**#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).

**facebook**.com/newton.institute | **twitter**.com/NewtonInstitute | **instagram.com**/isaacnewtoninstitute

The links for each puzzle are below and will become live on the relevant day:

- Teaser Puzzle £8.19
Two players play a game.

- They each start with an unlimited number of coins of denominations: 1p, 5p, 10p, 20p, 50p and 100p.
- They take it in turns putting coins into a pot one at a time.
- The winner is the person who places the final coin into the pot reaching the target total of £8.19.
- A player automatically loses if they exceed the target total.

Given that they are both perfect logicians and strategists, who wins?

- Thursday 1 August - 100 Pebbles
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:

- In the first round they must collect 100 pebbles, place them in a straight line, and put a red sticker on each.
- In the 2nd round they must turn over every 2nd pebble so that the sticker is no longer showing.
- In the 3rd round they must turn over every 3rd pebble so that if the sticker was previously showing it no longer shows and vice versa.
- In the 4th round they turn over every 4th pebble, in the 5th round they turn over the every 5th pebble and so on up to the end of the 100th round.

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:**- Junyang Wang
- Andrea Chlebikova
- Aaron Naidu
- Jason Curran
- Patrick Luo
- Soham Sud
- Emily Lies & Jan Ricken
- Kanha Seksaria
- Charlotte Harries-Harris
- Jacob
- Team Cabin
- FiveTriangles
- Fin Scott
- Ian Slagle
- Akshay Sant
- Yulia and Christopher
- Bethan
- Phil Milner
- Alex Austin
- Alex Bell
- Nora Howie
- Meg North
- Ben Katz
- Oree Golan

- Friday 2 August - Buckets of Fun
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:

- Fill the 12 litre bucket
- Pour the water from the 12 litre bucket into the 5 litre bucket until the smaller bucket is full.
- The amount remaining in the 12 litre bucket is then the desired 7 litres.

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:**- Evgeny Mirkes
- Andrea Chlebikova
- Lindsay Woodhead
- Nora Howie
- Jennifer Evans
- Kanha Seksaria
- Fin Scott
- Ali Raad
- Soham Sud
- Patrick Luo
- Niall Moran
- Emily and Jan
- James Arthur
- Alihan Huyuk
- Isaac
- Rosamund Riordan
- Kalin Krishna
- Georges Sbback
- Austin family
- Aaron Naidu
- David Martí-Pete
- Hana Dampf
- Cong
- Meg North
- Ben Katz
- Oree Golan

- Monday 5 August - Over and Out
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.

- Before tea Alice got 5 wickets for 10 runs and then in the next innings Bianca got 8 wickets for 50 runs.
- After tea Alice got 10 wickets for 150 runs and then Bianca got 2 wickets for 50 runs.

- Who got the better bowling average before tea?
- Who got the better bowling average after tea?
- Who got the better bowling average across the whole match?
- Who is the better bowler?

**Entry to this puzzle is now closed. Congratulations to the following for some great answers:**- Ali Raad
- Veera
- Evgeny Mirkes
- Akshay Sant
- Krishi
- Niall and Jenny
- Sarisha
- Charlotte
- Jan and Emily
- Fin
- Soham Sud
- Meg North
- Ben Katz

*Special congratulations to***Alihan Huyuk***who was the first person to identify that this problem is an example of Simpson’s paradox.*- Tuesday 6 August - Square Pegs in Round Holes
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:**- Niall and Jenny
- Ali Raad
- Veera
- Krishi
- Sarisha
- Evgeny Mirkes
- Nora Howie
- Georges Sabback
- Patrick Luo
- Austin and Knobloch families
- Emily Lies, Jan Ricken and Tobias Beyer
- Soham Sud
- Sittichoke Som-am
- Lindsay Woodhead
- Meg North
- Ben Katz

- Wednesday 7 August - Encoding Game
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.

- There are three participants: a Host, a Partner, and a Volunteer.
- The Partner is in a soundproof room.
- The Host gives the Volunteer six blank cards, five white and one blue.
- The Volunteer writes a different integer from 1 to 125 on each card, as the Host is watching.
- The Volunteer keeps the blue card.
- The Host arranges the five white cards in some order and passes them to the Partner.
- The Partner then announces the number on the blue card.

