Two letters of the alphabet are chosen at random. What is the probability that the bricks that correspond to those letters are adjacent in this wall:
The fraction 1174/5063 cannot be reduced further in the usual way, however it can be expressed exactly using only six digits instead of eight.
The line begins in a vertical downward direction, transitions into an arc of radius 60, then into another arc with a tighter radius, which just touches the horizontal dashed line. Then there is an arc of radius 50 in a clockwise direction, then a horizontal line. At each transition between line and arc or arc and arc the tangents are continuous.
What is the missing radius?
Each of the 25 squares contains either a building of height 1, 2, 3 or 4, or an empty lot. Each row and each column contains exactly one of each of those.
The numbers in the twenty ‘vantage points’ around the edge denote how many buildings you can see looking along that particular row or column, so for instance if the 2 building was closest, followed by 1, 4 and 3 in that order, the vantage point would say 2, as you can see the 2-building and the 4-building, whereas the 1-building and 3-building are hidden behind taller buildings.
Here’s an example (with only 1, 2 and 3 height buildings) which might make it clearer, followed by the puzzle itself:
You might be familiar with the famous ‘Nine Point Circle’ theorem, which says that for ANY triangle the 3 midpoints (shown below in red), the 3 perpendicular feet (shown in green), and the 3 points midway between the each of the vertices and the orthocentre (shown in blue), will all lie on a circle. (Of course for an equilateral or isosceles triangle some of these points will coincide). But I have discovered there is something quite special about the 9 points of a triangle with angles 45, 60 and 75.
What is so special?
The figure is to scale so you may be able to guess the answer.
I’m standing in a rectangular field. I am 34m from the top left corner and 62m from the bottom right corner. I am an equal distance from the other two corners: what is that equal distance?
The AREA of each of the eight squares is 26. What is the area of the rectangle that surrounds them?
There are 16 teams in the knockout stages of a football tournament. For the sake of argument, say that each team has a distinct ranking from 1 (best) to 16 (worst), and that in any individual match, the better team will always win and progress to the next round.
The first round matches are randomly arranged, and the eight winners of those matches are randomly arranged in four matches in the quarter final etc.
Which of the following scenarios is more likely:
The 2nd ranked team is knocked out before the semi-finals, or
The 7th ranked team progresses at least as far as the semi-finals?
Reassemble this word wall using the bricks provided. Unfortunately the bricks that go in the positions marked with an asterisk are missing, and must be reconstructed.
It may actually be useful to print and cut out the pieces.
The large isosceles triangle can be dissected into 12 smaller isosceles triangles as shown.
What is the angle at the top of the large isosceles triangle?
A squad of 20 players (3 goalkeepers and 17 outfield players) splits into two equal teams for a training match (with any unchosen players reduced to spectating). Each team must have exactly one goalkeeper, but otherwise the players are entirely interchangeable.
It is calculated that there are a little over 5 million distinct* ways to select the teams.
*(so, for instance, if it were three-a-side, ABC v DEF would be the same as FDE v BCA).
How many players on each team?
Seven identically-shaped jigsaw pieces fit into the grid to form words in horizontal and vertical directions. Unfortunately one of the seven pieces has been lost. Can you place the given pieces in their correct positions within the grid and reconstruct the final missing piece?
I have a 6 digit number ‘abcdef’ such that the digits a, b and c are all odd and d, e and f are all even. If I multiply my number by 5½ I get ‘defabc’.
What is my number?
It’s been a while since I made a word puzzle, so here is one I hope you’ll like:
Via a series of steps, change from one word to another, firstly changing to an anagram of the previous word, secondly to a word that rhymes with the previous word, and subsequently alternating between the two types of changes.
For instance if you wanted to change from CHAR to CHARM you could change CHAR to ARCH (anagram), then ARCH to MARCH (rhyme), then MARCH to CHARM (anagram).
Can you change from THING to PRIEST:
(x + 1/x) = 7
(x^3 + 1/x^3) ?
This being a puzzle rather than just a maths problem, there is of course a way to solve it without having to go through the messy business of working out what x is equal to. You might even want to let (x + 1/x) = A and then express (x^3 + 1/x^3) purely in terms of A.
Eight ants are positioned at the eight corners of a cube. At once they all move along one of the edges to another adjacent corner, each choosing at random from the three other corners available.
What is the probability that the ants will perform this manoeuvre without any of them having to pass another coming the other way along the same edge or ending up at the same corner as another ant?
I’ve discovered this interesting way of making sequences. Start off with any three numbers, then any subsequent number is the sum of the previous number and the number before that, minus the number before that.
