# Logic contest (Day 2)

Lets continue the logic contest. New rules:

1. Marking scheme for $i^{th}$ question is $3^{i^{log 1}}$.

2. Now the time limit for each question is $4$ hours.

3. Anybody can post a question if the previous question is answered and within 15 minutes, next question is not posted

4. The Q16 of day 1 is labelled as Q19 of today.

Best of luck

Prince: 18 marks

Sharky: 15 marks

AA: 6 marks

Kaustubh:6 marks

Jesse: 3 marks

Agnishom: 3 marks

Hung Woei Neoh: 3 marks

Note by Prince Loomba
4 years, 4 months ago

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Problem 29 Quacky found a paper in which product of two non-zero digits( 1 to 8 ) is given. Similarly Mario got sum of the same two numbers and then they are having a conversation. 1. Quacky says "I don't know the numbers". Mario says "I don't know the numbers". 2. Quacky says "I don't know the numbers". Mario says "I don't know the numbers". 3. Quacky says "I don't know the numbers". Mario says "I know the numbers". What are the numbers?

- 4 years, 4 months ago

The numbers are either $(3, 4)$ or $(2, 6)$. This is quite easy to break down. The proof is rather exhaustive but I'll type it up.

We originally have the following pairs:

[1,1], [1,2], [1,3], [1,4], [1,5], [1,6], [1,7], [1,8],

[2,2], [2,3], [2,4], [2,5], [2,6], [2,7], [2,8],

[3,3], [3,4], [3,5], [3,6], [3,7], [3,8],

[4,4], [4,5], [4,6], [4,7], [4,8],

[5,5], [5,6], [5,7], [5,8],

[6,6], [6,7], [6,8],

[7,7], [7,8],

[8, 8]

Since Quacky doesn't know the numbers, it means that the numbers must be among the combinations which have the same product. With that, we can eliminate a lot of cases to get:

[1,4], [2,2]

[1,6], [2,3]

[1,8], [2,4]

[2,6], [3,4]

[2,8], [4,4]

[3,8], [4,6]

Since Mario doesn't know the numbers, it means that the numbers among the available combinations must have the same sum. This process repeats again between Quacky and Mario again, and the Quacky only to get the final 2 possible combinations as [2,6] and [3,4].

- 4 years, 4 months ago

Right. Post the next one too

- 4 years, 4 months ago

Problem 23

There are 2 prisoners who have both been sentenced to life in prison. They are each going to be placed in solitary confinement (separate cells: they cannot communicate or see each other at all).

There are 100 guards working at this prison. Every day, each cell is guarded by one of the guards. If possible, both cells must be guarded. Also, a guard would prefer not to guard the same cell on consecutive days.

Lucky for the prisoners, there is a way for them to be released from prison.

Each guard has an astrological sign (there are 12 astrological signs). If either prisoner can successfully guess the sign of the guard that is currently guarding them, then both will be freed. If either guesses wrong, they both die.

Every single day, the prisoners can do one of three things: 1) Ask the guard what his sign is. If the prisoner does this, the guard will tell him, but will also become very offended and will never guard this prisoner's cell again. 2) Guess the guard's astrological sign. If the prisoner guesses correct, they both go free. Otherwise, they both die. 3) Do nothing.

As is usual with these kinds of problems, the prisoners have time beforehand to discuss their strategy. Can you think of a strategy that the prisoners can use to guarantee their freedom?

Clarifications 1) The priority of rules for the guards is as follows (first being highest priority): ----If a guard has been asked his sign by a prisoner, he will never guard that prisoner's cell again.

----If there is no other guard available to guard a cell, he will guard it.

----If the guard has guarded that cell the day before, he will not guard it today.

2) It is possible for a prisoner's cell to be completely unguarded on a day. For example, if the prisoner has talked to all 100 guards, then there will be no more guards left who are willing to guard him. This also means that the prisoner can no longer guess because there are no guards available to listen to him.

3) A prisoner is only allowed to guess the sign of the guard that is currently guarding him.

- 4 years, 4 months ago

Both prisoners assign a value for each sign between 1 and 12. Now prisoner one eliminated 99 guards, sums up the values of each sign and divides it by 12. Remembering the remainder as R1. If R1 is 0 he will instead remember 12. Now prisoner one sees the same guard one everyday. Prisoner two will never again see guard one until prisoner one chooses.