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:**- Cerulean O’Toole
- Ben Xiao
- Ben Katz
- Lindsay Woodhead
- Soham Sud
- Richard Hibbs
- Iam Paczek

- Thursday 8 August - 8 Octopuses
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.

- Octopus A’s watches on its 8 arms numbered 1..8 are set to 1, 2, 3, 4, 5, 6, 7 and 8 respectively.
- Octopus B’s watches on its 8 arms numbered 1..8 are set to 2, 4, 6, 8, 10, 12, 2 and 4 respectively.
- October C’s watches on its 8 arms numbered 1..8 are set to 3, 6, 9, 12, 3, 6, 9 and 12 respectively.
- Octopus D’s watches on its 8 arms numbered 1..8 are set to 4, 8, 12, 4, 8, 12, 4, 8 and 12 respectively.
- Octopus E’s watches on its 8 arms numbered 1..8 are set to 5, 10, 3, 8, 1, 6, 11 and 4 respectively.
- …and so on….

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:**- Emily Lies
- Jan Ricken and Tobias Beyer
- Cerulean O’Toole
- Fin and Rebecca Scott
- Lindsay Woodhead
- Meg North
- Patrick Luo
- Ben Katz

- Monday 12 August - Martian Birthday Paradox
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:**- Evgeny Mirkes
- Fin Scott
- Lindsay Woodhead
- Nora Howie
- Krishi (Kanha Seksaria)
- Soham Sud
- Ali Raad
- Georges Sabback
- Emily Lies and Jan Ricken
- Akshay Sant
- Niall Moran and Jenny McSharry
- Patrick Luo
- Ben Katz

- Tuesday 13 August - Logical Logistics
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.

- Of the cyclist and the person who arrived at 9:10am, one is from the University of St Andrews and the other is Professor Gowse.
- The person who arrived by car is either the person from the University of Auckland or the person who arrived at 9:50am.
- Professor Lovelylaces isn’t from the University of St Andrews.
- The person from Kyoto University arrived 20 minutes after Dr Livelyknees.
- The person on a scooter arrived 30 minutes after Professor Gowse.
- The person from Heidelberg University arrived 10 minutes before the person in the taxi.
- Dr Livelyknees was either the person from the University of Auckland or the person who arrived in a taxi.
- Prof du Chattels arrived at 9:30am.

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:**- Lindsay Woodhead
- Isaac
- Emily Lies and Jan Ricken
- Junyang Wang
- Fin Scott
- Akshay Sant
- Soham Sud
- Meg North
- Krishi
- Niall Moran and Jenny McSharry
- Evgeny Mirkes

- Wednesday 14 August - Distance Dilemma
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.

- AA Architects and BB Bank lie at opposite ends of the same (straight) road.
- Archie works at AA Architects and Bianca works at BB Bank.
- At 1pm Archie leaves AA Architects to drop off a letter at BB Bank and at the same time Bianca sets off from BB Bank to drop off a letter at AA Architects.
- They meet for the first time at a distance
*a*metres from AA Architects and for the second time at a distance*b*metres from BB Bank. - Assuming that they each walk at a constant speed, how far, in terms of
*a*and*b*, is AA Architects from BB Bank?

**Entry to this puzzle is now closed. Congratulations to the following for some great answers:**- Junyang Wang
- Ali Raad
- Niall Moran and Jenny McSharry
- Benjamin Dadoun
- Emily Lies and Jan Ricken
- Soham Sud
- Chengyuan Li
- Evgeny Mirkes
- Patrick Luo
- Ben Katz
- Ben Xiao

- Thursday 15 August - A Prize and 2 Goats
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?

- Friday 16 August - Office Ambles
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?

- Monday 19 August - Pirate Payouts
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…

- Tuesday 20 August - adnRmo Ordering
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?

- Wednesday 21 August - Consecutive Numbers
How many numbers from 1 up to 2019 can be expressed as

- The sum of 4 consecutive whole numbers?
- The product of 4 consecutive whole numbers?

