[b]p1.[/b] Find all triples of positive integers such that the sum of their reciprocals is equal to one.


[b]p2.[/b] Prove that $a(a + 1)(a + 2)(a + 3)$ is divisible by $24$.


[b]p3.[/b] There are $20$ very small red chips and some blue ones. Find out whether it is possible to put them on a large circle such that
(a) for each chip positioned on the circle the antipodal position is occupied by a chip of different color;
(b) there are no two neighboring blue chips.


[b]p4.[/b] A $12$ liter container is filled with gasoline. How to split it in two equal parts using two empty $5$ and $8$ liter containers?


PS. You should use hide for answers.  Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].
1. Let \(a, b, c\) be positive integers such that \(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 1\). Without loss of generality, assume \(a \leq b \leq c\).
2. Since \(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 1\), we have:
   \[
   \frac{3}{a} \geq \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 1 \implies a \leq 3
   \]
3. Also, since \(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 1\), we have:
   \[
   1 = \frac{1}{a} + \frac{1}{b} + \frac{1}{c} > \frac{1}{a} \implies a > 1
   \]
   Therefore, \(2 \leq a \leq 3\).

4. For \(a = 2\):
   \[
   \frac{1}{b} + \frac{1}{c} = 1 - \frac{1}{2} = \frac{1}{2}
   \]
   Since \(b \leq c\), we have:
   \[
   \frac{2}{b} \geq \frac{1}{b} + \frac{1}{c} = \frac{1}{2} \implies b \leq 4
   \]
   Also, \(\frac{1}{b} + \frac{1}{c} = \frac{1}{2}\) implies:
   \[
   \frac{1}{b} + \frac{1}{c} > \frac{1}{b} \implies b > 2
   \]
   Therefore, \(3 \leq b \leq 4\).

   - For \(b = 3\):
     \[
     \frac{1}{c} = \frac{1}{2} - \frac{1}{3} = \frac{3 - 2}{6} = \frac{1}{6} \implies c = 6
     \]
     Thus, \((a, b, c) = (2, 3, 6)\).

   - For \(b = 4\):
     \[
     \frac{1}{c} = \frac{1}{2} - \frac{1}{4} = \frac{2 - 1}{4} = \frac{1}{4} \implies c = 4
     \]
     Thus, \((a, b, c) = (2, 4, 4)\).

5. For \(a = 3\):
   \[
   \frac{1}{b} + \frac{1}{c} = 1 - \frac{1}{3} = \frac{2}{3}
   \]
   Since \(b \leq c\), we have:
   \[
   \frac{2}{b} \geq \frac{1}{b} + \frac{1}{c} = \frac{2}{3} \implies b \leq 3
   \]
   Also, \(\frac{1}{b} + \frac{1}{c} = \frac{2}{3}\) implies:
   \[
   \frac{1}{b} + \frac{1}{c} > \frac{1}{b} \implies b > 2
   \]
   Therefore, \(b = 3\).

   - For \(b = 3\):
     \[
     \frac{1}{c} = \frac{2}{3} - \frac{1}{3} = \frac{1}{3} \implies c = 3
     \]
     Thus, \((a, b, c) = (3, 3, 3)\).

6. To sum up, we have the following ordered triples:
   \[
   (2, 3, 6), (2, 6, 3), (3, 2, 6), (3, 6, 2), (6, 2, 3), (6, 3, 2), (2, 4, 4), (4, 2, 4), (4, 4, 2), (3, 3, 3)
   \]

The final answer is \( \boxed{ (2, 3, 6), (2, 6, 3), (3, 2, 6), (3, 6, 2), (6, 2, 3), (6, 3, 2), (2, 4, 4), (4, 2, 4), (4, 4, 2), (3, 3, 3) } \)