[b]p1.[/b] Solve the system of equations 
$$xy = 2x + 3y$$ 
$$yz = 2y + 3z$$ 
$$zx =2z+3x$$


[b]p2.[/b] For any integer $k$ greater than $1$ and any positive integer $n$ , prove that $n^k$ is the sum of $n$ consecutive odd integers.


[b]p3.[/b] Determine all pairs of real numbers, $x_1$, $x_2$  with $|x_1|\le 1$ and $|x_2|\le 1$ which satisfy the inequality: $|x^2-1|\le |x-x_1||x-x_2|$ for all $x$ such that $|x| \ge 1$.


[b]p4.[/b] Find the smallest positive integer having exactly $100$ different positive divisors. (The number $1$ counts as a divisor).


[b]p5.[/b] $ABC$ is an equilateral triangle of side $3$ inches. $DB = AE = 1$ in. and $F$ is the point of intersection of segments $\overline{CD}$ and $\overline{BE}$ . Prove that $\overline{AF} \perp \overline{CD}$.
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PS. You should use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].
To find the smallest positive integer having exactly 100 different positive divisors, we need to consider the prime factorization of the number. The number of divisors of a number \( n \) with prime factorization \( n = p_1^{e_1} p_2^{e_2} \cdots p_k^{e_k} \) is given by:

\[
(e_1 + 1)(e_2 + 1) \cdots (e_k + 1)
\]

We need this product to equal 100. Let's explore the possible factorizations of 100:

1. \( 100 = 100 \times 1 \)
2. \( 100 = 50 \times 2 \)
3. \( 100 = 25 \times 4 \)
4. \( 100 = 20 \times 5 \)
5. \( 100 = 10 \times 10 \)
6. \( 100 = 25 \times 2 \times 2 \)
7. \( 100 = 10 \times 5 \times 2 \)
8. \( 100 = 5 \times 5 \times 4 \)
9. \( 100 = 5 \times 5 \times 2 \times 2 \)

We need to find the smallest number by assigning these factorizations to the exponents of the prime factors.

### Case 1: \( 100 = 100 \times 1 \)
\[
n = p_1^{99}
\]
The smallest number is \( 2^{99} \), which is extremely large.

### Case 2: \( 100 = 50 \times 2 \)
\[
n = p_1^{49} p_2^1
\]
The smallest number is \( 2^{49} \cdot 3 \), which is still very large.

### Case 3: \( 100 = 25 \times 4 \)
\[
n = p_1^{24} p_2^3
\]
The smallest number is \( 2^{24} \cdot 3^3 \).

### Case 4: \( 100 = 20 \times 5 \)
\[
n = p_1^{19} p_2^4
\]
The smallest number is \( 2^{19} \cdot 3^4 \).

### Case 5: \( 100 = 10 \times 10 \)
\[
n = p_1^9 p_2^9
\]
The smallest number is \( 2^9 \cdot 3^9 \).

### Case 6: \( 100 = 25 \times 2 \times 2 \)
\[
n = p_1^{24} p_2^1 p_3^1
\]
The smallest number is \( 2^{24} \cdot 3 \cdot 5 \).

### Case 7: \( 100 = 10 \times 5 \times 2 \)
\[
n = p_1^9 p_2^4 p_3^1
\]
The smallest number is \( 2^9 \cdot 3^4 \cdot 5 \).

### Case 8: \( 100 = 5 \times 5 \times 4 \)
\[
n = p_1^4 p_2^4 p_3^3
\]
The smallest number is \( 2^4 \cdot 3^4 \cdot 5^3 \).

### Case 9: \( 100 = 5 \times 5 \times 2 \times 2 \)
\[
n = p_1^4 p_2^4 p_3^1 p_4^1
\]
The smallest number is \( 2^4 \cdot 3^4 \cdot 5 \cdot 7 \).

Among these, the smallest number is \( 2^4 \cdot 3^4 \cdot 5 \cdot 7 \).

\[
2^4 = 16, \quad 3^4 = 81, \quad 5 = 5, \quad 7 = 7
\]
\[
16 \cdot 81 = 1296
\]
\[
1296 \cdot 5 = 6480
\]
\[
6480 \cdot 7 = 45360
\]

Thus, the smallest positive integer having exactly 100 different positive divisors is \( \boxed{45360} \).