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?

- Thursday 22 August - Sweet Enough
A group of 5 children are discussing how they should divide up the ridiculously large amount sweets that they were given at Halloween.

- One of the children suggests that they could share them equally between themselves but then there would be 3 left over.
- Another of the children suggests that they could share them equally with the rest of their football team (ie between 11 children) but they realise that there would still be 6 sweets left over.
- Observing that there were almost as many sweets as there are days in the year, their parents decided that they should share them equally between all of their class (ie 31 children) and give the remaining 17 sweets to their teacher.

- How many sweets were there altogether?
- In the end, how many sweets did they each get?

p.s. Any answer is wonderful but we’d love it if you used the Chinese Remainder Theorem to solve this one 🙂

- Friday 23 August - Pot Luck Pet Prizes
Lady Luck is about to give out the prize envelopes following the Pet Show at the Annual Village Fete.

- In the Tortoise Class, 1 of the 3 tortoises was awarded Highly Commended and the others were awarded Commended.
- In the Cat Class, 2 of the 4 cats were awarded Highly Commended and the others were awarded Commended.
- And in the Dog Class, 3 of the 5 dogs were awarded Highly Commended and the others were awarded Commended.

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?

- Monday 26 August - Reporting Obfuscation
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:

- There were 6 UK-based participants;
- 12 participants were female;
- 15 participants were early career researchers (ECRs).

As he handed over the report he commented that of the 25 participants in the building over the weekend:

- Only 1 was UK based but neither female nor an ECR;
- There were 4 female ECRs, and 2 UK-based male ECRs;
- 6 of the particpants were female but neither UK-based nor an ECR.

How many of the 25 participants in the building were UK-based female ECRs and how many were non-UK-based male non-ECRs?

- Tuesday 27 August - Pagination Poser
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?

- Wednesday 28 August - Doors to your Destiny
- You are imprisoned in a land where analytical reasoning is prized more highly than guilt or innocence. The prison guard brings you to 2 green doors A and B with signs on them and truthfully tells you that for all pairs of green doors you may encounter, the sign on one door is true and the sign on the other is false.

- Having successfully avoided death in the first challenge you proceed along the corridor only to find yourself facing a pair of blue doors. The guard truthfully tells you that for blue doors, the signs are either both true or both false.

Which door do you choose?

- Having avoided death once more you proceed along the corridor only to find yourself facing another pair of blue doors.

Again, which door do you choose?

**Entry to this puzzle is now closed. Congratulations to the following for some great answers:**- Patrick Luo
- Krishi
- Alex Bell
- Richard Hibbs
- Mella Gallagher
- Evgeny Mirkes
- Ali Raad
- Emily Lies and Jan Ricken
- Manuja Isaac
- The Koenig Family
- Soham Sud

- Thursday 29 August - More Doors to your Destiny
*[Apologies for initial non-determinism in Q4. Now removed.]*You continue in your quest for freedom (see yesterday’s puzzle).

- You next encounter 3 yellow doors. The guard tells you that for yellow doors, one door will always lead you on the path to freedom and the other two will lead you to your death. He adds that at most one of the signs is true.

Which door do you choose?

- Having successfully avoided once more death you proceed along the corridor only to find yourself facing a set of 3 red doors. The guard tells you that one takes you to your freedom, the second to your death, and the third to eternal imprisonment. He adds that the sign on the door on the path to freedom is true, the sign on the door that leads you to your death is false, and the sign on the door to eternal imprisonment may be true or false.

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:**- Patrick Luo
- Richard Hibbs
- Krishi Isaac
- Emily Lies and Jan Ricken
- Fin Scott
- Evgeny Mirkes
- Ali Raad
- Manuja
- Soham Sud

- Friday 30 August - Finitely Infinite or Infinitely Finite?
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:**- Krishi
- Emily Lies and Jan Ricken
- Richard Hibbs
- Evgeny Mirkes
- Ali Raad
- Linsday Woodhead
- Danielle Amouzou-Akue
- Patrick Luo
- The Koenig Family