Prisoner two wait 99 days from the beginning and begins his eliminating 98 guards, with summing and dividing the same to get R2. Prisoner one and prisoner two will now each see their guards each day, guard one and guard two respectively.

Prisoner one having waited 98 days after his purging, now waits R1 days and finally eliminates guard one. Prisoner two does nothing now except count the days. (Note: that prisoner two will be guaranteed to see guard two at least once in this time.) Prisoner two will see guard one the day after, according to the priority rules. Having seen guard two and being able to tell guard one and guard two apart, he will have figured out R1. Since he counted the days since his last purging stopped. Since S1 is the modulo sum of every guard's sign except guard one, and S2 is the modulo sum sign of every guard's sign except guard one and guard two, their difference is the number corresponding to guard two's sign.

With this info prisoner two can correctly guess guard two's sign the following day.

- 4 years, 4 months ago

Great strategy. And of course great solution.

- 4 years, 4 months ago

can the prisoners tell when the other has left?

- 4 years, 4 months ago

Because if they can, the first prisoner can ask 99 guards. The one remaining guard switches between cells. The first prisoner figures out the guard's symbol by process of elimination. The second prisoner figures out the guard's symbol and leaves. They are both free.

- 4 years, 4 months ago

Wrong. This will not be the right answer

- 4 years, 4 months ago

Problem 24

A man walking on the beach finds two girls. He says, "You two must be twins" because they where almost identical. "Nope" they both replied. "In fact, one of us is 6 years older than the other. Can you guess which of us is the older?" He then asks them one question. "Have either of you ever traveled out of this country?" Whereupon they both say "No." He was able then to very easily tell who was the older of the two. How did he know?

Edit: Suppose it was a random question which was not discussed by the sisters earlier

- 4 years, 4 months ago

That means A is older.

The older person knows that the younger person never went out, so he could answer the question at first. The younger person was sure only after he knew that the older person did not go out.

That feels weird: Have those two brothers never talked to each other before?

- 4 years, 4 months ago

This is the right answer. They are sisters...

- 4 years, 4 months ago

Problem 25

Probably an overtold problem. You have an analytical balance that gives you the exact mass of the things in the pan.

There are 10 stacks of 10 coins available to you numbered from 1 to 10. All of the coins in all stacks are identical. However, there is a stack which contains counterfeit coins only. The counterfeit coins are known to be lighter.

What is the most efficient algorithm (minimum number of weighings) that correctly identifies the counterfiet stack?

- 4 years, 4 months ago

1 weighing. Take 1 from stack 1, 2 from stack 2,... 10 from stack 10. If the weight is lighter by suppose 8 grams, than the expected one then stack 8 would be damaged. I am supposing the difference to be 1 gram

- 4 years, 4 months ago

Haha this was my next problem

- 4 years, 4 months ago

Problem 27

This word for a living thing is the both the first and the last (at least in some books)?

Clarification: First in some and last in some or both in some

Hint for one of the books

I saw an unusual book:

The foreword comes after the epilogue; The end is in the first half of the book; The index comes before the introduction. Name the book.

- 4 years, 4 months ago

The book is a dictionary. In the dictionary, the word 'foreword' is indexed after 'epilogue' and 'index' comes before 'introduction'. Also, the word 'end', in most dictionaries (Some dictionaries with volumes might be different) appears in the first half of it.

Apparently, the word is zygote (from Prince Loomba), which I disagree with. This word is not the very last word in my dictionary. I have appendices in mine and also, my dictionary has the word zymurgy. Also, the other book is biology (still don't understand how), and zygote is the stage when lifeforms are created (just after fertilisation of an egg). I would have thought, even with biology, then the word is cyanobacteria to describe in such an aspect since life originated from cyanobacteria.

- 4 years, 4 months ago

I think you're right. But of course justify why to be complete anyway.

- 4 years, 4 months ago

@A A Better?

- 4 years, 4 months ago

Of course. That's what I thought you had in mind when you said "dictionary" haha.

- 4 years, 4 months ago

Dictionary is right but later answer is wrong.

