Pot | First can | Second can | Ladle |
---|---|---|---|
It is evident that we were able to put 6 liters into the first can without using the second one and thus we can repeat the process with the second can.
We will now prove that it is always possible to create 2n-1 integers. For n=1, we have one integer (e.g. 5) and 21-1 = 1. For n>1, we will append a digit to all integers with n-1 digits in such a way, that the new integers will have the odd number of occurences of digit 5. In this way, we will create 2n-1 integers which will differ in at least 2 digits. There could have been 2n-1 numbers on the Sherlock's notice.
B. In this case, the answer is yes. Just consider this sequence of operations:
C. Take an arbitrary necklace N which can be constructed from "RR" and mark the numbers or red beads between blue beads as ci for i from [1,2n], where 2n is the number of beads on the necklace. Choose c1 arbitrarily and choose arbitrary orientation as well. We will prove, that alternating sum
If we chose a different c1, the remainder after division S(N) by 3 would change only from 1 to 2 or vice versa - S(N)s would differ in the sign. For a necklace with all red beads, we define S(N) equal to S(N) of any necklace, which can be constructed from it by one operation (divisibility by 3 will not be ambiguous).
S(N) mod 3 is 0 for the model "BB", and S(N) mod 3 is 2 for the model "RR". If there had been a sequence of operations leading from "RR" to "BB", there would have been an operation in this sequence, which would have resulted in the model with S(N) mod 3=0. As we have seen, such operation would have not been legal and such sequence doesn't exist.
Proof by contradiction. Let's assume there exists a case which Mulder was talking about. The square is divided into several parts. Perimeters of these parts consist of lines inside the square and perimeter of the large square. Each line belongs to at most 2 perimeters (some lines don't belong to any perimeter). Then, the sum of all perimeters is at most 40 (40=2.18+4). The sum of areas of parts is 1. Take one of these parts. Its perimeter is O and area S. In case its shape is not convex, its perimeter is not larger than the perimter of the smallest circumscribed rectangle. The area of this part must be smaller than the area of the circumscribed rectangle (see picture). If the rectangle has sides a and b, we get: