# Puzzle of the Week #186 - Prime of my Life

I celebrated my birthday this week. Right now, my age, and those of my wife, my daughter and my son, are all prime numbers. After my daughter celebrates her next birthday in a few months we will not all be prime ages until 30 years from now. And after that (if we are all still around) it would be another 30 years.

My daughter is about four and a half years older than my son, and I am about one and a half years older than my wife. How old are we all?

# Puzzle of the Week #185 - How Long is a Piece of String?

I have a loop of string N centimetres long. I can form this loop into a square with n cm along each side (n being exactly a quarter of N). Not surprisingly I find that if I draw diagonals from opposite corners of this square, they cross at right angles. I also find that I can tilt the square into a rhombus shape, while still having n cm along each side, the diagonals still cross at right angles. Still not surprising.

However, I discover that if I move some of the corners so that instead of all being n cm long, the four edges of the quadrilateral are n, n-5, n+1 and n+4, then the diagonals STILL cross at right angles.

How long in my loop of string?

# Puzzle of the Week #184 - Random Code

Replace each letter in this code either by the letter preceding it or following it in the alphabet.

# Puzzle of the Week #182 - The Rickety Carousel

There is a rickety old carousel, with 24 horses equally spaced around the edge of a disc. There are at least two things wrong with it: one is that the horses in positions A, B, C, D, E, F and G are broken and not safe to ride; another is that the carousel needs to be perfectly balanced around the centre of the disc, otherwise it won’t go round. This means that when there are two children wishing to ride, they would have to be positioned in diametrically opposite positions, eg, A and M, in order that the centre of gravity is at the centre of the disc. If there were three children they could be at positions A, I and Q.

On this particular day, nine children wish to ride the carousel. Where do they need to be positioned such that the centre of gravity is at the centre of the disc, and that positions A to G are unoccupied?

# Puzzle of the Week #181 - Number Hunt: Four Digit Product

Find a four digit number ABCD such that ABCD x ABC x AB x A = 369,457,740

# Puzzle of the Week #180 - Letter Wall

Two letters of the alphabet are chosen at random. What is the probability that the bricks that correspond to those letters are adjacent in this wall:

# Puzzle of the Week #179 - Reducing the Irreducible

The fraction 1174/5063 cannot be reduced further in the usual way, however it can be expressed exactly using only six digits instead of eight.

How?

# Puzzle of the Week #178 - Tangent Curve

The line begins in a vertical downward direction, transitions into an arc of radius 60, then into another arc with a tighter radius, which just touches the horizontal dashed line. Then there is an arc of radius 50 in a clockwise direction, then a horizontal line. At each transition between line and arc or arc and arc the tangents are continuous.

What is the missing radius?

# Puzzle of the Week #177 - Skyscrapers and Empty Lots

Each of the 25 squares contains either a building of height 1, 2, 3 or 4, or an empty lot. Each row and each column contains exactly one of each of those.

The numbers in the twenty ‘vantage points’ around the edge denote how many buildings you can see looking along that particular row or column, so for instance if the 2 building was closest, followed by 1, 4 and 3 in that order, the vantage point would say 2, as you can see the 2-building and the 4-building, whereas the 1-building and 3-building are hidden behind taller buildings.

Here’s an example (with only 1, 2 and 3 height buildings) which might make it clearer, followed by the puzzle itself:

# Puzzle of the Week #176 - Nine Point Circle

You might be familiar with the famous ‘Nine Point Circle’ theorem, which says that for ANY triangle the 3 midpoints (shown below in red), the 3 perpendicular feet (shown in green), and the 3 points midway between the each of the vertices and the orthocentre (shown in blue), will all lie on a circle. (Of course for an equilateral or isosceles triangle some of these points will coincide). But I have discovered there is something quite special about the 9 points of a triangle with angles 45, 60 and 75.

What is so special?

The figure is to scale so you may be able to guess the answer.

# Puzzle of the Week #175 - Out Standing In My Field

I’m standing in a rectangular field. I am 34m from the top left corner and 62m from the bottom right corner. I am an equal distance from the other two corners: what is that equal distance?

# Puzzle of the Week #174 - Rectangle Area

The AREA of each of the eight squares is 26. What is the area of the rectangle that surrounds them?

# Puzzle of the Week #173 - Tournament

There are 16 teams in the knockout stages of a football tournament. For the sake of argument, say that each team has a distinct ranking from 1 (best) to 16 (worst), and that in any individual match, the better team will always win and progress to the next round.

The first round matches are randomly arranged, and the eight winners of those matches are randomly arranged in four matches in the quarter final etc.

Which of the following scenarios is more likely:

The 2nd ranked team is knocked out before the semi-finals, or

The 7th ranked team progresses at least as far as the semi-finals?

# Puzzle of the Week #172 - Wordwall

Reassemble this word wall using the bricks provided. Unfortunately the bricks that go in the positions marked with an asterisk are missing, and must be reconstructed.

It may actually be useful to print and cut out the pieces.

# Puzzle of the Week #171 - Isosceles Triangle

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?

# Puzzle of the Week #170 - Football Squad

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?