Numpty >> mercredi janvier 4 - 18:48

This is a separate thread for solutions to the Brain Teasers to avoid posting spoilers in the main thread.  

https://rockingsoccer.com/en/soccer/forum/home-en/1?topic=87884

Numpty >> mercredi janvier 4 - 18:49, Modifié mercredi janvier 4 - 18:50

#1. "Make 25"

Using the numbers 2 4 6 and 8 (only once each) and any of the mathematical symbols + (plus) - (minus) * (times) / (divide) and left/right brackets (), produce an expression that equals 25. 

Solution

(6 + 2/8) * 4 =  25

Numpty >> mercredi janvier 4 - 18:52, Modifié vendredi janvier 6 - 10:16

#2. "Raffle tickets"

Siege has ticket number 1, Pii has number 2, Taavi has 6, Onuelver has 55, Stockfish has 101, David has 1006, Alex Seymour has 1060. What ticket number does Vincent have? 

Solution

The number of each raffle ticket is calculated as the sum of all the roman numerals in each name, ignoring any other letters. 

So Vincent = 5 + 1 + 100 = 106

The first correct answer was by @onuelver

Numpty >> mercredi janvier 4 - 18:55, Modifié mercredi janvier 4 - 19:41

#3. The Most Intelligent Princess

A king wants his daughter to marry the smartest of all the extremely intelligent young princes in the kingdom. The 4 very best candidates are shorlisted:

• Siege
• Awful
• dj_speed
• You

[Yes I know, but every puzzle has it's flaws, lol.] 

The king's wise men devise the following intelligence test. 

The 4 princes are to be gathered into a room and seated, facing one another, and are shown 4 blue hats and 3 yellow hats. They are blindfolded, and 1 hat is placed on each of their heads, with the remaining hats hidden in a different room.

The king tells them that the first prince to deduce the colour of his hat without removing it or looking at it will marry his daughter. A wrong guess will mean death. The blindfolds are then removed. Each prince must provide a written answer with either of the two colours or blank if they don't wish to answer. This avoids giving any information to the other 3 candidates.

You know that your competitors are very intelligent and want nothing more than to marry the princess. You also know that the king is a man of his word, and he has said that the test is a fair and equal test of intelligence and bravery.

But there's a twist. 

The princess is seriously unimpressed by all 4 candidates and doesn't wish to be the prize in such a competition. In fact, she finds them so disgusting that she decides to interfere in the test in such a way that all 4 princes will give the wrong answer and will be put to death. 

Unbeknown to the king and the princes she manages to switch the hats. 

What colour hats are the 4 princes wearing when the blindfolds are removed?

Solution

The key to solving this is in the sentence:

"the king is a man of his word, and he has said that the test is a fair and equal test of intelligence and bravery"

Consider the following. In the event the test was conducted fairly as promised:

•  3 yellows and one blue hat

It's clearly not a fair test if the princes were wearing 3 yellow and 1 blue hat, because any prince wearing the blue hat would immediately see 3 yellow hats and know he must be wearing blue. 

But this prince would smell a rat because he would realise it wasn't a fair test. He would be wise not to answer, as he might legitimately wonder whether it was in fact a test of integrity as well as intelligence.  

Whereas the other 3 princes would see 2 yellows and a blue hat and for the same reason know that using 3 yellow hats wouldn't be a fair test so they must be wearing blue.  

Following similar logic, and avoiding a very lengthy explanation, there are only 2 ways of conducting a fair and equal test. 

• 2 yellow hats and 2 blue hats

This would be a fair test. 2 princes would see 2 yellow hats and 1 blue hat, and the other 2 would see 1 yellow hat and 2 blue hats. Those seeing 2 yellow hats would answer blue and those seeing 2 blue hats would answer yellow. 

• 4 blue hats

This would be a fair test. They all see 3 blue hats and the correct answer in each case would of course be blue. 

Now for the twist

The princess needs to do 2 things with her selection of hats:

• the princes have to believe that the test is equal and fair or they might refuse to answer
• they all need to give the wrong answer

The way to do this is for the princess to give the princes 3 blue hats and 1 yellow hat. 

• One prince, wearing yellow, would see 3 blue hats and since he knows it's a fair test, would reason that they must all be wearing blue hats so would answer blue. 
• The other 3 princes, wearing blue, would see 2 blue hats and 1 yellow and since they know it's a fair test would reason there must be 2 yellow and 2 blue hats in use. So they would all answer yellow.  

