Spot The Difference
Challenge

It is impossible. Assume the opposite and consider the smallest cube on the bottom, C1. Since all cubes surrounding it are larger, they must “wall in” its top face. We repeat the same argument for the smallest cube which lies on the top face of C1, call it C2. If all surrounding cubes are larger, they must wall in its top face. Thus, we can create an infinite sequence of cubes with decreasing sizes lying on top of each other: C1, C2, C3, etc. Since the initial cube is split into a finite number of smaller cubes, we get a contradiction.

First, we show that for every four pirates, there exists a lock which cannot be opened only by them and can be opened by everyone else. We choose an arbitrary group of four pirates. If they can open every lock, then they can access the treasure without the need of a majority. If any of the remaining pirates cannot open that lock, then he, together with the initial group of four still cannot access the treasure. Thus, the claim is proved and to each group of four pirates we can assign a unique lock. These are
(
(
4
9
)=126 locks in total. Finally, every pirate should get keys for
(
(
4
8
)=70 of these locks, one for each group of four additional pirates he can be a group of.

Notice that the N-th fish in the stream survives if and only if it is the fastest among the first N fish. The probability of this event happening is equal to 1/N. Since the expected number of fish that survive is equal to the sum of the survival probabilities for each of them, the answer is 1+1/2+1/3+…+1/N.
For more details on the last claim, consider reading our blog post “How Many Times on Average?”

Let the entering point is A, the leaving point is B and the center of the apple is C. Consider the plane P containing the points A, B and C and project the worm’s tunnel on it. Since 99 < 2×50, the convex hull of the tunnel’s projection will not contain the center C. Therefore we can find a line L through C, such that the tunnel’s projection is entirely in one of the semi-planes of P with respect to L. Now cut the apple with a slice orthogonal to P passing through the line L and you are done.

Once the warden writes the numbers on the prisoners’ foreheads, they form a mental circle, arranging themselves alphabetically in it (or according to any other order they agree on). They include the warden in this mental circle and imagine he has infinity written on his forehead.
Then, each prisoner examines the sequence of 100 numbers written on the foreheads of the others, and computes the number of inversions, i.e. the pairs which are not in their natural order. The prisoners which count an even number of permutations put black hats on, and the prisoners that count an odd number of permutations put white hats on. The infinity symbol is treated as the largest number in the sequence.
For example, if a prisoner sees the sequence {2, -6, 15.5, ∞, -100, 10}, then he counts seven inversions, which are the pairs (2, -6), (2, -100), (-6, -100), (15.5, -100), (15.5, 10), (∞, -100), (∞, 10), and puts a white hat on.
Next, we prove that the devised strategy works. We consider two prisoners, P1 and P2, who are adjacent in the final line the warden forms. These two prisoners split the mental circle in two arcs: A and B.
The number of inversions P1 counts is:
I
1
=I(A)+I(B)+I(A,B)+I(A,P
1
)+
where
I(X) denotes the number of inversions in a sequence
X, and
I(X,Y) denotes the number of inversions in a pair of sequences
(X,Y):
I(X)=∥x
i
,x
j
∈X:i<j,x
i
>x
j
∥
I(X,Y)=∥x∈X,y∈Y:x>y∥
Similarly, the number of inversions P2 counts is:
I
2
=I(A)+I(B)+I(B,A)+I(B,P
2
)+
We sum
I
1
and
I
2
to get:
I
1
+I
2
=2I(A)+2I(B)+I(A,B)+I(B
+I(A,P
1
)+I(P
2
,A)+I(P
1
,B)+
=2(I(A)+I(B))+∥A∥∥B∥+∥A
Since
∥A∥+∥B∥=99 is an odd number, we see that
I
1
+I
2
is also odd. Therefore, one among P1 and P2 would have counted an even number of inversions, and one would have counted an odd number of inversions. Thus, their hats have alternating colors.

The answer is NO. In order to show this, assume they can and consider their reverse movement. Now the grasshoppers start at the vertices of some square, say with unit length sides, and end up at the vertices of a smaller square. Create a lattice in the plane using the starting unit square. It is easy to see that the grasshoppers at all times will land on vertices of this lattice. However, it is easy to see that every square with vertices coinciding with the lattice’s vertices has sides of length at least one. Therefore the assumption is wrong.

The first player has a winning strategy.
His strategy is with each turn to keep the graph connected, until a single connected component of 6 or 7 nodes is reached. Then, his goal is to make sure the graph ends up with either connected components of 8 and 2 nodes (8-2 split), or connected components of 6 and 4 nodes (6-4 split). In both cases, the two players will have to keep connecting nodes within these components, until one of them is forced to make the graph connected. Since the number of edges in the components is either
C
2
8
+C
2
2
=29, or
C
2
6
+C
2
4
=21, which are both odd numbers, Player 1 will be the winner.
Once a single connected component of 6 or 7 nodes is reached, there are multiple possibilities:
The connected component has 7 nodes and Player 2 connects it to one of the three remaining nodes. Then, Player 1 should connect the remaining two nodes with each other and get an 8-2 split.
The connected component has 7 nodes and Player 2 connects two of the three remaining nodes with each other. Then, Player 1 should connect the large connected component to the last remaining node and get an 8-2 split.
The connected component has 7 nodes and Player 2 makes a connection within it. Then, Player 1 also must connect two nodes within the component. Since the number of edges in a complete graph with seven nodes is
C
2
7
=21, eventually Player 2 will be forced to make a move of type 1 or 2.
The connected component has 6 nodes and Player 2 connects it to one of the four remaining nodes. Then, Player 1 should make a connection within the connected seven nodes and reduce the game to cases 1 to 3 above.
The connected component has 6 nodes and Player 2 connects two of the four remaining nodes. Then, Player 1 should connect the two remaining nodes with each other. The game is reduced to a 6-2-2 split which eventually will turn into either an 8-2 split, or a 6-4 split. In both cases Player 1 will win, as explained above.

First, we choose an axis, so that Napoleon and the two policemen lie on a single parallel. Then, the strategy of the two policemen is to move with the same speed as Napoleon, keeping identical latitudes as his at all times, and squeezing him along the parallel between them.
In order to claim 25% of the planet’s surface, Napoleon must travel at least 90°+90°=180° in total along the magnitudes. Therefore, during this time the policemen would travel 180° along the magnitudes each and catch him.

Split 4 of the loaves into 3 pieces and the other 3 loaves into 4 pieces. Then, give each person one of both types of pieces.

First, we note that if the number of people is a power of 2, then the first person will survive every round. The greatest power of 2 that is less than 1000 is 512. Therefore, after 488 people die, there will be 512 remaining and the first one to kill the 489-th person will survive. This person has number 1+2×488=977.