Puzzle of the Week #373 - Letter Swap: IDLE

Here is a 4x4 word square, which has four words reading across, and four others reading downwards, except that it is all wrong!

Precisely none of the letters is in the correct place in the grid. However, they have not been placed randomly: eight pairs of letters need to be swapped in order to solve the grid. So, for instance if you decided that the top left letter belongs in the bottom right square, that would also mean that the bottom right letter belongs in the top left square.

I can also tell you that none of the swaps are within a row or column: each letter will end up in a different row AND column from where it started.

Beware of repeated letters: even if you know for sure that a particular letter belongs in a certain square, if there are several copies of that letter in the grid, you have to choose the correct one, otherwise the other letter of the swapped pair would end up in the wrong place.

Because it is very difficult to get started with this puzzle, I will tell you one of the solution words but you’ll have to decide where it goes.

Good luck!

Puzzle of the Week #372 - Totient Trouble

Allow me to introduce you to Euler’s Totient function, phi(n). It is the number of numbers less than a number that don’t share any factors with that number. For instance, when n is 6, phi(n) is 2, because there are only 2 numbers less than 6 that are coprime with 6 (1 and 5).

There is a shortcut way of finding the totient function of a number: first list all of the prime factors of the number, then go through them one by one, if you see a prime factor for the first time, subtract 1 from it, but if it’s one you’ve already seen, leave it as is. Then multiply the (some now modified) factors back together. For example, the totient function of 24:

 

24 = 2x2x2x3, phi(24) = (1)x2x2x(2) = 8.

 

Now after that crash course it’s going to get even more complicated as we consider doing it in reverse. The inverse totient function lists all of the numbers n for which phi(n) equals a particular value. We have seen that phi(24)=8, but for what other values of n is phi(n)=8?

The full list is 15, 16, 20, 24 and 30. Only one of these numbers is odd (15). This is no accident, and brings us around (finally!) to the question I want to ask you.

 

To simplify things slightly, I ONLY want to consider cases where n is odd, and phi(n) is a power of 2, for example:

phi(1)=1, phi(3)=2, phi(5)=4, phi(15)=8, phi(17)=16, phi(51)=32, phi(85)=64, etc.

 

In each case there is exactly one odd value for which phi(n) is equal to a particular power of 2. However this pattern doesn’t last forever, and eventually we will find that some powers of 2 are not the totient function of ANY odd numbers.

 

What is the first such power of 2?

 

 

Puzzle of the Week #371 - Letter Swap: CAVA

Here is a 4x4 word square, which has four words reading across, and four others reading downwards, except that it is all wrong!

Precisely none of the letters is in the correct place in the grid. However, they have not been placed randomly: eight pairs of letters need to be swapped in order to solve the grid. So, for instance if you decided that the top left letter belongs in the bottom right square, that would also mean that the bottom right letter belongs in the top left square.

I can also tell you that none of the swaps are within a row or column: each letter will end up in a different row AND column from where it started.

Beware of repeated letters: even if you know for sure that a particular letter belongs in a certain square, if there are several copies of that letter in the grid, you have to choose the correct one, otherwise the other letter of the swapped pair would end up in the wrong place.

Because it is very difficult to get started with this puzzle, I will tell you one of the solution words but you’ll have to decide where it goes.

Good luck!

Puzzle of the Week #370 - Points of Rotation

I have two identical 4 x 4 squares as shown, identical except for orientation and position.

Clearly it is possible to move the tilted square to the position of the upright square by first moving it so that one of its corners coincides with that of the upright square, and then to perform a rotation so that the other three corners also coincide. You might also choose to perform a rotation first, and then a move to get the squares to line up.

Anyone who’s ever studied group theory will know that the result of rotation plus a translation can always be done using ONLY a rotation - it’s just a question of figuring out where the centre of rotation needs to be.

Since the shape we are using, a square, has order 4 rotational symmetry itself, there are in fact four possible centres of rotation, depending on which of the four sides of the tilted square eventually coincides with the base of the upright square. For instance, a 30 degree clockwise rotation around the top rotation point will align the squares such that the bottom right side of the tilted square becomes the base, whereas a 60 degree anti-clockwise rotation around the lowest rotation point will also align the squares, but now the bottom left side of the tilted square coincides with the base of the upright square.

As an aside it is interesting to note that these four points are collinear.

This week’s challenge is to find the co-ordinates of two of these four centres of rotation. Or for extra kudos, the coordinates of all four.

Puzzle of the Week #369 - Alphabet Crossword

Your task is to reconstruct this crossword. One of each of the 26 letters of the alphabet have been removed from the crossword and needs to be put back in. In addition, the black squares need to be added in.

The final pattern of black squares is symmetrical in the horizontal axis and in the vertical axis, but NOT along the diagonal axes. There is at least one black square in every row and every column, and all words are at least three letters long. Every word in the final solution contains both given (black) letters and alphabet (red) letters.

Puzzle of the Week #367 - Come Close But Don't Touch

If you have five unit fractions (a unit fraction is a fraction with 1 on top and a positive whole number on the bottom, such as 1/2, 1/3 or 1/1000), and you add them together, how close can you get to a sum of 1 without actually equalling 1 or going beyond it?

 

For instance, if the first four fractions were each 1/5, the most the last one could be is 1/6, which would make the sum 29/30. Close to 1, but not equal, as we are seeking. You can get even closer of course, but you can’t get arbitrarily close; there is a maximum.

