[b]p1.[/b] In the group of five people any subgroup of three persons contains at least two friends. Is it possible to divide these five people into two subgroups such that all members of any subgroup are friends?


[b]p2.[/b] Coefficients $a,b,c$ in expression $ax^2+bx+c$ are such that $b-c>a$ and $a \ne 0$. Is it true that equation $ax^2+bx+c=0$ always has two distinct real roots?


[b]p3.[/b] Point $D$ is a midpoint of the median $AF$ of triangle $ABC$. Line $CD$ intersects $AB$ at point $E$. Distances $|BD|=|BF|$. Show that $|AE|=|DE|$.


[b]p4.[/b] Real numbers $a,b$ satisfy inequality $a+b^5>ab^5+1$. Show that $a+b^7>ba^7+1$.


[b]p5.[/b] A positive number was rounded up to the integer and got the number that is bigger than the original one by $28\%$. Find the original number (find all solutions).


[b]p6.[/b] Divide a $5\times 5$ square along the sides of the cells into $8$ parts in such a way that all parts are different.


PS. You should use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].
Let the original number be \( x \). According to the problem, when this number is rounded up to the nearest integer, the result is \( 1.28x \). Let \( n \) be the integer to which \( x \) is rounded up. Therefore, we have:
\[ n = 1.28x \]

Since \( n \) is the nearest integer greater than or equal to \( x \), we can write:
\[ n - 1 < x \leq n \]

Substituting \( n = 1.28x \) into the inequality:
\[ 1.28x - 1 < x \leq 1.28x \]

Let's solve the left part of the inequality:
\[ 1.28x - 1 < x \]
\[ 1.28x - x < 1 \]
\[ 0.28x < 1 \]
\[ x < \frac{1}{0.28} \]
\[ x < \frac{100}{28} \]
\[ x < \frac{25}{7} \]
\[ x < 3.5714 \]

Now, let's solve the right part of the inequality:
\[ x \leq 1.28x \]
\[ 0 \leq 0.28x \]
\[ 0 \leq x \]

Combining both parts, we get:
\[ 0 \leq x < 3.5714 \]

Since \( n \) is an integer, we need to find the possible integer values of \( n \) such that \( n = 1.28x \) and \( x \) is within the range \( 0 \leq x < 3.5714 \).

Let's consider the possible integer values for \( n \):
1. If \( n = 1 \):
\[ 1 = 1.28x \]
\[ x = \frac{1}{1.28} \]
\[ x = \frac{25}{32} \]

2. If \( n = 2 \):
\[ 2 = 1.28x \]
\[ x = \frac{2}{1.28} \]
\[ x = \frac{25}{16} \]

3. If \( n = 3 \):
\[ 3 = 1.28x \]
\[ x = \frac{3}{1.28} \]
\[ x = \frac{75}{32} \]

4. If \( n = 4 \):
\[ 4 = 1.28x \]
\[ x = \frac{4}{1.28} \]
\[ x = \frac{25}{8} \]

Therefore, the possible original numbers are:
\[ \boxed{\frac{25}{32}}, \boxed{\frac{25}{16}}, \boxed{\frac{75}{32}}, \boxed{\frac{25}{8}} \]