- 4 years, 4 months ago

Porblem 28

A 120 wire cable has been laid firmly underground between two telephone exchanges located 10km apart.

Unfortunately after the cable was laid it was discovered to be the wrong type, the problem is the individual wires are not labeled. There is no visual way of knowing which wire is which and thus connections at either end is not immediately possible.

You are a trainee technician and your boss has asked you to identify and label the wires at both ends without ripping it all up. You have no transport and only a battery and light bulb to test continuity. You do have tape and pen for labeling the wires.

What is the shortest distance in kilometers you will need to walk to correctly identify and label each wire?

- 4 years, 4 months ago

20 km Is it correct?

- 4 years, 4 months ago

Yes, it is correct.

- 4 years, 4 months ago

At 1 end, connect wires in group of 1, 2, 3… 15 Now, walk to the other end(10 Km), and using the battery and bulb, identify the groups 1 to 15. Then number the wires as : wire 1 in group 1, wire 2 and 3 in group 2, wire 4, 5 and 6 in group 3 and so on… Then connect the first wire from each group to wire 1, second wire from each group to wire 2, third wire from each group to wire 3 and so on… Travel back to the first end(another 10 Km) and no. each of the wires as follows: no. group 1 wire as 1. Then from each group, identify the wire connected to wire 1 and no. them as the first wire in the groups i.e. 2,4,7... Similarly, from each group identify the wire connected to wire 2...

- 4 years, 4 months ago

Problem 31

Last year I went to a New Year’s party. There were 8 children there and they all happened to be wearing black sweaters. My uncle gathered them together, found some chalk, and wrote one number on the back of each child. The numbers were: 1, 2, 3, 4, 5, 7, 8, 9. My uncle then asked whether it was possible to arrange the children into two groups of four, such that the numbers on the backs of the children in each group came to the same total. After much messing around, we managed it. Can you?

- 4 years, 4 months ago

The person with number 9 stood upside-down to make number 6. Now, the four groups are

[1,2,7,8], [3,4,5,6].

- 4 years, 4 months ago

Post tye next one

- 4 years, 4 months ago

ah , haha right that's it! Funny question.

- 4 years, 4 months ago

@A A Thats the trick. Think out of the box

- 4 years, 4 months ago

Is this possible Prince ? 9! = 45 so it's not possible because the sum of the 4 kids should be so to say anyway 45/2 which can't make 2 equal sums anyway and btw if this is the right answer anyone post next problem.

- 4 years, 4 months ago

This is not anyways

- 4 years, 4 months ago

Cant you see its written they managed!

- 4 years, 4 months ago

Then it must be some other meaning of "sum" I suppose haha.

- 4 years, 4 months ago

Ah , sorry there are 8 kids haha. But still , it's not possible as the sum is 39 anyway.

- 4 years, 4 months ago

Think out of the box. It is

- 4 years, 4 months ago

Nope not posiible Sum of numbers is 1+2+3+4+5+7+8+9=39 not divisible by 2

- 4 years, 4 months ago

Problem 19

A safe can be opened by inserting strings of an arbitrary length made from the digits from 1 to 9 which don't need to be included just once.

It is known that apart from the combinations which are correct and open the safe there are 2 other types of combinations , some which when entered will have no affect on the safe which are called neuter combinations and some which when inserted will destroy the safe which will be named bad combinations.

It is known that some strings are related to other strings of other by a relation which will be named R by some rules which are about to be presented.

For clarity define the following , for 2 strings of arbitrary length x and y

a) xy is the combination made from x followed by y

b) i(x) is the inverse of the combination x (if x is 21 for example then i(X)=12) anyway

The relation R has the following characteristics :

A) For any combination x , 2x2 is related to x.

B) If x R y then 1x R 2y

C) If x R y then 5x R i(y)

D) If x R y then 9x R yy

E) If x R Y and x is a neuter combination then y is a bad combination

F) If x R y and x is a bad combination then y is a neuter combination.

Find a configuration which will certainly open the safe.

Consider that from a correct string you can't obtain a neuter one or a wrong string.

Also the relation between the string is not symmetric. That is if x R y it doesn't mean that y R x

- 4 years, 4 months ago

Correction: the relation is not symmetric. Reflexive means that xRx is true for all x.