Puzzle of the Week #365 - Twenty Coloured Balls

I asked my 10-year-old son Austin the question: if you had four balls in a bag and you wanted to colour them such that a randomly drawn pair of balls would have a 50% chance of matching in colour, how would you colour them? I was fully expecting him to go for the obvious but wrong answer of two of one colour, two of another. Instead, Austin, quick as a flash, came up with the correct answer of three of one colour and one of a second colour. I was intrigued as to how he did it so quickly; he explained that he visualised the four balls in a 2x2 grid. That way it was obvious to him that with a (3,1) colouring, half the rows half the columns and half the diagonals were matching in colour. Nice one Austin!

 

I now have 20 coloured balls of various colours in a bag, such that if you pick two balls at random the chance that the two chosen balls match is 50%. For this one there probably isn’t a neat visual shortcut, so arithmetic is likely needed.

 

What does the arrangement of the twenty coloured balls need to be?

Puzzle of the Week #363 - Zero to Pi(ish)

I discovered an interesting fact the other day, that you can start with 0, and reach ANY positive rational number by selectively using two different actions:

The two actions are as follows:

a)    Adding 1, such that x becomes (x+1)

b)    Taking the reciprocal, such that x becomes (1/x)

 

For example, trying to reach 2/5:

 

starting with 0,

Action a makes it 1

Action a makes it 2

Action b makes it 1/2

Action a makes it 1 1/2

Action a makes it 2 1/2

Finally action b makes it 2/5

 

Pi is famously not a rational number, however 355/113 is a great approximation to it. Starting with 0, how many steps will it take to reach 355/113?

 

Puzzle of the Week #361 - Round and Round

Choose a value x such that:

You begin with 3x, and round to the closest whole number.

Multiply this by 4x, and round to the closest whole number

Finally multiply this by 5x, and you will have the answer 273.

 

For example:

Let x = 7/4,

3x = 21/4 = 5.25, round to 5,

Multiply by 4x (7) = 35, round to 35.

Finally multiply by 5x (35/4) to get 306.25

As 306.25 is not 273, this is not the correct value of x. But what is?

 

 

Puzzle of the Week #360 - Isosceles Nine-Point Circle part 2

Following on from part 1, I have a similar situation, examining the eight distinct points of a nine-point circle of an isosceles triangle.

 

However, they now form an IRREGULAR octagon. In particular, the two sides of the octagon that meet at the midpoint of the base are one length, and the other six sides are a second length.

 

Now what is the angle at the apex of the triangle?

 

(For uniqueness, let the angle be less than 90 degrees).

 

Puzzle of the Week #359 - Isosceles Nine-Point Circle

It is well known that for any triangle, the following nine points:

a) the three midpoints,

b) the three feet of the altitudes (the altitudes are the lines from each vertex to the opposite side, perpendicular to that side),

c) and the three points midway between the vertices and the orthocentre (the orthocentre is the point where the three altitudes cross),

all lie on a circle, known as the Nine-Point Circle.

Sometimes some of those points coincide with one another, for instance the midpoint of the base of an isosceles triangle is also the foot of the altitude from the apex down to the base.

I have an isosceles triangle, and so since two of the nine points coincide, there are now eight distinct points. In my triangle these eight points are equally spaced, forming the vertices of a regular octagon.

 

What is the angle at the apex of the isosceles triangle?

 

For bonus points, what is a second possible value for the apex angle?

Puzzle of the Week #358 - Power Dates

Using the DDMMYY format, you can change a date to a six digit number, for instance 15th August 2022 would be 150822, and 1st November 2023 would become 011123. Some of these dates can be perfect powers, for instance 11th February 2024 would be 110224, which is 332 squared, and 9th November 2025 would be 091125 which is 45 cubed. You can also make 4th and 5th powers, and also one more beyond that. What is the only date that becomes a power beyond the fifth power?

 

Puzzle of the Week #356 - Counter Game 2

In a similar vein to my previous puzzle Counter Game, there are a number of counters on the table, but this time you can either remove 2 or 7 of them on each turn. For instance if there were 11 counter there are three solutions: 2,2,7; 2,7,2; 7,2,2; (as before removing 7 then 2 is distinct from removing 2 then 7). Some numbers of counters have no solution, for instance if there are 5 counters you cannot remove them all.

If the number of counters is high enough, there are always more different ways to remove all the counters than to leave one counter behind, but this is not true for some smaller numbers, for instance if there were 12 counters there is only one way of removing them all (2,2,2,2,2,2), but as we have seen, three ways of removing 11 and leaving 1 behind.

 

What is the highest number of counters, for which there are fewer ways to remove them all than to leave one remaining?

Puzzle of the Week #354 - Counter Game

I have 16 counters. On each turn I either remove 1, 2 or 3 counters until there are no counters left.

I could remove 2 counters eight times, or I could remove five lots of 3 counters then a single counter, or I could remove 3, 2, 2, 1, 3, 1, 1, 1, 1, 1, or any number of different ways.

Well in fact not ‘any’ number of different ways. Your task is to find out exactly how many ways there are. Note: the same combination but occurring in a different order counts as different ways, for instance, there is only one way using 2 eight times, but there are six ways involving five 3s and a 1.

I’ve chosen the number of counters to be 16 because the total number of ways is the square of a prime number. This fact doesn’t particularly help you, except for checking whether or not you are correct.

How many different ways are there?