We have written the numbers from 1 to 12 on the faces of a regular dodecahedron. Then, we have written on each vertex the sum of the five numbers on the faces incident with it. Is it possible that 16 of these 20 sums are the same?
SOLUTION
We color the vertices in five colors as shown in the image below, such that each face of the dodecahedron has 1 vertex of each color. Then, the sum of the four numbers from each color must be equal to the sum of all numbers written on the faces: 1 + 2 + … + 12 = 78.
If 16 of the vertices have the same number written on them, then by the pigeonhole principle there will be 4 vertices with identical colors and identical numbers. Since 78 is not divisible by 4, we conclude that this is impossible.
Your goal is to switch the positions of the three white knights with the positions of the three black knights. What is the least number of moves required to do this?
SOLUTION
The least number of moves required to switch the positions of the knights is 16. An example is shown below.
Next, we prove that it is impossible to switch the positions of the knights with less than 16 moves. Since the white knights occupy 2 black and 1 white squares, they need to end up on 2 white and 1 black squares, and each knight must make at least 2 moves in order to get to the opposite side, the total number of moves for the white knights should be an odd number, larger or equal to 2+2+2=6. The same applies to the total number of moves for the black knights. Therefore, the only possible way to get a total number of moves less than 16 is if both the white knights and the black knights move exactly 7 times.
We assume it is possible to switch the positions with 7+7=14 moves in total. Then, the white knight on A2 and one of the white knights on A1, A3 should make 2 moves each, and the third white knight should make 3 moves and land on a white square, either D1 or D3. Without loss of generality, we assume the knight on A1 makes 3 moves: A1-B3-C1-D3. Then, the knight on A2 should make 2 moves: A2-C3-D1, and the knight on A3 should make 2 moves: A3-B1-D2.
We make the same argument for the black knights. Since it is impossible that the white knight on A1 moves along the trajectory A1-B3-C1-D3 and also the black knight on D3 moves along the trajectory D3-C1-B3-A1, we conclude that the black knight on D1 should make 3 moves: D1-C3-B1-A3, the black knight on D2 should make 2 moves: D2-B3-A1, and the black knight on D3 should make 2 moves: D3-C1-A2.
This is only possible if:
the knight on A1 moves to C1after the knight on D3 moves to A2
the knight on D3 moves to A2 after the knight on A2 moves to C3
the knight on A2 moves to C3 after the knight on D1 moves to B1
the knight on D1 moves to B1 after the knight on A3 moves to D2
the knight on A3 moves to D2 after the knight on D2 moves to B3
the knight on D2 moves to B3 afterthe knight on A1 moves to C1
We get a contradiction which means that the least number of moves is 16.
You are chasing a criminal riding a bicycle, and you find his tracks left in the dirt. By investigating the tracks, can you determine which direction the criminal has fled to: left or right?
SOLUTION
The front wheels of the bicycle are traveling more distance and making sharper turns. Thus, the red tracks (as shown on the image below) correspond to the front wheel, and the green tracks correspond to the back wheel.
Then, since the back wheel is always pointing towards the front wheel, we conclude that the bicycle is moving towards the right.
Partition the grid into disjoint “creatures”, according to the following rules:
Each creature is defined as a shape of 4 connected branches that are each 1 cell wide.
For each creature, one of the branches ends up with a HEAD (always clued) and the other three branches end up with LIMBS (whenever clued, their directions matter).
A creature can never occupy a 2×2 region of cells and can never touch itself.
Examine the first example, then solve the other three puzzles.
Can you find a triple of three-digit numbers that sum up to 999 and collectively contain all digits from 1 to 9 exactly once? How many such triples are there? What if the sum was 1000?
SOLUTION
There are exactly 180 such triples that sum up to 999 and none that sum up to 1000.
In order to see that, notice that the sum of the first digits of the numbers can be no more than 9. Since the sum of all digits is 45, the sum of the middle and the sum of the last digits should be both no more than 9+8+7=24, and no less than 45-9-24=12. We then see that the sum of the last digits should be exactly 19 and the sum of the middle digits should be exactly 18. The sum of the first digits should be 45-19-18=8.
There are 2 ways to get 8 using unique digits from 1 to 9: 1+2+5 and 1+3+4.
If the first digits are {1, 2, 5}, the options for the middle digits are {3, 6, 9}, {3, 7, 8}, and {4, 6, 8}. The last digits end up {4, 7, 8}, {4, 6, 9}, and {3, 7, 9} respectively.
If the first digits are {1, 3, 4}, the options for the middle digits are {2, 7, 9} and {5, 6, 7}. The last digits end up {5, 6, 8} and {2, 8, 9} respectively.
Since the set of the first digits, the set of the middle digits, and the set of the last digits of the numbers can be permuted in 6 ways each, we get a total of 5×6×6×6=1080 solutions, or 180 up to permutation of the 3 three-digit numbers.
In order to see that we cannot get a sum of 1000, we note that since the sum of the digits from 1 to 9 is divisible by 9, then the sum of the 3 three-digit numbers should be divisible by 9 as well. Since 1000 is not divisible by 9, the statement follows.
Add the 25 letters between A and Y to the grid below. Each one should appear exactly once. When you have finished, you must be able to spell the following words moving horizontally, vertically or diagonally around the grid:
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