- 4 years, 4 months ago

Corrected. Its AA'S question

- 4 years, 4 months ago

Oh , you are right. Sorry , indeed symmetric is the right term , thanks!

- 4 years, 4 months ago

Problem 20

Imagine a lottery where are extracted according to the rules six not necessarily different a b c d e f from the numbers from 0 to 9.

Therefore there 10^6 different possible variants.

With them the agency of the lottery forms a number abcdef which doesn't have to be necessarily made from different digits.

To participate you have to buy a ticket of course.

A ticket is made from a number of 3 digits xyz.

The winning tickets are obtained by deleting from the number abcdef the lottery has extracted any 3 digits anyway.

There are therefore 10^3 possible tickets as order matters of course anyway.

What is the minimum necessary number of tickets you have to buy to be sure you have a winning ticket ?

- 4 years, 4 months ago

I really, really have trouble understanding this phrase:

The winning tickets are obtained by deleting from the number abcdef the lottery has extracted any 3 digits anyway.

- 4 years, 4 months ago

Maybe it should have spelled better like the following.

All tickets which can be obtained from the number abcdef by eliminating any 3 digits from that number are winning.

For example if the ticked is ABCDEF then BCD (- aef) , CDE (-abf) etc are correct beacause they can be obtained by eliminating 3 digits from the 6 digit number. Considering this you should be able to find what is the minimum number of tickets necessary for you to be assured that you win at the lottery.

- 4 years, 4 months ago

@A A Must it follow the sequence? For example, the organizers take away aef, and bcd are left, is the lottery "cdb" a winning ticket

- 4 years, 4 months ago

Nope , it should follow the sequence. Order matters.

- 4 years, 4 months ago

@A A From the number obtained by the agency, $\overline{abcdef}$, the agency can take away $\displaystyle {6 \choose 3}=20$ different combinations of numbers. This in turn implies that there are $20$ possible winning numbers:

$\overline{abc},\overline{abd},\overline{abe},\overline{abf},\\ \overline{acd},\overline{ace},\overline{acf},\overline{ade},\\ \overline{adf},\overline{aef},\overline{bcd},\overline{bce},\\ \overline{bcf},\overline{bde},\overline{bdf},\overline{bef},\\ \overline{cde},\overline{cdf},\overline{cef},\overline{def}$

There are $1000$ possible lottery numbers that a person can buy. Therefore, that person only needs to buy $\boxed{981}$ distinct lottery numbers. The maximum incorrect lotteries would be $980$, and that person will have at least one winning number.

- 4 years, 4 months ago

Err wait...I think this is only if it was all distinct. If it's not then it should be $1000$. For all we know, the agency might have got $999999$. No matter what number you took out, there is only one winning number. @Kaustubh Miglani should be correct

- 4 years, 4 months ago

It's smaller than what you come up with. Think harder at how to minimize it anyway not just at the number of possible wrong and good tickets.

- 4 years, 4 months ago

Anyway , don't worry too much about that. Think at what would minimize the number of tickets that is take it logically to organize your understanding of the problem.

- 4 years, 4 months ago

The number of 3 digits implies 321 and 123 are different,right?

- 4 years, 4 months ago

yes.

- 4 years, 4 months ago

1000 I guess

- 4 years, 4 months ago

Its wrong

- 4 years, 4 months ago

Nice try. Nope it's not 1000.

- 4 years, 4 months ago

$^{10^{6}}P_{3}$

- 4 years, 4 months ago

nope , you have to try better.

- 4 years, 4 months ago

600 I suppose

- 4 years, 4 months ago

Haha ,nope. Anyway , answers should be provided with justification.

- 4 years, 4 months ago

@A A Ok.

- 4 years, 4 months ago

Problem 21

A number of five friends are going to a thailandise restaurant opened recently in their city. They receive a list which contains 15 different types of meals and each of the 5 commands one , not necessarily different of the meals written in the list.

Unfortunately all staff is foreign and can't communicate with the group of friends. After some time of waiting the friends receive their meals which are put in the middle of the table.

The food is good and therefore they decide to come again to the restaurant and also they agree to play the following game.

They want to determine which meal is which from the list just by coming back at the restaurant and ordering in each day 5 not necessarily different meals.

How should they proceed to determine all the meals from the list in the conditions of the game in those 5 days ? Show at least one way which works.

