[b]p1.[/b] In a $5\times 5$ square $25$ integer numbers are written such that the product of numbers in every row is negative. Prove that there is at least one column with product of its numbers also negative.


[b]p2.[/b] $13$ kids sat around a round table and decided that boys will always lie to girls and tell truth to each other, and girls will always lie to boys and say truth to each other. One of them told to the right neighbor: “Most of us are boys”. Then the neighbor told to his (or her) right neighbor “Most of us are girls”. And so they went on, alternating the phrases, till the last kid told the first one “Most of us are boys”. How many boys are at the table?


[b]p3.[/b] The right triangles $\vartriangle ABC$ and $\vartriangle DAF$ on the picture below are congruent with $\angle BAC = \angle ADF = 90^o$. Prove that the triangle $\vartriangle ADB$ in the intersection is equilateral.
[img]https://cdn.artofproblemsolving.com/attachments/6/a/4f3e90d3c066b733da845531b5d5c90041ecdf.png[/img]


[b]p4.[/b] Among $9$ coins: $5$ quarters, $3$ dimes and a nickel there is one fake coin, it weighs less than it should. By using balance scale (no extra weights allowed) can you determine in two rounds which coin is fake?


PS. You should use hide for answers.
1. Consider the product of all the numbers in the $5 \times 5$ square. Since the product of the numbers in each row is negative, the product of all the numbers in the square is the product of five negative numbers, which is negative. Now, assume for contradiction that the product of the numbers in every column is non-negative. Then the product of all the numbers in the square would be the product of five non-negative numbers, which is non-negative. This contradicts our earlier conclusion that the product of all the numbers is negative. Therefore, there must be at least one column with a negative product.

\[
\boxed{\text{There is at least one column with a negative product.}}
\]

2. Let's denote boys by $B$ and girls by $G$. The first kid says "Most of us are boys." If the first kid is a girl, she is lying, so most of them are girls. The second kid says "Most of us are girls." If the second kid is a boy, he is lying, so most of them are boys. This alternation continues around the table. Since the number of kids is odd (13), the statement made by the last kid must be the same as the statement made by the first kid. Therefore, the first kid must be a boy, and the sequence must start with a boy. 

If there are more boys than girls, the sequence must be $BBGGBBGGBBGGB$. Counting the boys and girls, we find there are 7 boys and 6 girls. This sequence satisfies the conditions given in the problem.

\[
\boxed{7 \text{ boys and } 6 \text{ girls}}
\]

3. Given that $\triangle ABC$ and $\triangle DAF$ are congruent right triangles with $\angle BAC = \angle ADF = 90^\circ$, we have $AB = AD$ and $\angle ABC = \angle FAD$. Since $\triangle ABC \cong \triangle DAF$, we also have $BC = DF$ and $AC = AF$. 

In $\triangle ADB$, we have $AB = AD$ and $\angle BAD = 60^\circ$ (since $\angle BAC = 90^\circ$ and $\angle CAD = 30^\circ$). Therefore, $\triangle ADB$ is isosceles with $AB = AD$. Since $\angle BAD = 60^\circ$, $\triangle ADB$ is equilateral.

\[
\boxed{\text{The triangle } \triangle ADB \text{ is equilateral.}}
\]

4. To determine the fake coin in two weighings, follow these steps:
   1. Divide the 9 coins into three groups of 3 coins each: Group 1, Group 2, and Group 3.
   2. Weigh Group 1 against Group 2.
      - If the scale balances, the fake coin is in Group 3.
      - If the scale does not balance, the lighter group contains the fake coin.
   3. Take the group that contains the fake coin and divide it into three individual coins.
   4. Weigh two of these coins against each other.
      - If the scale balances, the third coin is the fake coin.
      - If the scale does not balance, the lighter coin is the fake coin.

\[
\boxed{\text{The fake coin can be determined in two weighings.}}
\]