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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?
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?
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.
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)?
I have a right cone, whose base diameter is equal to its height. Its surface area exceeds its volume, and the difference between surface area and volume is at a maximum. What is the surface area?
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?
a, b, c, d are positive integers.
The sum of products of triples is equal to the sum of products of pairs:
abc+abd+acd+bcd = ab+ac+ad+bc+bd+cd = X
What is the value of X?
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.
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.
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
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?
I start with the number 74.
I append just enough 3s to the start so that the new number 3…374 is divisible by 74.
Finally I append just enough 1s to the start so that the new number 1…13…374 is divisible by the previous number 3…374.
How many digits do I now have altogether?
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)?
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?
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?
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!
1/a + 1/b = 5/391
a and b are positive whole numbers, what are they?
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!
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?
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!
