[b]p1.[/b] Is the last non-zero digit of the number $1 \cdot 2\cdot 3 \cdot ...\cdot  2011 \cdot  2012$ even or odd? Justify your answer.


[b]p2.[/b] Seven coins lie along a circle HEADS up. Can you make them TAILS up if you are only allowed to turn over any
a) five successive coins
b) four successive coins
at a time?


[b]p3.[/b] In a geometry lesson Jay and Bob drew geometrical tadpoles, outlined in bold on the right. (The four circles in the figure have the same radius, the triangle is equilateral, and its horizontal side is a diameter of the circle.) Which of the two tadpoles has greater area?
[img]https://cdn.artofproblemsolving.com/attachments/d/0/1f27c5f83b9d164cbae194d68dadd48cd20a78.png[/img]


[b]p4.[/b] Three pawns are placed on vertices of a pentagon. Your goal is to have one pawn in its original position and have the other two switch their places. If you are allowed to move a pawn along sides of the pentagon to any free vertex, then the goal is clearly impossible. Can you achieve the goal if it is allowed to move pawns along diagonals of the pentagon instead of its sides?


PS. You should use hide for answers.
1. To determine whether the last non-zero digit of the number \(1 \cdot 2 \cdot 3 \cdot \ldots \cdot 2011 \cdot 2012\) is even or odd, we need to consider the prime factorization of \(2012!\).

2. The factorial \(2012!\) is the product of all integers from 1 to 2012. The number of trailing zeros in \(2012!\) is determined by the number of times 10 is a factor in the product. Since \(10 = 2 \times 5\), we need to count the number of pairs of factors 2 and 5.

3. The number of times a prime \(p\) divides \(n!\) is given by:
   \[
   \left\lfloor \frac{n}{p} \right\rfloor + \left\lfloor \frac{n}{p^2} \right\rfloor + \left\lfloor \frac{n}{p^3} \right\rfloor + \cdots
   \]
   For \(p = 5\) and \(n = 2012\):
   \[
   \left\lfloor \frac{2012}{5} \right\rfloor + \left\lfloor \frac{2012}{25} \right\rfloor + \left\lfloor \frac{2012}{125} \right\rfloor + \left\lfloor \frac{2012}{625} \right\rfloor = 402 + 80 + 16 + 3 = 501
   \]
   So, there are 501 factors of 5 in \(2012!\).

4. For \(p = 2\) and \(n = 2012\):
   \[
   \left\lfloor \frac{2012}{2} \right\rfloor + \left\lfloor \frac{2012}{4} \right\rfloor + \left\lfloor \frac{2012}{8} \right\rfloor + \left\lfloor \frac{2012}{16} \right\rfloor + \left\lfloor \frac{2012}{32} \right\rfloor + \left\lfloor \frac{2012}{64} \right\rfloor + \left\lfloor \frac{2012}{128} \right\rfloor + \left\lfloor \frac{2012}{256} \right\rfloor + \left\lfloor \frac{2012}{512} \right\rfloor + \left\lfloor \frac{2012}{1024} \right\rfloor = 1006 + 503 + 251 + 125 + 62 + 31 + 15 + 7 + 3 + 1 = 2004
   \]
   So, there are 2004 factors of 2 in \(2012!\).

5. Since there are more factors of 2 than factors of 5, the number \(2012!\) has \(501\) trailing zeros, and the remaining factors of 2 will ensure that the last non-zero digit is even.

6. Therefore, the last non-zero digit of \(1 \cdot 2 \cdot 3 \cdot \ldots \cdot 2011 \cdot 2012\) is even.

The final answer is \(\boxed{\text{Even}}\).