In mathematical terms a(n) = a(n-1)+a(n-2)-a(n-3)
If you start with, say, 1, 1, 1, the sequence continues with 1s forever.
If you start with 1, 2, 3, the sequence continues to count up through all of the integers in turn.
I plugged in some random numbers as the first three terms and looked way down the list that was produced:
The 243rd number in the sequence is 1092.
The 302nd number in the sequence is 1355.
The 577th number in the sequence is 2595.
What were the first three numbers in the sequence?
Sorry to still be banging on about this triple coin flip game (see POTW #160 and #161), but the maths behind it is particularly interesting. Today’s puzzle has a very long preamble, but I feel it’s worth the journey and the puzzle itself is quite good fun!
For a given array there will be a particular number of coins, and a particular number of possible moves. For instance, a 4 by 4 grid will have 16 coins and 24 moves.
There are four directions of moves: Horizontal, Vertical, Downhill and Uphill. For the purposes of notation I will place a letter H, V, D or U at the point in the grid corresponding to the centre of the three coins being flipped. So the grid to the right shows all 24 possible moves.
The solution to POTW #161 is particularly interesting, as the number of coins and the number of moves are the same. This means that for any given starting arrangement, there is exactly one set of moves that will solve it. (We can safely say that performing the same move twice is the same as not performing it at all, so we don’t need to worry about repeating a move; a move is either performed or not).
For any arrangement larger than the solution to POTW #161, there will be some sets of moves, which I will call ‘null sets’, that cancel each other out (the simplest of these to understand would be three horizontal moves and three vertical moves within a 3 by 3 area of the array – each of the 9 coins will be flipped twice and end the same as it began). You can even work out how many ‘null sets’ that will exist:
For example, a 3 by 4 grid has 12 coins, and 14 moves. This means there are 2^12 (4096) possible arrangements and 2^14 (16384) possible sets of moves. This means there are 4 times as many ‘move sets’ as there are arrangements. It follows from that that there are 4 ‘null sets’ (although one of those null sets is the empty set – no moves at all).
I’ll now introduce the concept of an ‘independent null set’. That is a null set that cannot be formed by combining other null sets you’ve already found (so technically any null set might be an independent null set if it happens to be one of the first you find). For the 3 by 4 grid there will be 2 independent null sets (let’s call them a and b), and the total of 4 null sets can be formed by a binary combination of these (that is, ‘a and b’, ‘a only’, ‘b only’, and ‘the empty set’). Below is a possible ‘a’ and ‘b’ for the 3 by 4 grid:
This actually simplifies things a little, as the number of independent null sets will simply be the difference between the number of moves and the number of coins. So back to the 4 by 4 grid which will be the basis of this week’s puzzle. As mentioned before, there are 16 coins and 24 moves, so there will be 8 independent null sets. I have found six of them for you. It may initially look as if reflections of e and f will also be independent, but in fact they can be formed by combining each of them with all four of a, b, c and d.
The final two independent null sets are reflections or rotations of each other. Can you find them?
So to recap in case you are a little lost, find a set of moves within the 4 by 4 grid (with no individual move repeated) that cancels itself out, and that cannot be formed by combining any of the other six such sets of moves above.
Addendum: to explain how to 'combine' existing sets of moves: you superimpose all of the moves in the move sets you want to combine. Any moves that occur an even number of times cancel out, any that occur an odd number will be in the combined move set.
Some numbers can be the hypotenuse of an integer triangle, and some cannot. For instance, 24 cannot, neither can 27, but 25 and 26 both can (for example 7,24,25 and 10,24,26). In fact 25 and 26 is the first case where two consecutive numbers can each be hypotenuses. The first case of three numbers in a row which can be hypotenuses is 39, 40 and 41 (eg 15,36,39; 24,32,40; 9,40,41).
Which numbers are the first example of four in a row that could each be hypotenuse?
Following on from last week’s coin flipping puzzle, where a horizontal or vertical line of three coins could be flipped together, with the object being to end with all the coins heads up. I’ll introduce a new move now – a diagonal (45 degree) line of three coins can also be flipped together. It turns out that with this new rule, as long as the arrangement of coins is big enough, any initial arrangement can be solved. But the question is: how big is big enough?
Find the arrangement with the fewest coins, such that any starting arrangement of heads or tails can be made all heads by a series of three-in-a-row flips in horizontal, vertical or diagonal directions.
(The coins must lie in a rectangular grid pattern. If there are spaces in the grid without coins, then a triplet of coins you wish to flip cannot bridge across the gaps).