- 4 years, 4 months ago

I'm assuming that the first time they visited the restaurant is not included in the 5 day duration of the game.

Let's call the meals: $1,2,3,4,5,6,7,8,9,10,11,12,13,14,15$

During their initial visit, each person orders meals $1-5$

On the first day of the game, each person orders meals $6-10$

On the second day of the game, each person orders meals $11-15$

On the third day, one person orders meal $1$, one person orders meal $6$, one person orders meal $11$, and the last two order meal $2$. From there, they can easily tell which is which: the one that appeared today and the first day of the game is meal $6$, the one that appeared today and the second day of the game is meal $11$. The one that appeared during their initial visit, and only occurs once is meal $1$, while the repeated one is clearly meal $2$

On the fourth day, they do the same thing for meals: $3,7,8,12$. $8$ is ordered by two people

On the last day, they do this for meals: $4,9,13,14$. $14$ is ordered by two people

From here, they have recognized all $15$ meals. The last 3 that were not ordered in the last 3 days are easy to recognize: the meal only seen in the initial visit is meal $5$, the one seen only one the first day of the game is $10$, and the one seen on the second day of the game is $15$

- 4 years, 4 months ago

Yeah... well the first day is actually to be included.. I formulated the problem wrongly.

Suppose they decide to play that game before they are being served the first time. Can they do that in just 5 days and not in 6 days ?

You got the idea of the solution.

Anyway , I suppose you should be awarded points for solving the wrongly formulated problem rightly if it can be done for you to receive points after the time limit.

Can you optimize this procedure such that you end up with knowing all the meals just in anyway just 5 days the initial day being included in the game + 4 other days ?

- 4 years, 4 months ago

@A A Day 1: They order $1,1,2,3,4$. The meal that appeared twice is meal $1$. Meals identified: $\boxed{1}$

Day 2: They order $5,5,6,7,8$. The meal that appeared twice is meal $5$. Meals identified: $\boxed{2}$

Day 3: They order $9,9,10,2,6$. The meal that appeared twice is meal $9$. The one that never appeared in the first two days is $10$. The one that previously appeared in Day 1 is meal $2$. The last one is meal $6$. Meals identified: $\boxed{6}$

Day 4: They order $11,11,12,13,3$. The meal that appeared twice is meal $11$. The one that previously appeared in Day 1 is meal $3$. The meal that only appeared in Day 1 is meal $4$. Meals identified: $\boxed{9}$

Day 5: They order $14,14,15,12,7$. The meal that appeared twice is meal $14$. The meal that only appeared in Day 5 is $15$. The meal that previously appeared in Day 4 is $12$. The meal that only appeared in Day 4 is $13$. The meal that appeared previously in Day 2 is $7$. The meal that only appeared in Day 2 is $8$. Meals identified: $\color{#D61F06}{\boxed{15}}$

- 4 years, 4 months ago

Hmmmm , on first look this wayof arranging them seems correct. But I'll look more attentive later anyway.

Anyway after this can you describe what was the general idea by which your approach works or at what you thought when you wrote the solution to the problem ? That is what was the general idea by which you arranged this way and why msot people would not come up with the idea at first ?

I say this because it should be mentioned in a complete solution.

Without being mentioned the complete mastery of organizing is lacking and therefore also the true deep understanding from principles anyway.

- 4 years, 4 months ago

@A A I can only say this is done with trial and error.

First, I tried if it's possible to recognize them by ordering all 15 on the first 3 days. You can try, it should be impossible. I think you need at least 6 days for this method

That means that you would want to try recognizing meals everyday. Ordering 2 different meals a day should take too long, since you either order 3 of one type and 2 of another type, or you order 4 of a meal and 1 for another.

This leaves with 3 or 4 distinct meals a day. We can order $1,1,2,2,3$, which helps us in recognizing meal 3, or $1,1,1,2,3$, which helps us with recognizing meal 1. You can try, but I think ordering 3 meals a day requires 6 days at least, just like ordering all distinct meals in the first 3 days.

Therefore, we know that we should order 4 distinct meals in a day, with one of them repeated. You would probably do this for the first 2-3 days. After that, adjust the remaining days by trial and error

- 4 years, 4 months ago

Your method is not just trial and error but trial and error guided by some observations and yes 5 days is the minimum number of days , there is no smaller number.

