#SummerMathsPuzzles
Each weekday throughout August 2019 we will be publishing a new maths-based puzzle. They won’t require sophisticated maths to solve, but equally they won’t be easy. Discussing your ideas might help.
We hope that you’ll enjoy solving them with family, friends and colleagues and we ask that you share them either by forwarding this page, or via our social media (see below).
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The links for each puzzle are below and will become live on the relevant day:
Two players play a game.
Given that they are both perfect logicians and strategists, who wins?
A family are on holiday. The parents want to spend the morning lying peacefully on the beach. The children want to go surfing.
The parents give the children a pack of 100 red stickers and tell them that before they can go surfing they should play a game with 100 rounds and the following rules:
The parents say that they will take the children surfing when they tell them whether the 100th pebble has its sticker showing or not at the end of the game and how many times it was turned over during the game.
The children go away and a few minutes later come back and tell their parents the correct answers. What were their answers and how did they solve it?
Entry to this puzzle is now closed. Congratulations to the following for some great answers:
The parents decide that yesterday’s activity was such fun that they’re going to set the children another challenge today. They tell the children that they’ll take them sea kayaking once they’ve solved a bucket and spade (or rather just buckets) puzzle.
The children object strongly and say that they’re far too old for buckets and spades!
Undaunted, and having spotted that each of the (rather large!) buckets has its capacity marked on the bottom, the parents set the children their next challenge. The challenge is to make each whole number volume from 1 up to 12 using their three buckets which are of capacity 5 litres, 8 litres and 12 litres.
To explain what they mean, they show the children how to make 7 litres using the buckets:
They show the children that this can be written down as follows: 0-0-12 5-0-7 0-0-7
Today’s problem is to solve this latest challenge set by the parents, to figure out how to make all the whole number volumes from 1 up to 12 litres using the 3 buckets provided, so that the children get to go sea kayaking.
Entry to this puzzle is now closed. Congratulations to the following for some great answers:
Today, the parents decided to take the children to see a cricket match. They explained that in cricket, a bowling average is defined to be:
Number of runs conceded / Number of wickets taken
Obviously, the lower the score the better the bowler.
The match was between local teams the Aardvarks and the Bisons. Each side played two innings, one before and one after tea. Alice is the star bowler for the Aardvarks and Bianca is the star bowler for the Bisons.
Entry to this puzzle is now closed. Congratulations to the following for some great answers:
Special congratulations to Alihan Huyuk who was the first person to identify that this problem is an example of Simpson’s paradox.
As the kids are taking down their tent they strike up a philosophical conversation about square pegs in round holes and thence round pegs in square holes. They ask their parents the following question:
Which fits better: a square peg in a round hole or a round peg in a square hole?
Can you help the parents solve this problem? Consider a square inscribed within a circle and circle inscribed within a square as illustrated below. In each case, find the area of the shaded section as a percentage of the whole.
Entry to this puzzle is now closed. Congratulations to the following for some great answers:
IBM has been setting its challenging Ponder This puzzles every month since 1998. Up until 2005 these puzzles were set by cryptographer and mathematician Puzzle Master Don Coppersmith. This is the first part of a puzzle from that era:
A game has the following rules.
How does the Host use the white cards alone to communicate the number on the blue card to the Partner?
Entry to this puzzle is now closed. Congratulations to the following for some great answers:
This puzzle, also taken from the first part from one of IBM’s Ponder This collection, seemed particularly apt for today.
An octopus has 8 arms which we can number from 1 to 8.
We have eight octopuses named A,B,C,D,E,F,G and H. Each octopus has a watch on each of its arms set to the times shown below. These are special watches which show only the hour.
To adjust the hour on the watches, we can move all the watches of a single octopus (octopus C for example) one hour ahead, or move all the watches on a specific arm of all the octopuses (arm #3 for example) one hour ahead.
What adjustments do you need to make to fix it so that over 2/3 of the 64 watches show either 3, 6, 9 or 12?
Entry to this puzzle is now closed. Congratulations to the following for some great answers:
The oldest child, who wants to be an astronaut, remains obsessed with all things relating to Mars and this inspires the parents for today’s challenge.
The Birthday Paradox, the parents explain, famously asks what is the smallest number of people that you need in order for the probability of two of them sharing a birthday to be greater than 50%. The surprising answer to this question is 23.
[If you haven’t seen this problem before then read up about it online or watch a video tutorial – see eg https://www.youtube.com/watch?v=KtT_cgMzHx8].
The parents tell the children first to go and research how many days there are in a Martian Year and then to calculate how many Martians there would have to be at a party to make it more likely than not that two of them share a birthday?
Help the children solve this challenge. Their reward will be to go to visit the planetarium.
For fun, since you know the answer to the regular Birthday Paradox, why not try estimating the answer for the Martian Birthday Paradox before you do your calculation.
Bonus question for which we don’t know the answer: is reality consistent with the statistics? Given that 23 people is equivalent to a referee and 2 teams of 11 football players each, did at least two people share a birthday for over half the Premier League games last season?
