Puzzle of the Week #382 - Zipline Futoshiki

You must place the number 1 to 16 in the grid such that:

An even number N must be placed in the same row or column as N-1.

An odd number M must be placed in the same diagonal as M-1.

(So if you were to trace the paths from 1 to 2, on to 3, etc all the way to 16, you would be alternating between rook moves and bishop moves).

All of the inequalities are correctly observed.

If the inequality is red, the numbers concerned are consecutive numbers.

A red X means that one or other of the pairs of numbers diagonally adjacent to the X are consecutive numbers.

Puzzle of the Week #381 - Sorted!

I am going to show you a sorting algorithm. In this example there are 9 letters, three each of A, B and C. They begin in a collated order: ABCABCABC, but they must end in a sorted order: AAABBBCCC in as few steps as possible. A ‘step’ involves selecting some portion of the string and reversing the order of the letters within it.

For our example this is possible in only three steps:

A[BCABCA]BC > A[ACBACB]BC

AA[CBA]CBBC > AA[ABC]CBBC

AAAB[CCBB]C > AAAB[BBCC]C

 

Now to the actual puzzle. This time you have 12 letters, which start collated and must finish sorted. I will let you decide how many different letters there are in the 12: either 2 each of 6 different letters, 3 each of 4 different letters, 4 each of 3 different letters, or 6 each of 2 different letters. Which of the following collated strings will take the fewest steps to sort?

 

ABCDEFABCDEF

 

ABCDABCDABCD

 

ABCABCABCABC

 

ABABABABABAB

 

Puzzle of the Week #380 - Prime Balance

I have some identical coins positioned around the edge of a flat disc, such that:

 

The coins are in several distinct piles, and the piles are equally spaced around the edge of the disc.

Each pile contains a prime number of coins.

No two piles are the same size.

The centre of gravity of the coins is in the exact centre of the disc.

 

What is the minimum number of coins I could have?

 

Puzzle of the Week #378 - Star Balance

You have a supply of two different sizes of blocks: some weigh 1kg, others are heavier than 1kg but not as much as 2kg (you get to decide the exact weight of these heavier blocks but they must all weigh the same).

 

If none of the five positions is allowed to be left empty, and each of the five piles is different, what is the minimum number of blocks you can arrange at the five points of a regular penta-star so that the entire arrangement is balanced around the centre?

 

Bonus: would it make a difference if one of the positions can be left empty (but only one as otherwise two ‘piles’ would be the same)?

Puzzle of the Week #377 - Cyberpunk Number

Consider the letters of the word CYBERPUNK. There are nine letters without repeats, so therefore there are 9! (=362880) different ways of arranging them.

If you place those 362880 ways in alphabetical order and number them: 1=BCEKNPRUY, 2=BCEKNPRYU, 3=BCEKNPURY, … 362880=YURPNKECB, which number is CYBERPUNK?

 Bonus question: which string of letters will be 100,000th in the sequence?

 

Puzzle of the Week #376 - Balancing Act

I start with a regular pentadecagon (15-sided shape) with has a kilogram weight on each of its 15 vertices. Clearly the centre of gravity will be in the exact centre of the shape.

Someone comes along with a 16th kilogram weight and places it at vertex ‘O’.

How can I take the eight weights from vertices A to H and redistribute them amongst vertices A to H so that the overall centre of gravity is once more in the exact centre of the shape? The weights cannot be subdivided, and must be placed on those vertices, not elsewhere on the shape.

Bonus question: if you wanted to put a 17th weight somewhere on the left-hand side (positions I to O), such that the eight weights on the right-hand side (A to H) could again be rearranged to balance the system, where would the 17th weight need to be placed?

Puzzle of the Week #375 - Letter Swap: OMIT

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 #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?