In kindergarden, there are 32 children and three classes: Judo, Agriculture, and Math. Every child is in exactly one class and every class has at least one participant.

One day, the teacher gathered 6 children to clean up the classroom. The teacher counted and found that exactly 1/2 of the Judo students, 1/4 of the Agriculture students and 1/8 of the Math students are cleaning.

How many children are in each class?
1. Let \( j \) be the number of children in the Judo class, \( a \) be the number of children in the Agriculture class, and \( m \) be the number of children in the Math class. We know that:
   \[
   j + a + m = 32
   \]
   This equation represents the total number of children in the kindergarten.

2. The teacher gathered 6 children to clean up the classroom. According to the problem, exactly \(\frac{1}{2}\) of the Judo students, \(\frac{1}{4}\) of the Agriculture students, and \(\frac{1}{8}\) of the Math students are cleaning. Therefore, we have:
   \[
   \frac{j}{2} + \frac{a}{4} + \frac{m}{8} = 6
   \]

3. To simplify the second equation, we can multiply through by 8 to clear the denominators:
   \[
   4j + 2a + m = 48
   \]

4. Now we have a system of linear equations:
   \[
   \begin{cases}
   j + a + m = 32 \\
   4j + 2a + m = 48
   \end{cases}
   \]

5. Subtract the first equation from the second equation to eliminate \( m \):
   \[
   (4j + 2a + m) - (j + a + m) = 48 - 32
   \]
   Simplifying this, we get:
   \[
   3j + a = 16
   \]

6. Now we have two equations:
   \[
   \begin{cases}
   j + a + m = 32 \\
   3j + a = 16
   \end{cases}
   \]

7. Solve for \( a \) in terms of \( j \) from the second equation:
   \[
   a = 16 - 3j
   \]

8. Substitute \( a = 16 - 3j \) into the first equation:
   \[
   j + (16 - 3j) + m = 32
   \]
   Simplify this equation:
   \[
   j + 16 - 3j + m = 32
   \]
   \[
   -2j + 16 + m = 32
   \]
   \[
   m = 32 + 2j - 16
   \]
   \[
   m = 16 + 2j
   \]

9. Now we have:
   \[
   \begin{cases}
   a = 16 - 3j \\
   m = 16 + 2j
   \end{cases}
   \]

10. Since \( j, a, m \) must be natural numbers and every class has at least one participant, we need to find \( j \) such that \( a \) and \( m \) are also natural numbers and positive:
    \[
    1 \leq j \leq 5
    \]
    (since \( a = 16 - 3j \) must be positive, \( j \) must be less than or equal to 5).

11. Check the possible values of \( j \):
    - For \( j = 4 \):
      \[
      a = 16 - 3(4) = 16 - 12 = 4
      \]
      \[
      m = 16 + 2(4) = 16 + 8 = 24
      \]

12. Verify the solution:
    \[
    j + a + m = 4 + 4 + 24 = 32
    \]
    \[
    \frac{j}{2} + \frac{a}{4} + \frac{m}{8} = \frac{4}{2} + \frac{4}{4} + \frac{24}{8} = 2 + 1 + 3 = 6
    \]

Thus, the solution satisfies both equations.

The final answer is \( \boxed{ (j, a, m) = (4, 4, 24) } \).