Puzzle of the Week #290 - Special Sequence

This is a special sequence:

3 1 2 1 3 2

For each number ‘k’ from 1 to 3, the number of numbers between the pair of ‘k’s is equal to k.

In other words, there is one number between the pair of 1s, two numbers between the pair of 2s and three numbers between the pair of 3s.

Can you create a similar sequence containing pairs of numbers but using numbers 4, 5, 6 etc up to n (n can be whatever you need it to be to make a sequence that has no gaps), where for each number k, the number of numbers between the pair of ‘k’s is k (so there will be four numbers between the pairs of 4s, five numbers between the pair of 5s, etc.)?

 

 

Puzzle of the Week #288 - Round Table

A family that consists of parents and 6 children sit around a table in age order: father, mother, eldest child etc, so the youngest child sits next to the father.

The gender of each of the children is male or female, with 50% probability of each.

Knowing only this information, what is the probability that all the males are seated together?


Puzzle of the Week #286 - The Cube and the Cup

This came to me whilst trying to balance a Rubik’s cube on the circular rim of a coffee cup. There are three ways of orienting a cube on top of a cup, which I’ll refer to as face-down, edge-down (resting on the rim in four places) and point-down (resting on the rim in three places. For these latter two cases the cube extends below the level of the rim of the cup. Depending upon the particular cube and the particular cup, the cube might extend deeper into the cup when it is edge-down or when it is point-down.

cubes.jpg

Assuming that the edge-down and point-down orientations are possible for the given cube and cup, how can you tell, very quickly and without using any measuring devices, only looking at and manipulating the cube and the cup, which of those two orientations extends deeper into the cup?

Puzzle of the Week #285 - Left Right Pairs

This puzzle was inspired by a real-life problem I had in my job as an engineer. Faced with a number of lengths I needed to come up with a set of left parts and a set of right parts which could combine to form every length on the list. For instance, the length 52 could be made up of 26L + 26R, or 20L + 32R, or any number of different combinations. For reasons of making and stocking the individual left and right parts it was important to minimise the total number of different parts I needed.

For this puzzle I have massively shortened and simplified the list but the idea remains the same: what is the least number of distinct parts you will need, some left-handed, some right-handed (parts that are the same length but different hands count as two different parts), that will fit together in left-right pairs to form any length on this list:

20 23 28 44 46 48 50 52 56 58 70 74


Puzzle of the Week #282 - Heptagon Revisited

You may remember this heptagon figure from a few weeks ago. One consequence of the solution was that the circle shown here, whose diameter endpoints are also extended intersections of pairs of sides of the heptagon, also passes through two midpoints of sides of the heptagon.

As it happens, for a regular heptagon there is another distinct way in which you can draw a circle whose diameter endpoints are extended intersections of the sides, and which also passes through midpoints of sides.

By distinct I mean not just merely a rotation of the circle shown. The other circle will have a different diameter to the one shown.

If the circle shown has a diameter of 6296, what is the diameter of the other possible circle (to the nearest unit)?

Alternatively, if you don’t want to do the numbers, just send me a sketch of the alternative circle.

heptagon revisited puz.JPG

Puzzle of the Week #268 - Dice Game

This puzzle is based on a dice game my son and I invented, involving 12-sided, 10-sided, 8-sided and 6-sided dice. (As with modern convention I will be using the word ‘dice’ to refer to both the plural and the singular, rather than the more archaic ‘die’).

In round 1, both players each have a 12-sided dice. They roll them and the player with the lowest number loses that round (if the numbers are equal they both simply roll again). The losing player exchanges their 12-sided dice for a 10-sided dice for the second round.

The game continues such that at each round the loser changes their dice for one with two fewer sides. The final round occurs when one of the players has to roll a 6-sided dice. The winner of this round is the overall winner of the game (even if they lost all of the previous rounds).

Since the game is symmetrical, the probability of each of the players winning is 50%, but the puzzle question is, how important is the first round? What is the probability that you win the overall game if you win/lose the initial round?

Puzzle of the Week #267 - Tetrahedron Game

In this game there is a grid of 22 triangles, four of which are painted. You may initially place a tetrahedron on any triangle you want but from there forward you move by ‘rolling’ the tetrahedron over one of its edges so that it will then be on an adjacent triangle on the grid and resting on a different face of the tetrahedron. If you roll onto a painted triangle, the paint transfers to that face of the tetrahedron (unless that face is already painted), and when a painted face of the tetrahedron lands on an unpainted triangle, the paint is transferred from the tetrahedron to the triangular grid.

The object of the game is to continue to roll around the grid until all four faces of the tetrahedron are painted, however only one of the four grids shown can actually be solved. The other three are not solvable regardless of where the tetrahedron is initially placed.

Which is one that can be solved?

tetrahedron puzzle.JPG

Puzzle of the Week #266 - Base Four Code

I have taken a quotation, and I have replaced each of the letters with the numbers that denote their position in the alphabet. However, I have used the base 4 number system. 

base4.JPG

Be careful, as some sequences of numbers could lead to several words, for instance 31110 could mean CAT (3,1,110), but could equally mean MAD (31,1,10). 

132121333211  1132033  10311033100103  3011110232213213  21103  333010,  11320111102011102 1101131132110121  33102  11211320110121.  132121333211  1132033  231111100103  3011110232213213 1031101121103  121331113213.  1102011  1310211111011103110  11020213213  12133111  3132  1033  21103 231111100  12133111102  31213210  121331113213.