Numpty >> mercredi janvier 4 - 20:04, Modifié mercredi janvier 4 - 20:06

#4. 100 Gold Hryvnia's 

Five mercenaries have obtained 100 gold hryvnia coins and have to divvy up the loot. The mercenaries are all intelligent, treacherous, violent and greedy.

The commander proposes a distribution of the loot and all of the mercenaries vote on the proposal. If at least half of them agree the loot is divided as proposed, as no-one would be willing to take on the commander without superior numbers on their side.

If the commander fails to obtain the support of at least half his men (which includes himself), he faces a mutiny, and they will turn against him and kill him. The remainder start again with the next highest in rank taking over command.

What is the maximum number of gold coins the commander can keep without risking his life?

Solution

The correct way to reason this is to work out what would happen in the event there were only 2 mercenaries left and then add 1 more in each time. That allows them all to look ahead and know what would happen in the following rounds. 

For convenience let's call them M5, M4, M3, M2 and M1 in order of seniority. 

The 3 potential future rounds would work out like so:

1. If M1 and M2 are left in the final round, then M2 simply proposes that he keeps all 100 coins and since M1 can't outvote him he takes the lot and M1 gets nothing. Zero.
2. In the penultimate round M3 would propose giving 1 coin to M1 and keeping 99 for himself.  M1 would take 1 coin in preference to getting nothing in the final round. In this round M3 gets 99, M2 gets 0 and M1 gets 1. 
3. M4 proposes giving 1 coin to M2 and keeping 99 for himself. M2 would vote for that because he knows he would get zero in the next round. The vote is 2 for, 2 against - but that's enough. M4 gets 99, M3 gets 0, M2 gets 1 and M1 gets 0. 

So now the commander, M5, knows what would happen in the next round. And so do all the others. 

He proposes that he keeps 98 coins and gives 1 coin to M3 and 1 coin to M1. The vote is carried 3-2.

Balthazor >> mercredi janvier 4 - 20:41

But what would solve this test with 4 blue hats if king said it previously that all have equally distributed? Nothing. It would be like "police academy test" in several jokes. All princes were extremely intelligent and would give right answer. People with not English as first language would that key sentence interprete quite differently like "it tests equally intelligence and bravery" - yep, at level zero because you know answer before.

Numpty >> mercredi janvier 4 - 21:04, Modifié mercredi janvier 4 - 21:41

Yes, they should all give the correct answer. But the first one to submit the correct answer would win.

"The king tells them that the first prince to deduce the colour ... "

As for the second part, you make a fair point. There is a slight ambiguity in the way in which it's written. It could perhaps have been worded a little better.

The meaning was that the test was meant to be fair and equal between all 4 competitors - and what was being tested was intelligence and bravery.

But the idea that somehow intelligence and bravery could be tested equally seems to be bordering on the philosophical. It would be difficult to imagine a test where these could be treated equally - and certainly not with 4 hats. At least for me that removes the possibility of any ambiguity in the wording. 

Numpty >> jeudi janvier 5 - 10:33, Modifié jeudi janvier 5 - 10:34

#5. Going round in circles or a wild goose chase?

Does this rather strange looking series have a simple solution or will you be chasing it forever? If so, what's the answer?

• 1 - 1/3 + 1/5 - 1/7 + 1/9 - 1/11 + 1/13 - 1/15 ...

Hint: Multiplying by 4 might make it a little easier

Solution

If you multiply the sum of this infinite series by 4 you get π.  

So the answer is π/4.

But the series converges very slowly which makes it a little difficult to guess that's where it's going. However, since it's an alternating series you can make an estimate halfway between the peaks and troughs, and this would give you 3.14... after the first 10 terms and 3.142...  after about 20 terms. 

This is known as the Liebniz formula for π. Mathematical proof can be found here:

https://en.wikipedia.org/wiki/Leibniz_formula_for_π

Numpty >> vendredi janvier 6 - 10:59, Modifié vendredi janvier 6 - 11:49

#6. On The Run

A column of soldiers 4 miles long marches at a constant speed for 3 miles. At the same time a runner runs at a constant speed from the back of the column to the front and back again to the back of the column.

How far does the runner run?

How fast is the running speed in comparison to the marching speed?

Bonus questions 

1. Consider the effect of changing the numbers for both the column length and the marching distance.

Do they both matter, and if so, which matters most or do they contribute equally to the total running distance. 

2. Is it theoretically possible for the runner to achieve this with any length of column and marching distance and still be able to run at a constant speed, assuming there are no speed limitations on the runner. Or can the column length be too long in relation to the distance being marched?

Solution

If d is the total distance that is run and the total time taken is t, then the running speed is d/t and the marching speed is 3/t. The relative running speeds determine how quickly the 4 mile gap between the front and rear of the column can be closed. 