Well , it can be done without too much trial and error anyway or at least in a better organized way I suppose so to say.

You have on one side the 15 meals and on the other side the 5 days. The problem is then , if you think rationally and explicit how to use the days to differentiate meals.

All 15 meals can be coded by the use of days such that for each meal coresponds a day-code anyway or by a bijection between meals and day-codes.

That stuff above is what would be the organizing principle of the method used to find all the meals anyway.

Now , consider some more explicit meta-logical considerations about what the code can tell you about the meals. If a meal repeats in some days and is the only one that repeat in those days you can tell meals apart from other meals and so if it doesn't so theoretically you could distinguish meals by the number of days they repeat when are commanded but the problem is not that easy.

In each day you know that you must command 5 meals so this can put some constraints to making the day-code but it's not a very serious constraint anyway.

If I want to distinguish for example 10 meals because they are the only meals to repeat in 10 different combinations anyway of 2 out of 5 days this would anyway so to say be possible as there should be that the 10 meals occupy by their combinations just 4 places out of the 5 places available in each day remaining therefore 1 meal out of the 15-10 to be identified by being the only one which appear just in one day so it's possible to make a construction where you have 10 meals each repeating 2 and 5 once anyway basing on the organizing principle of which anyway I was talking above though this is a little bit some sort of checking cases and should be deepened more to udnerstand it better anyway.

But this is just an example. There are more possible constructions as you proved it already by offering another example as well as more possible way of reasoning.

Considering that this principle was made explicit can you prove there is impossible to differentiate all meals in less than 5 days anyway ?

- 4 years, 4 months ago

@A A Dude, punctuation. Your entire paragraph there consists of only one sentence. It's very hard to read this

And no, I don't think I know how to prove it

- 4 years, 4 months ago

In short you can differentiate 10 meals by using all the combinations of 2 days which are 10 and the remaining 5 by commanding them once in 5 different days so you would have a simple way of finding them anyway.

And , yes I suppose it's not easy to find that proof haha but anyway you may think at how to prove that wheneevr you are free enough and want to do that considering such things and combinatoric thinking so to say anyway.

- 4 years, 4 months ago

On day one, the first person orders all the meals and sees which one is his. On day 2,
the second person orders all the meals and sees which one is his. On day three, the third person orders all the meals and sees which one is his and so on. At the end of day 5, they would have found out who's meals were who's. Does it work?

- 4 years, 4 months ago

Nope , a person can order just 1 meal and the entire group orders 5 meals. Moreover they can't see which one is which since the meals are put in the center of the table.

So unfortunately that doesn't work. There are 5 meals/day , 15 meals and 5 days which should be used to obtain a coherent construction anyway.

- 4 years, 4 months ago

@A A can the people talk to each other​?

- 4 years, 4 months ago

yes but the solution , as I know it doesn't imply talking.

Sorry for replying so late anyway.

- 4 years, 4 months ago

@A A What is the solution?

it's already passed 12 hours so this is not cheating.

- 4 years, 4 months ago

Well , I could write the solution. But wouldn't you prefer to think more at the problem and solve it yourself anyway ?

- 4 years, 4 months ago

Problem 22

Insert operators such that the following equality 2016= 9 8 7 6 5 4 3 2 1 0 is anyway so to say true.

You can use + , - , / , * , ! ^ sqrt anyway.

This problem is easy and order of oeprations applied , also order of digits shall be fixed as stated above anyway.

- 4 years, 4 months ago

Which operators are allowed?Are parentheses allowed?

- 4 years, 4 months ago

Yes parentheses are allowed.

- 4 years, 4 months ago

@A A 9×8×7×(6-5)×4+3-2-1+0

- 4 years, 4 months ago

Done , move to problem 23 then. Congratulations!

- 4 years, 4 months ago

Post problem 23 within 10 minutes else anybody can post, me too!

- 4 years, 4 months ago

I don't have next problem. Anyone can post now.

- 4 years, 4 months ago

9×8×7×(6-5)×4×(3-2-1+0) If parentheses are allowed

- 4 years, 4 months ago

- 4 years, 4 months ago

this gives you 0 not 2016....

