[b]p1.[/b] Arrange $9$ fruits whose weights are $2, 3, . . . , 9, 10$ lbs into $3$ light suitcases, $3$ in each, such that each suitcase weighs less than $20$ lbs and no suitcase contains a pair of fruits whose weights divide one another.


[b]p2.[/b] Find all integers $n$ such that $3^n + 4^n = 5^n$.


[b]p3.[/b] Build the regular $12$-gon with side length $1$ using squares with side length $1$ and equilateral triangles with side length $1$. You can stick the figures together along their sides, but no overlaps.


[b]p4.[/b] There are $14$ identical true coins and $2$ identical fake ones. It is not known if the fake coins are heavier or lighter than the true ones. Using an ordinary balance scale can you divide the $16$ given coins into two piles of the same weight in just $3$ weighings?


PS. You should use hide for answers.
To solve the equation \(3^n + 4^n = 5^n\), we need to find all integer values of \(n\).

1. **Check small values of \(n\):**
   - For \(n = 0\):
     \[
     3^0 + 4^0 = 1 + 1 = 2 \quad \text{and} \quad 5^0 = 1
     \]
     Thus, \(3^0 + 4^0 \neq 5^0\).

   - For \(n = 1\):
     \[
     3^1 + 4^1 = 3 + 4 = 7 \quad \text{and} \quad 5^1 = 5
     \]
     Thus, \(3^1 + 4^1 \neq 5^1\).

   - For \(n = 2\):
     \[
     3^2 + 4^2 = 9 + 16 = 25 \quad \text{and} \quad 5^2 = 25
     \]
     Thus, \(3^2 + 4^2 = 5^2\).

2. **Consider \(n > 2\):**
   - By Fermat's Last Theorem, there are no positive integers \(a, b, c\) and \(n > 2\) such that \(a^n + b^n = c^n\). In our case, \(a = 3\), \(b = 4\), and \(c = 5\). Therefore, there are no solutions for \(n > 2\).

3. **Alternative approach for \(n > 2\):**
   - Divide both sides of the equation \(3^n + 4^n = 5^n\) by \(5^n\):
     \[
     \left(\frac{3}{5}\right)^n + \left(\frac{4}{5}\right)^n = 1
     \]
   - The left-hand side is a strictly decreasing function of \(n\) because \(\left(\frac{3}{5}\right) < 1\) and \(\left(\frac{4}{5}\right) < 1\). Since the function is strictly decreasing, there can be at most one \(n\) such that the left-hand side equals 1. We already know that \(n = 2\) works, so there can be no other \(n\) that satisfies the equation.

Conclusion:
The only integer solution to the equation \(3^n + 4^n = 5^n\) is \(n = 2\).

The final answer is \( \boxed{ n = 2 } \).