• The relative running speed while running to the front = d/t - 3/t
• The relative running speed while returning to the back = d/t + 3/t

• Time taken to reach the front = 4 / [d/t - 3/t]
• Time taken to reach the back = 4 / [d/t + 3/t] 

Total time t = time to reach the front + time to reach the back 

t = 4 / [d/t - 3/t] + 4 / [d/t + 3/t]

Solving this for d gives us d = 9, so the solution is 9 miles. 

Since the runner covers 9 miles during the same time that the soldiers march 3 milles then he is running at 3 times their marching speed. It doesn't matter how fast they are marching or how long they take, he just has to run 3 times faster.

So the relative running speeds are twice as fast on the way to the front and 4 times faster on the return. This means he takes 2/3 of the time to reach the front during which they have marched 2 miles and he has run 6 miles. He takes the remaining 1/3 of the time to return to the back, after which they have marched one mile further and so he runs 3 miles while closing the gap making 9 miles in total. 

Solution to Bonus questions

Let's call the column length L, the marching distance M, and the total running distance D.  Using the same method as above we get a general solution where: 

• D = L + √[L^2 + M^2]

1. We can see from the general solution that the column length L is what matters most. It doesn't necessarily contribute more if it's a very short column, but always contributes unequally.

We can easily see this with a few examples, if they don't move then M = 0 and the running distance is simply 2 times the length of the column. If the column length is very short and the marching distance is long then the running distance is simply the marching distance plus a few metres. 

If we reverse the numbers in the main question, then the running distance becomes 8 miles rather than 9 miles. So the column length has the greater effect.

2. We can see from the general solution that any values of L and M are able to give a value for D. There are no impossible solutions. 

Numpty >> vendredi janvier 6 - 12:44

#7. What's The Number?

What number should replace the question mark?

• 204 981 864 416 636 1010?

Solution

Each number is made up of two distinct values. A value followed by it's square.  2^2=4, 9^2=81 and so on. 

10^2 = 100, so the final digit is a 0. 

First correct answer was from @Kupus

Numpty >> vendredi janvier 6 - 14:41, Modifié vendredi janvier 6 - 15:06

#8. The Washing Machine 

The five mercenaries in question #4 are joined by another more senior officer, called Siege, making 6 of them.  

They are told that they are going to have to retreat but they are given leave to go on the rampage and see what else they can pillage and plunder before leaving. They are very angry and frustrated when they return with just one washing machine.  

The same rules apply as before, 

Siege proposes a distribution of the loot and all of the mercenaries vote on the proposal. If at least half of them agree the loot is divided as proposed.. 

If he fails to obtain the support of at least half his men (which includes himself), he faces a mutiny, and they will turn against him and kill him. The remainder start again with the next highest in rank taking over command.

They are so angry, they now value in priority order: 

1. Their lives
2. The washing machine
3. Killing the other mercenaries

So if given the choice between two equal outcomes, in which they either get the loot or they don't, they'd choose the outcome where they get to see more of the others die.

But there are 6 of them and just one washing machine.

Can Siege save his skin, and if so, how?

Solution

We now have Siege and 5 others called M5, M4, M3, M2 and M1 in order of seniority. Following the same method as #4, the 4 potential future rounds would work out like so:

1. If M1 and M2 are left in the final round, then M2 simply proposes that he keeps the w/m and M1 gets nothing. 
2. In the penultimate round M3 proposes giving the w/m to M1 as this is the only way to save himself. M1 gets the w/m, M2 and M3 get nothing.
3. M4 has a choice, he can propose M2 or M3 gets the w/m and either option will save him. M2 and M3 have a 50% chance of the w/m, M1 gets nothing and M4 gets nothing. 
4. M5 has no good choices. He can only improve the outcome for 1 of the others so he is guaranteed to be outvoted and faces certain death. Whatever he proposes loses and it will go to the next round.  

Surprisingly then Siege has a choice. M5 can't let it go the next round, so he will vote for any proposal Siege makes because it's the only way to ensure his survival.  

Siege can propose that either M1 or M4 gets the w/m as they will get nothing if it goes 2 more rounds. He can even propose M2 or M3 gets it as a certain outcome is better than a 50% chance in 2 rounds time. 

Numpty >> vendredi janvier 6 - 19:04, Modifié vendredi janvier 6 - 19:15

#9. The Monkey and the Coconut

Ten managers are punished with a forum ban and marooned on a deserted island to serve their sentence. There they find lots of coconuts and a resident monkey, who is feisty and aggressive. During their first day they gather coconuts and put them all in a community pile. After working all day they decide to sleep and divide them into ten equal piles the next morning.