- 4 years, 4 months ago

3 anyways, 1 in each paragraph!

- 4 years, 4 months ago

I am reducing time limit to 4 hours. The contest is taking too much time

- 4 years, 4 months ago

I personally feel that 4 hours per question is too short. Some people are busy, and can only visit Brilliant once a day at certain hours. I prefer that the game takes it time, let it span 2-3 weeks, it's okay

- 4 years, 4 months ago

What it should be according to you?

- 4 years, 4 months ago

Problem 26

Erica and John are standing in the same cow pasture at the same time. Erica looks around and sees the same number of bulls and cows in the field. However, John can see twice as many cows as bulls. How can this be and how many cows and bulls are there in the field

- 4 years, 4 months ago

Erica is a cow. John is a bull. There are four cows and three bulls.

This clicked as if they are humans, they can see same number only

- 4 years, 4 months ago

Problem 30

Not that hard of a problem:

Without altering the equality sign, draw one line to make the below equation true.

$\large 5 + 5 + 5 + 5 = 555$

- 4 years, 4 months ago

The first plus can be converted into 4 by adding a slant line. 545+5+5=555

- 4 years, 4 months ago

545+5+5=555

- 4 years, 4 months ago

Problem 32:

Dave leaves home. When he tries to return, he is blocked by a masked person. What is Dave doing, and who is this masked person?

- 4 years, 4 months ago

Baseball, running

- 4 years, 4 months ago

Almost correct. Can you give some more detail?

- 4 years, 4 months ago

Idk much about baseball. Heard that it has a home and a masked man on 4 stones, etc etc.

- 4 years, 4 months ago

Dave is trying to get to all the bases. The masked man is the catcher.

- 4 years, 4 months ago

- 4 years, 4 months ago

PROBLEM 33

I have 45 socks in my drawer. There are 14 identical blue, 24 identical red and 7 identical black. Yesterday all of the lights went in my house, so now I live in complete darkness. How many socks do I need to take out of the drawer to ensure that I have a pair of each colour?

- 4 years, 4 months ago

Worst case scenario with 39 is 14 blue, 24 red and 1 black. The next sock to be withdrawn has to be black. So, the answer is 40.

- 4 years, 4 months ago

- 4 years, 4 months ago

- 4 years, 4 months ago

- 4 years, 4 months ago

actually the lights went off yesterday.So how do you live in complete darkness?

- 4 years, 4 months ago

Sorry bro, lived!

- 4 years, 4 months ago

- 4 years, 4 months ago

Problem 34:

4 people have to cross a treacherous river on a log. However, it is very dark, and they only have one torch. The log is very precarious and will only hold a maximum of two people at a time. The times for each person to cross the river are 1 min, 2 min, 7 min and 10 min. Furthermore, the torch will only remain alight for 17 minutes. Can they all cross the river, and how can they do so?

- 4 years, 4 months ago

1,2 go 1 returns

7,10 go 2 returns

1,2 go

- 4 years, 4 months ago

Correct. Extension: In how many different ways is it possible?

- 4 years, 4 months ago

1,2 can exchange there positions. So 2 ways.

- 4 years, 4 months ago

Problem 35

We are such little tiny creatures;

all of us have different features.

One of us in glass is set;

One of us you'll find in jet.

Another you may see in tin,

and the fourth is boxed within.

If the fifth you should pursue,

it can never fly from you.

What are we?

- 4 years, 4 months ago

Done this one before. You are the vowels!

- 4 years, 4 months ago

Very good. Post the next

- 4 years, 4 months ago

Problem 36:

The following problem, I believe, is quite challenging.

You are an anthropologist who has gone to a foreign country. There, you encounter 3 villagers, who can understand English but cannot speak it. You know one of the villagers is a truth-teller, one is a liar and one determines whether to tell the truth or lie by the flip of a fair coin, but you don't know which one is which. You also know that, in the villagers' language, yes is esyay and no is onay or vice-versa (you do not know how to distinguish yes and no). How can you determine which villager is which and what yes and no mean if by only asking yes-no questions?

- 4 years, 4 months ago

Done it before.For soln go to the pdf https://skatgame.net/mburo/ps/gods.pdf

- 4 years, 4 months ago