a + b = 28$

Average of two largest numbers = 28, so $e + f = 56$

Goal: Find the maximum possible average of all six numbers.

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The average of all six numbers is:

$\text{Average} = \frac{a + b + c + d + e + f}{6}$

Since we know $a + b = 28$ and $e + f = 56$, we can rewrite this as:

$\text{Average} = \frac{28 + c + d + 56}{6} = \frac{84 + c + d}{6}$

Key Insight: To maximize the average, we need to maximize $c + d$.

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Since all numbers are distinct natural numbers and in ascending order:

$a < b < c < d < e < f$

All numbers are positive integers

No two numbers can be equal

Since $e + f = 56$ and $e < f$, we think about the maximum possible value of $e$.

If $e$ is too large, then $f$ would have to be even larger, which might create problems with our ordering constraint $d < e$.

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To maximize $c + d$, we want $e$ to be as large as possible so $d$ can be close to $e$.

Choose $e$ and $f$ such that $e + f = 56$

We try $e = 27$ and $f = 29$.

Check: $27 + 29 = 56$

These are distinct natural numbers

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Maximize $d$ under the constraint $d < e = 27$

Since $d$ must be less than 27 and distinct from other numbers:

$d = 26$ (the largest natural number less than 27)

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Maximize $c$ under the constraint $c < d = 26$

Since $c$ must be less than 26 and distinct from other numbers:

$c = 25$ (the largest natural number less than 26)

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We need $a + b = 28$ where $a < b < c = 25$.

We can choose $a = 1, b = 27$... Wait! This doesn't work because $b = 27 = e$.

We try $a = 13, b = 15$:

Check: $13 + 15 = 28$

Check: $13 < 15 < 25$

All numbers are distinct

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Our six numbers are: $13, 15, 25, 26, 27, 29$

Maximum average:

$\text{Average} = \frac{13 + 15 + 25 + 26 + 27 + 29}{6} = \frac{135}{6} = 22.5$

Therefore, the maximum possible average is 22.5.