Entry to this puzzle is now closed. Congratulations to the following for some great answers:
Five mathematicians arrived at the Euclid Institute between 9am and 10am yesterday. Given the information provided below, can you determine what time each person arrived, their mode of transport, and the name of their home institution.
Using the grid below, can you find out the order in which the mathematicians arrived, and their respective home institutions and modes of transport.
Please submit your answer in grid form as shown below.
Entry to this puzzle is now closed. Congratulations to the following for some great answers:
This morning, the middle child foolishly asked the parents their least favourite question: “Are we nearly there yet…..?”. This inspired them to set the children the following challenge.
Entry to this puzzle is now closed. Congratulations to the following for some great answers:
This morning the children were having an argument. This reminded the parents of the Monty Hall problem, infamous for having caused major arguments amongst some leading mathematicians. It goes like this:
In a game show there are three doors, two of which have a goat behind them and one of which has a major prize. The contestant chooses a door but doesn’t open it. The game show host then opens one of the other two doors to reveal a goat before asking the contestant whether they want to stick with their original choice or swap and open the other door.
So the question is, should the contestant stick with their original choice and open that door and win the prize behind it, or should they swap and open the remaining door and win the prize behind that. The (somewhat surprising) answer, probabilistically speaking, is that they should swap: they are twice as likely to win the prize if they swap than if they don’t.
Having explained the Monty Hall problem to the children, the parents asked a more generalised question: suppose there are n doors and g goats, should they still swap? What is the probability that they win if they swap? Is there a relationship between n and g that determines whether they should or shouldn’t swap?
Can you help the children answer this question?
The parents had to respond to a couple of work related emails this morning and, to buy themselves some time, they set the children the following problem:
A 2-storey office block has 8 offices, one at each corner A, B, C, D, E, F, G and H of the building. Corridors run along the edges but not diagonally across the building and each office is connected vertically by a lift (elevator) with the office directly above or below it.
Every Monday morning the janitor, carrying with him 8 pot plants, enters the office block via the front door which leads directly into Office A. He leaves a pot plant in Office A and then travels randomly from office to office via the lifts and corridors leaving a plant (or at times a second, third plant or fourth plant) in the office he has just entered.
What is the probability that he places the 8th pot plant in office G having placed left one pot plant in each of the other offices along the way?
Can you help the children solve the problem?
2019 pirates are on board the Pirate Ship Ptolomy with a treasure chest containing 10,000 gold pieces. The pirates are ranked from 1 (the most junior) to 2019 the senior pirate. The most senior pirate must propose a plan for how to divide the gold. The pirates will then vote on the plan. If the majority (or exactly half if there is an even number as the most senior pirate has the casting vote) agree then the plan is accepted and that is how the gold is divided. If the plan is not accepted then the most senior pirate is thrown overboard and it’s up to the next most senior pirate to propose their plan. The pirates, although ruthless, are perfectly rational and will abide by the rules.
How much should the Senior Pirate offer and to whom in order to secure the most gold for herself.
Hint: Consider first the case when there are two pirates 1 and 2 and use this to predict what would happen when there are 3 pirates 1, 2 and 3 and so on…
6 cards numbered from 1 up to 6 are placed at random in a straight line.
What is the probability that for each pair of neighbouring cards, one is a multiple of the other?
How many numbers from 1 up to 2019 can be expressed as
Bonus question: by considering multiples of 4 or otherwise, can you prove that no number is both the sum and product of 4 consecutive whole numbers?
A group of 5 children are discussing how they should divide up the ridiculously large amount sweets that they were given at Halloween.
p.s. Any answer is wonderful but we’d love it if you used the Chinese Remainder Theorem to solve this one 🙂
Lady Luck is about to give out the prize envelopes following the Pet Show at the Annual Village Fete.
Unfortunately, the prize envelopes have been muddled up. Lady Luck decides to go ahead and select envelopes at random (fortunately she knows that the yellow envelopes are for the tortoises, the green envelopes are for the cats, and the blue envelopes are for the dogs).
What is the probability that everyone gets the right prize?
The Director of the Euclid Institute asked her Deputy to prepare a report on the diversity of participants who spent their weekend working in the building. He produced the following report:
As he handed over the report he commented that of the 25 participants in the building over the weekend:
How many of the 25 participants in the building were UK-based female ECRs and how many were non-UK-based male non-ECRs?
The editor of a mathematics journal asks her publisher what the page limit is. The publisher replies that every page must be numbered but that she can use at most 2019 digits in total to number the pages.
What is the page limit for the volume?
Which door do you choose?
Again, which door do you choose?
Entry to this puzzle is now closed. Congratulations to the following for some great answers:
[Apologies for initial non-determinism in Q4. Now removed.]
You continue in your quest for freedom (see yesterday’s puzzle).
Which door do you choose?
This is your final choice. Which door do you choose?
Entry to this puzzle is now closed. Congratulations to the following for some great answers:
On this final day of our month of puzzles we ask you simply for the name of the shape that has finite volume but infinite surface area, and a description of how it is formed.
Entry to this puzzle is now closed. Congratulations to the following for some great answers: