[b]p1.[/b] One out of $12$ coins is counterfeited. It is known that its weight differs from the weight of a valid coin but it is unknown whether it is lighter or heavier. How to detect the counterfeited coin with the help of four trials using only a two-pan balance without weights?


[b]p2.[/b] Below a $3$-digit number $c d e$ is multiplied by a $2$-digit number $a b$ . Find all solutions $a, b, c, d, e, f, g$ if it is known that they represent distinct digits.

$\begin{tabular}{ccccc}
 &  & c & d & e \\
x &  &  & a & b \\
\hline
&  & f & e & g \\
+  & c & d & e &  \\
\hline
 & b & b & c & g \\
\end{tabular}$


[b]p3.[/b] Find all integer $n$ such that $\frac{n + 1}{2n - 1}$is an integer.


[b]p4[/b]. There are several straight lines on the plane which split the plane in several pieces. Is it possible to paint the plane in brown and green such that each piece is painted one color and no pieces having a common side are painted the same color?


PS. You should use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].
To find all integer \( n \) such that \(\frac{n + 1}{2n - 1}\) is an integer, we start by letting the expression equal \( k \), where \( k \) is an integer. Thus, we have:

1. \[
k = \frac{n + 1}{2n - 1}
\]

2. Multiply both sides by \( 2n - 1 \) to clear the fraction:
\[
k(2n - 1) = n + 1
\]

3. Distribute \( k \) on the left-hand side:
\[
2nk - k = n + 1
\]

4. Rearrange the equation to isolate terms involving \( n \):
\[
2nk - n = k + 1
\]

5. Factor out \( n \) on the left-hand side:
\[
n(2k - 1) = k + 1
\]

6. Solve for \( n \):
\[
n = \frac{k + 1}{2k - 1}
\]

For \( n \) to be an integer, \(\frac{k + 1}{2k - 1}\) must also be an integer. This implies that \( 2k - 1 \) must divide \( k + 1 \). We will now check the possible values of \( k \) that satisfy this condition.

7. Consider the equation:
\[
\frac{k + 1}{2k - 1} = m
\]
where \( m \) is an integer. This implies:
\[
k + 1 = m(2k - 1)
\]

8. Expand and rearrange to solve for \( k \):
\[
k + 1 = 2mk - m
\]
\[
k - 2mk = -m - 1
\]
\[
k(1 - 2m) = -m - 1
\]
\[
k = \frac{-m - 1}{1 - 2m}
\]

For \( k \) to be an integer, \(\frac{-m - 1}{1 - 2m}\) must be an integer. We will now check the possible values of \( m \) that satisfy this condition.

9. Consider the possible values of \( m \):
\[
1 - 2m \text{ must divide } -m - 1
\]

10. Check the values:
\[
1 - 2m = \pm 1, \pm 3
\]

11. Solve for \( m \):
\[
1 - 2m = 1 \implies m = 0 \quad (\text{not valid since } m \text{ must be an integer})
\]
\[
1 - 2m = -1 \implies 2m = 2 \implies m = 1
\]
\[
1 - 2m = 3 \implies 2m = -2 \implies m = -1
\]
\[
1 - 2m = -3 \implies 2m = 4 \implies m = 2
\]

12. Substitute \( m \) back to find \( k \):
\[
m = 1 \implies k = \frac{-1 - 1}{1 - 2} = 2
\]
\[
m = -1 \implies k = \frac{-(-1) - 1}{1 - 2(-1)} = 0
\]
\[
m = 2 \implies k = \frac{-2 - 1}{1 - 2(2)} = 1
\]

13. Substitute \( k \) back to find \( n \):
\[
k = 2 \implies n = \frac{2 + 1}{2(2) - 1} = \frac{3}{3} = 1
\]
\[
k = 0 \implies n = \frac{0 + 1}{2(0) - 1} = \frac{1}{-1} = -1
\]
\[
k = 1 \implies n = \frac{1 + 1}{2(1) - 1} = \frac{2}{1} = 2
\]

Thus, the integer values of \( n \) that satisfy the given condition are \( n = 2, 1, 0, -1 \).

The final answer is \( \boxed{ n = 2, 1, 0, -1 } \).