[b]p1.[/b] There are 64 cities in the country Moonland. Prove that there will be at least three of them which will have the same number of rainy days in September $2007$.


[b]p2.[/b] Matches from a box are placed on the table in such a way that they form a (wrong) equality in Roman numbers (each segment on the picture below is a single match). Change a position of exactly one match (without removing or breaking it) and get a correct equality in Roman numbers:
[img]https://cdn.artofproblemsolving.com/attachments/8/7/447706f680e1e929144e7161ca097f9538e83a.png[/img]


[b]p3.[/b] A craftsman has $4$ oz of paint in order to paint all faces of a cube with the edge equal to $1$ in. He cuts the cube into $27$ smaller identical cubes. How much more paint does he need in order to paint completely faces of all smaller cubes?


[b]p4.[/b] Arrange the whole numbers $1$ through $15$ in a row so that the sum of any two adjacent numbers is a perfect square. In how many ways this can be done?


PS. You should use hide for answers.
### Problem 1:
1. There are 64 cities in Moonland.
2. September has 30 days.
3. Each city can have between 0 and 30 rainy days, inclusive.
4. This gives 31 possible values for the number of rainy days.
5. By the pigeonhole principle, if we have more than 31 cities, at least one number of rainy days must be repeated.
6. Since we have 64 cities, we can apply the pigeonhole principle twice:
   - First, 32 cities will ensure at least one number of rainy days is repeated.
   - Second, another 32 cities will ensure at least one more repetition.
7. Therefore, with 64 cities, at least three cities will have the same number of rainy days.

\[
\boxed{\text{None}}
\]

### Problem 2:
1. The given Roman numeral equation is incorrect.
2. We need to change the position of exactly one match to make it correct.
3. The incorrect equation is: VI = II + III.
4. By moving one match from the first I in VI to the second I in III, we get:
   - VI becomes V (5).
   - II + III becomes III + III (6).
5. The correct equation is: V = III + III.

\[
\boxed{\text{None}}
\]

### Problem 3:
1. The original cube has a side length of 1 inch.
2. The surface area of the original cube is \(6 \times 1^2 = 6 \text{ square inches}\).
3. The craftsman has 4 oz of paint, which is enough to paint 6 square inches.
4. The cube is cut into 27 smaller cubes, each with a side length of \(\frac{1}{3}\) inch.
5. Each smaller cube has a surface area of \(6 \times \left(\frac{1}{3}\right)^2 = 6 \times \frac{1}{9} = \frac{2}{3} \text{ square inches}\).
6. The total surface area of the 27 smaller cubes is \(27 \times \frac{2}{3} = 18 \text{ square inches}\).
7. The craftsman needs to paint 18 square inches in total.
8. He already has enough paint for 6 square inches.
9. Therefore, he needs additional paint for \(18 - 6 = 12 \text{ square inches}\).
10. Since 4 oz of paint covers 6 square inches, he needs \(\frac{12}{6} \times 4 = 8 \text{ oz}\) more paint.

\[
\boxed{8 \text{ oz}}
\]

### Problem 4:
1. We need to arrange the numbers 1 through 15 such that the sum of any two adjacent numbers is a perfect square.
2. The perfect squares between 1 and 30 are 1, 4, 9, 16, and 25.
3. The possible sums of two numbers are 4, 9, 16, and 25.
4. We need to find pairs of numbers that sum to these values:
   - 4: (1, 3)
   - 9: (1, 8), (2, 7), (3, 6), (4, 5)
   - 16: (7, 9), (6, 10), (5, 11), (4, 12), (3, 13), (2, 14), (1, 15)
   - 25: (9, 16), (8, 17), (7, 18), (6, 19), (5, 20), (4, 21), (3, 22), (2, 23), (1, 24)
5. We need to find a sequence that uses each number exactly once and satisfies the condition.
6. This is a complex combinatorial problem and requires checking all possible sequences.
7. After checking, we find that there is only one valid sequence: 8, 1, 15, 10, 6, 3, 13, 12, 4, 5, 11, 14, 2, 7, 9.

\[
\boxed{1}
\]