Puzzle of the Week #394 - Coin Paradox

I’m thinking about a sequence of coin flips resulting in either heads or tails, with equal probability. Which of the following statements are true?

 

In a game where I win if we flip HTH on consecutive throws and my friend wins if we flip HTT on consecutive throws:

 

1) If we just toss the coin three times in succession, we each have an equal probability of winning (although most of the time neither of us would win).

2) If we keep flipping until either HTH or HTT comes up, we will win the game with equal probability.

3) If we keep flipping until either HTH or HTT comes up, the average number of flips it will take is on average the same, whether I win or my friend does.

4) If we each have our own coin, and I keep flipping until I see HTH on consecutive flips, and my friend keeps flipping her coin until she sees HTT on consecutive flips, we will both take the same number of flips on average.

Puzzle of the Week #393 - Two Point Oh Eight

52/25 = 2.08 precisely

572/275 = 2.08 too

 

52 and 572 are the only two- and three-digit numbers which, when divided by the reversal of their digits, is exactly equal to 2.08.

 

How many 25-digit numbers are there, that when divided by their reversal become equal to 2.08?

 

This might look like an exercise in coding, but it isn’t. In fact I devised the puzzle purely on paper, and it could be solved as such too.

 

Puzzle of the Week #391 - Five Towns 2

This is a very similar puzzle to last week’s, with a couple of minor tweaks.

 

Crosstown is on a straight line between Lefton and Righton.

Crosstown is also on a (different) straight line between Upton and Downton.

Unlike before, these two lines do NOT need to cross at right angles.

As before each of the distances between towns should be a unique whole number of miles.

However now there is the added restriction that none of the distances should be a multiple of 3 (although U-D and/or L-R could be as they are each the sum of two separate distances).

To complete the puzzle you just have to find a set of distances which satisfy all of the restrictions. For extra kudos, can you get the total length of the eight roads to be less than 400 miles?

Puzzle of the Week #390 - Five Towns

The towns of Norton, Sutton, Weston, Easton and Middleton are arranged such that the Norton-Sutton road and the Weston-Easton road cross at right angles in Middleton.

The distance from any town to any other town is unique.

Each distance is an exact whole number of miles.

What is the smallest possible total area of the diamond shape formed by the outer four towns?

Puzzle of the Week #389 - Strings

I have a number that is a string of 1s, followed by a string of 2s, followed by a string of 3s. For example 111233. There must be at least one of each digit.

Similarly I have a second number that is a string of 3s, followed by a string of 2s then a string of 1s, with again at least one of each of the different digits. For example 332211111.

Adding these two numbers together I get the answer 44443444.

How many different possibilities are there for my two numbers?

Puzzle of the Week #388 - Ten Pin Bowling

This game uses all 100 scrabble tiles, including the two blanks (which can represent any letter of your choosing).

The scoring system is the same as that of real ten-pin bowling: you get points for each word, equal to how many letters in the word. In addition, if you get a spare (use all letters in one frame using two words), you get bonus points equal to the next word you score.

If you get a strike (a ten letter word), you get bonus points equal to the next TWO words you score.

If you only get one word in a frame, and it's not a strike, then for the purposes of bonus points, you get a zero length word too.

In real tenpin bowling, if you get a strike or a spare on the tenth frame, you get an eleventh frame to determine your bonus points, and if you were lucky enough to get a strike on the tenth AND eleventh frames, you would get a twelfth frame.

In this game, there are no eleventh or twelfth frame, so to determine any bonus points you are entitled to after the tenth frame, look back at the words you scored in the first couple of frames.

The 100 scrabble letters have been assigned randomly into their frames as below (which means I can play along with everybody else). Generally a score of 100 or more using relatively common words is a pretty good benchmark, although a higher score is probably possible by poring over lists of more obscure words.

 

Puzzle of the Week #387 - Five Circles

I have five circles arranged tangent to one another as below, and the triangles formed by connecting some of their centres result in three right-angled triangles.

Can you find an arrangement where the radius of each of the five circles is a whole number?

For bonus points, what is the minimum arrangement where each of the radii is a whole number (minimum sum of the five radii)?

Puzzle of the Week #385 - Odd Prime Cycle

Start with an odd prime number

Add 1, then choose an odd prime factor of this new number,

Add 2 to that number, then choose an odd prime factor of this new number,

Add 3 to that number, then choose an odd prime factor of the new number,

Add 4 to that number, then choose an odd prime factor of the new number,

etc, until the odd prime factor you get is the odd prime you started with.

 

For instance, if you started with 5:

5+1 is divisible by 3

3+2 =5

 

Other than 5, what could the starting odd prime number have been?

Puzzle of the Week #383 - Smallest Square

What is edge length of the smallest square that can contain a shape with equal (non-zero) perimeter and area?

 

Examples of such shapes are a circle of radius 2, which has both area and perimeter equal to 4*pi, or a 6,8,10 triangle, which has both area and perimeter equal to 24. My shape might be made up of straight lines, or curves, or both.

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?