CLICK TO PRINT MAZE

]]>

This is a non-transitive dice set, i.e. every dice in it is weaker than some other dice. Can you design a non-transitive set with only 3 dice?

*Remark: "Weaker" means that it loses more often than it wins.*

The simplest solution is given by:

2, 2, 4, 4, 9, 9;

1, 1, 6, 6, 8, 8;

3, 3, 5, 5, 7, 7.

Another solution is given by the so-called "Miwin's dice". They are as follows:

1, 1, 3, 5, 5, 6;

2, 3, 3, 4, 4, 5;

1, 2, 2, 4, 6, 6.

]]>

Coming soon.

]]>

SLOTH + HARROW - ARROW = SLOTH

FERRULE - RULE + VINE + GARRET - VINEGAR = FERRET

]]>

You should choose the second barber. Since there are only two barbers, the chances are that they give each other haircuts. Also, probably the second barber's salon is dirtier because he has lots of work and does not have time to clean up properly.

]]>

In how many pieces at most can you cut a doughnut with only 3 slices?

The maximum number of pieces is 13, as shown on the image below (courtesy of HunkinsExperiments.com).

]]>

You have a drawer with 10 pairs of black socks and 10 pairs of white socks. How many times do you need to blindly reach inside the drawer and take out a sock, so that you get a matching pair?

Only 3 times. Once you have two socks of the same color, they already form a matching pair.

]]>

There are 25 horses and you want to find the fastest 3 among them. You can race any 5 of the horses against each other and see the final standing, but not the running times. If all the horses have constant, permanent speeds, how many races do you need to organize in order to find the fastest 3?

Let us label the horses H1, H2, H3, H4, ..., H24, H25.

We race H1 - H5 and (without loss of generality) find that H1 > H2 > H3 > H4 > H5. We conclude that H4, H5 are not among the fastest 3.

We race H6 - H10 and (without loss of generality) find that H6 > H7 > H8 > H9 > H10. We conclude that H9, H10 are not among the fastest 3.

We race H11 - H15 and (without loss of generality) find that H11 > H12 > H13 > H14 > H15. We conclude that H14, H15 are not among the fastest 3.

We race H16 - H20 and (without loss of generality) find that H16 > H17 > H18 > H19 > H20. We conclude that H19, H20 are not among the fastest 3.

We race H21 - H25 and (without loss of generality) find that H21 > H22 > H23 > H24 > H25. We conclude that H24, H25 are not among the fastest 3.

We race H1, H6, H11, H16, H21 and (without loss of generality) find that H1 > H6 > H11 > H16 > H21. We conclude that H16, H21 are not among the fastest 3.

Now we know that H1 is the fastest horse and only H2, H3, H6, H7, H11 could complete the fastest three. We race them against each other and find which are the fastest two among them. We complete the task with only 7 races in total.

]]>

CLICK TO PRINT MAZE

]]>

You have 20 strings of pasta left on your plate. You randomly start tying up their ends, until there are no loose ends anymore. What is the average number of loops which are created?

The expected (average) number of loops at the end of the procedure is equal to the expected number of loops created after the first tying, plus the expected number of loops created after the second tying, etc. After each tying, the number of non-loop strings decreases by 1, and then the probabilites to create a new loop are 1/19, 1/17, 1/15, etc. Therefore, the answer is the sum 1/19 + 1/17 + 1/15 + ... + 1/3 + 1/1 ~ 2.1.

]]>

Nothing like picking the best present for your mate.

*Concept by Puzzle Prime, art by Nizar Ilman.*

]]>

Bonus: Is the number 9991 prime?

Let the missing digit be

Bonus: 9991 = 10000 - 9 = 100

]]>

COLORS - SNAIL + NAIL + ADOBE + DRUG - BED - RUG = COLORADO

]]>

Since we can not place any more coins on the table, each point of it is at distance at most 2r from the center of some coin, where r is the raidus of the coin. Now shrink the entire table twice in width and length, then replace every shrunk coin with a full sized one. This way the small table will be completely covered, because every point of it will be at distance at most r from the center of some coin. Add three more of these smaller tables, covered with coins, to create a covering of the big table.

]]>

1. Nb1+ Kb3

2. Qd1+ Rc2

3. Bc1 axb6

4. Ra1 b5

5. Rh1 bxc4

6. Ke1 c3

7. Ng1 f3

8. Bf1 f2#

]]>

What is orange and sounds like a parrot?

The answer is "carrot".

]]>

Imagine you have three identical bricks and want to find the common length of their main diagonals. What is the easiest way to do this using just a measuring tape?

PLace the bricks as shown on the image below, then measure the marked segment.

]]>

Assume the opposite. Imagine the rings have zero thickness and reposition them in such a way, that two of them, say ring 1 and ring 2, touch each other in two points. These two rings lie either on a sphere or a plane which ring 3 must intersect in four points. However, this is impossible.

]]>

CLICK TO PRINT MAZE

]]>

SCOOP - COOP + NAIL = SNAIL

FOWL - OWL + OX = FOX

]]>