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?

Street Address

City, State, Zip

Phone Number

Your Custom Text Here

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:

If

(x + 1/x) = 7

what is

(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).

In case you weren't aware: this puzzle blog has a Facebook page:

https://www.facebook.com/elliottlinepuzzles/

Each Friday morning I post the link to the Puzzle of the Week to the Facebook page, and invite people to send me their answers as a message (to avoid spoilers).

Everyone is welcome to follow my page!

In this game, the object is to make it so that all the coins are facing heads up, but on each move you must flip three coins together in a line, horizontally or vertically. For instance you could flip the first three coins on the top row, or the last three coins in the second column, or the middle three coins in the fourth row.

As it stands, the game cannot be solved, however if you flip one individual coin from the middle nine before you start, then the game can be solved.

Which of those middle nine coins should you opt to flip?

A President discovered on a TV show that you can generate an infinity of numbers even if you only had a finite number of prime numbers, and so decided that prime numbers beyond a certain size could be dispensed with.

He decreed that only the first ten prime numbers (those up to and including 29) could remain, and any numbers that contained prime factors higher than those would simply be skipped over when counting. So if you had 30 of something, and someone gave you one more, you would now be said to have 32 under this new system.

I recently did some freelance work in that country but when it came time to be paid in gold coins I was dismayed to discover that I only received half the gold coins I had been expecting! The amount I had invoiced them for was a number they didn't recognise, and so I was paid a gold coin for each of the numbers up to that number that they did recognise.

How many gold coins did I actually receive?

I have a thin strip of paper 1 inch wide. I lay it out horizontally in front of me. Along the top edge I mark two points, at five inches and ten inches from the top left corner.

I fold the top left corner down and slightly to the right, such that the fold line passes through the bottom left corner of the strip and the ten inch mark on the top edge.

I then draw a straight line from the bottom left corner of the strip, passing through the new position of the five inch mark, and continuing on its shallow diagonal trajectory until it hits the top edge of the strip.

How far along is the point where the line meets the top edge?

Place a 0 or 1 in each of the empty cells so that in each row and column a pair of 5-digit binary numbers can be read (therefore with decimal values between 1 and 31), such that the product of the two numbers in a particular row or column is shown at the end of that row or column.

Below is the final standings of a World Cup first round group.

The total number of goals was 16

What were the results in each of the 6 matches played?

You have two unmarked jugs, one with a capacity of 700ml and another with capacity of 1100ml. There is also a sink with a tap, however the 1100ml jug is too large to fill up at the sink. How can you measure out exactly 500ml?

If someone in Port-au-Prince writes to someone in Zagreb, someone in Bratislava writes to someone in Bishkek, someone in Rabat writes to someone in Yerevan, and someone in Baku writes to someone in Cape Town, where does someone in Sofia send their letters?

Here is an interesting mathematical game to play. The idea is to reach a total by multiplying and/or subtracting positive integers, whilst minimising the sum of those integers. Rather than explain I’ll demonstrate:

For example if the target is 99 you could have:

11 x 3 x 3 = 99, giving a score of 17

Alternatively you could save a couple of points thus:

5 x 5 x 4 – 1 = 99, giving 15 points

But even this can be improved upon:

3 x 4 – 1 x 3 x 3 = 99, scoring 14 points.

Note that each calculation, whether multiplying or subtracting, is performed stepwise, rather than using the BODMAS order of operations.

So, the target I have chosen for the puzzle is 12345. What is the lowest score you can achieve?

I should point out that I am not confident that the answer that I have is the best achievable (twice before I thought I had the best answer and then found a better one). If your score is higher I’ll be able to tell you that a lower score is achievable, but I am fully prepared to discover that my score is not the best either!

Additionally, if anyone can find an algorithm that finds the best solution in the general case I’d love to hear of it. At present all I have been able to establish is a loose set of rules of thumb.