The sum of three consecutive one-digit, positive, odd integers is one-seventh of the product of the same three integers. What is the middle integer when the three integers are listed in ascending order?
Represent the three integers as $n-2$, $n$, and $n+2$, where $n$ is the middle integer.  The problem states that  \[
n(n-2)(n+2)=7(n+(n+2)+(n-2)),
\] which simplifies to $(n-2)(n+2)=21$.  Since $7\cdot3$ and $21\cdot1$ are the only representations of 21 as a product of two positive integers, we see that $n-2=3$ and $n+2=7$ which implies $n=\boxed{5}$.
5
Level 3
Algebra
Problem: The sum of three consecutive one-digit, positive, odd integers is one-seventh of the product of the same three integers. What is the middle integer when the three integers are listed in ascending order?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To solve the problem, we first identify the three consecutive one-digit, positive, odd integers. These integers can be represented as x - 2, x, and x + 2, where x is the middle integer. 

Next, we calculate the sum of these integers:
Sum = (x - 2) + x + (x + 2) = 3x.

Now, we calculate the product of these integers:
Product = (x - 2) * x * (x + 2).
Using the difference of squares, we can simplify this:
Product = x(x^2 - 4) = x^3 - 4x.

According to the problem, the sum is one-seventh of the product:
3x = (1/7)(x^3 - 4x).

To eliminate the fraction, we multiply both sides by 7:
21x = x^3 - 4x.

Rearranging this equation gives us:
x^3 - 25x = 0.

Factoring out x, we have:
x(x^2 - 25) = 0.

This gives us the solutions:
x = 0 or x^2 - 25 = 0. 

Solving x^2 - 25 = 0, we find:
x^2 = 25, leading to x = 5 or x = -5. 

Since we are looking for positive integers, we take x = 5. 

Thus, the middle integer, when the three integers are listed in ascending order, is 5.<|END_OF_LATENT|> The final answer is $5$.