[b]p1.[/b] A man is walking due east at $2$ m.p.h. and to him the wind appears to be blowing from the north. On doubling his speed to $4$ m.p.h. and still walking due east, the wind appears to be blowing from the nortl^eas^. What is the speed of the wind (assumed to have a constant velocity)?


[b]p2.[/b] Prove that any triangle can be cut into three pieces which can be rearranged to form a rectangle of the same area.


[b]p3.[/b]  An increasing sequence of integers starting with $1$ has the property that if $n$ is any member of the sequence, then so also are $3n$ and $n + 7$.  Also, all the members of the sequence are solely generated from the first nummber $1$; thus the sequence starts with $1,3,8,9,10, ...$  and $2,4,5,6,7...$ are not members of this sequence. Determine all the other positive integers which are not members of the sequence.


[b]p4.[/b] Three prime numbers, each greater than $3$, are in arithmetic progression. Show that their common difference is a multiple of $6$.


[b]p5.[/b] Prove that if $S$ is a set of at least $7$ distinct points, no four in a plane, the volumes of all the tetrahedra with vertices in $S$ are not all equal.


PS. You should use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].
To show that the common difference of three prime numbers, each greater than \(3\), in arithmetic progression is a multiple of \(6\), we proceed as follows:

1. **Representation of Primes Greater than 3**:
   Any prime number greater than \(3\) can be written in the form \(6k \pm 1\) for some integer \(k\). This is because any integer can be expressed as \(6k\), \(6k+1\), \(6k+2\), \(6k+3\), \(6k+4\), or \(6k+5\). Among these, \(6k\), \(6k+2\), \(6k+3\), and \(6k+4\) are not prime (since they are divisible by \(2\) or \(3\)). Thus, primes greater than \(3\) must be of the form \(6k \pm 1\).

2. **Arithmetic Progression of Primes**:
   Let the three primes in arithmetic progression be \(p_1\), \(p_2\), and \(p_3\) with a common difference \(d\). Thus, we have:
   \[
   p_2 = p_1 + d \quad \text{and} \quad p_3 = p_1 + 2d
   \]

3. **Modulo 6 Analysis**:
   Since \(p_1\), \(p_2\), and \(p_3\) are all of the form \(6k \pm 1\), we analyze their forms modulo \(6\):
   - \(p_1 \equiv \pm 1 \pmod{6}\)
   - \(p_2 = p_1 + d \equiv \pm 1 + d \pmod{6}\)
   - \(p_3 = p_1 + 2d \equiv \pm 1 + 2d \pmod{6}\)

4. **Ensuring All Primes**:
   For \(p_2\) and \(p_3\) to also be of the form \(6k \pm 1\), the common difference \(d\) must ensure that \(p_2\) and \(p_3\) are congruent to \(\pm 1 \pmod{6}\). This implies:
   \[
   \pm 1 + d \equiv \pm 1 \pmod{6} \quad \text{and} \quad \pm 1 + 2d \equiv \pm 1 \pmod{6}
   \]

5. **Solving for \(d\)**:
   - For \(p_2\) to be of the form \(6k \pm 1\):
     \[
     d \equiv 0 \pmod{6} \quad \text{or} \quad d \equiv \pm 2 \pmod{6}
     \]
   - For \(p_3\) to be of the form \(6k \pm 1\):
     \[
     2d \equiv 0 \pmod{6} \quad \text{or} \quad 2d \equiv \pm 2 \pmod{6}
     \]

   Since \(2d \equiv 0 \pmod{6}\) implies \(d \equiv 0 \pmod{3}\), and combining with the requirement \(d \equiv 0 \pmod{2}\), we get:
   \[
   d \equiv 0 \pmod{6}
   \]

Thus, the common difference \(d\) must be a multiple of \(6\).

\(\blacksquare\)

The final answer is \( \boxed{ 6 } \).