p1. Determine the smallest natural number that has the property that it's cube ends in $888$.


p2. Triangle $ABC$ is isosceles with $AC = BC$. The angle bisector at $A$ intercepts side $BC$ at the point $D$ and the bisector of the angle at $C$ intercepts side $AB$ at $E$. If $AD = 2CE$, find the measure of the angles of triangle $ABC$.


p3.In the accompanying figure, the circle $C$ is tangent to the quadrant $AOB$ at point $S$. AP is tangent to $C$ at $P$ and $OQ$ is tangent at $Q$. Calculate the length $AP$ as a function of the radius$ R$ of the quadrant $AOB$.
[img]https://cdn.artofproblemsolving.com/attachments/3/c/d34774c3c6c33d351316574ca3f7ade54e6441.png[/img]


p4.  Let $A$ be a set with seven or more natural numbers. Show that there must be two numbers in $A$ with the property that either its sum or its difference is divisible by ten. Show that the property can fail if set $A$ has fewer than seven elements.


p5. An $n$-sided convex polygon is such that there is no common point for any three of its diagonals. Determine the number of triangles that are formed such that two of their vertices are vertices of the polygon and the third is an intersection of two diagonals.
1. **Determine the smallest natural number that has the property that its cube ends in $888$.**

To solve this problem, we need to find the smallest natural number \( n \) such that \( n^3 \equiv 888 \pmod{1000} \).

First, note that \( n^3 \equiv 888 \pmod{1000} \) implies that the last three digits of \( n^3 \) are 888. We can start by checking small values of \( n \) to see if their cubes end in 888.

Let's try \( n = 12 \):
\[
12^3 = 1728 \quad \text{(last three digits are 728, not 888)}
\]

Next, let's try \( n = 22 \):
\[
22^3 = 10648 \quad \text{(last three digits are 648, not 888)}
\]

Next, let's try \( n = 32 \):
\[
32^3 = 32768 \quad \text{(last three digits are 768, not 888)}
\]

Next, let's try \( n = 42 \):
\[
42^3 = 74088 \quad \text{(last three digits are 088, not 888)}
\]

Next, let's try \( n = 52 \):
\[
52^3 = 140608 \quad \text{(last three digits are 608, not 888)}
\]

Next, let's try \( n = 62 \):
\[
62^3 = 238328 \quad \text{(last three digits are 328, not 888)}
\]

Next, let's try \( n = 72 \):
\[
72^3 = 373248 \quad \text{(last three digits are 248, not 888)}
\]

Next, let's try \( n = 82 \):
\[
82^3 = 551368 \quad \text{(last three digits are 368, not 888)}
\]

Next, let's try \( n = 92 \):
\[
92^3 = 778688 \quad \text{(last three digits are 688, not 888)}
\]

Next, let's try \( n = 102 \):
\[
102^3 = 1061208 \quad \text{(last three digits are 208, not 888)}
\]

Next, let's try \( n = 112 \):
\[
112^3 = 1404928 \quad \text{(last three digits are 928, not 888)}
\]

Next, let's try \( n = 122 \):
\[
122^3 = 1815848 \quad \text{(last three digits are 848, not 888)}
\]

Next, let's try \( n = 132 \):
\[
132^3 = 2299968 \quad \text{(last three digits are 968, not 888)}
\]

Next, let's try \( n = 142 \):
\[
142^3 = 2863288 \quad \text{(last three digits are 288, not 888)}
\]

Next, let's try \( n = 152 \):
\[
152^3 = 3511808 \quad \text{(last three digits are 808, not 888)}
\]

Next, let's try \( n = 162 \):
\[
162^3 = 4251528 \quad \text{(last three digits are 528, not 888)}
\]

Next, let's try \( n = 172 \):
\[
172^3 = 5088448 \quad \text{(last three digits are 448, not 888)}
\]

Next, let's try \( n = 182 \):
\[
182^3 = 6028568 \quad \text{(last three digits are 568, not 888)}
\]

Next, let's try \( n = 192 \):
\[
192^3 = 7077888 \quad \text{(last three digits are 888, which is correct)}
\]

Thus, the smallest natural number \( n \) such that \( n^3 \) ends in 888 is \( n = 192 \).

The final answer is \( \boxed{192} \).