That night one castaway wakes up hungry and decides to take his share early. After dividing up the coconuts he finds he is one coconut short of ten equal piles. He also notices the monkey holding one more coconut. So he tries to take the monkey's coconut to have a total evenly divisible by 10. However when he tries to take it the monkey conks him on the head with it and kills him.

Later another castaway wakes up hungry and decides to take his share early. On the way to the coconuts he finds the body of the first castaway, which pleases him because he will now be entitled to 1/9 of the total pile. After dividing them up into nine piles he is again one coconut short and tries to take the monkey's slightly bloodied coconut. The monkey conks the second man on the head and kills him.

One by one each of the remaining castaways goes through the same process, until the 10th person to wake up gets the entire pile for himself. What is the smallest number of possible coconuts in the pile, not counting the monkey's?

Solution

I'm going to start by borrowing the monkey's coconut and add it to the pile. Okay, he wouldn't let me have it, so I'll just pretend to borrow it.

That means that each pile can now be shared equally. So the total number of coconuts is a multiple of 10, 9, 8, 7, 6, 5, 4, 3, and 2. We need to find the smallest number and this is called the lowest common multiple.

We find this by multiplying together all the factors that make up the above numbers from 2 to 10, ignoring any duplicates. 

LCM = 2 x 2 x 2 x 3 x 3 x 5 x 7 = 2,520

Now we subtract the coconut we borrowed from the monkey, so the smallest number of possible coconuts in the pile is 2,519.

Numpty >> lundi janvier 9 - 00:11, Modifié lundi janvier 9 - 00:18

#10. The Rocking Soccer Prison

The jailer, Vincent, meets with 25 new managers - also known as prisoners - when they arrive. He tells them, "You may meet today and plan a strategy. But after today, you will be in isolated cells and will have no communication with one another."

"Hidden somewhere in the game is a secret switch room, which contains two light switches labeled A and B, each of which can be in either the on or the off position. I am not telling you their present positions. The switches are not connected to anything."

"After today, from time to time, whenever I feel so inclined, I will select one prisoner at random and escort him to the switch room. This prisoner will select one of the two switches and reverse its position. He must flip one switch when he visits the switch room, and may only flip one of the switches. Then he'll be led back to his cell."

"No one else will be allowed to alter the switches until I lead the next prisoner into the switch room. I'm going to choose prisoners at random. I may choose the same manager three times in a row, or I may jump around and come back. I will not touch the switches, if I wanted you dead you would already be dead."

"Given enough time, everyone will eventually visit the switch room lots of times and over a long duration the visits will be shared out evenly. At any time, anyone may declare to me, 'We have all visited the secret switch room.'"

"If it is true, then you will all be set free and may leave the game. If it is false, and somebody has not yet visited the switch room, you will all die horribly. You will be carefully monitored, and any attempt to break any of these rules will result in instant death to all of you."

What is the strategy they come up with so that they can be free?

Solution

They agree to use switch A as the 'counting switch' for counting the prisoner visits and switch B as a dummy switch which will be used for every visit when it's not necessary to flip switch A. So switch B is simply used to satisfy the rule that one switch must always be flipped and it's position is otherwise ignored. 

The group choose a leader to do the counting, and he is the only one who will eventually announce that everyone has visited the secret switch room.

All the other prisoners, except for the counter, will turn switch A 'on' at their first opportunity to do so. They can only do this if they find switch A in the 'off' position, if they find it already 'on' they flip switch B instead and wait for a future opportunity to flip switch A. 

The leader/counter, is the only person who can turn switch A 'off'. 

He counts how many times he has flipped switch A into the 'off' position and when he has counted 24 he knows that there are 2 possibilities. Either (1.) the switch started in the 'on' position and he had to turn it 'off' before starting the count and then he subsequently counted 23 prisoner visits, or (2.) the switch started in the 'off' position and there have been 24 different visitors. The difficulty is that he doesn't know for sure that he has counted 24 other visitors. 

So they decide to count everyone twice each. This means that all the other prisoners turn switch A 'on' at both their first opportunity to do so - and also at their second opportunity. Once they have turned switch A 'on' twice then on every subsequent visit they will always flip switch B instead. 

Once the leader has counted to 48 then he knows that they have all visited the secret switch room at least once.  Either (1.) the switch started in the 'on' position and he counted 23 prisoners who had visited at least twice each plus one who had only turned it 'on' once, or (2.) the switch started in the 'off' position and all 24 of them have visited at least twice. 

He can now announce 'We have all visited the secret switch room.'