Maths problems (interesting ones)
 

just going to post some random puzzles here, to keep people interested in maths!

Post your answers here, i might make it a quiz or something !
puzzle 1 )
4 astronauts must get from their pod to their ship in 12 mins

they have 1 oxygen tank which can support up to two people at a time

all astronauts travel at different speeds and if two astronauts share the oxygen take, they travel at the speed of the slowest astronaut
The speeds at which they can travel are as follows:

A:1
B:2
C:4
D:5

if A and D go and A returns, this is (Total 6 mins), then A and C go and A returns (Total 11 mins) then A and B go the total is 13 mins. How can it be done?


Puzzle 2)
2 fathers and 2 sons went into a restaurant and ordered 3 whole pies, and they ate 1 whole pie each (no! 1 whole pie, not pi, because 1 pi radian is only half of a pie). How come?

Puzzle 3)
How can you get 4 liters of water EXACTLY from using an unlimited supply of water and a 5 liter and 3 liter bucket???

More to follow if there is any interest......

Re: Maths problems (interesting ones)
 

1) Astronaught A and B cross first (2 minutes). A returns (1 minute; total 3). C and D cross (5 minutes; total 8). B takes the oxygen tank back (2 minutes; total 10) and then A and B go back (2 minutes; Total 12)



2) There was an old man (Person A), a middle aged man (Person B), and a child (Person C). Person A was a father to Person B, and a grandfather to Person C. Person B was a father to Person C and a son to Person A. Person C was a son to Person B.



3) Fill the 5 litre bottle. The pour what you can of that into the 3 litre bottle. You're now left with 2 litres of water in the 5 litre bottle. Empty the 3 litre bottle and pour those 2 litres into it. Fill the 5 litre bottle completely, and from that, fill up the 3 litre bottle to the top (1 litre). You're now left with 4 litres in the 5 litre bottl.e

Re: Maths problems (interesting ones)
 

yep thats right for the second two,
you need to know some discrete mathmatics (thats like ordering things etc) for first one....
hmm, do i put some more up?


ah you got the first one now

Re: Maths problems (interesting ones)
 

:P I was in the middle of figuring it out; I've seen the puzzle before (using a bridge instead) I just couldn't quite remember the answer, and I wanted to post the other two before someone beat me. :P

Yes yes more.

Re: Maths problems (interesting ones)
 

hehe ok
here are the next couple
4)
I live in a village and I am a shopkeeper. I sell different things and all of them are sold in kg's ( i.e 1 kg, 12 kg etc.). I only sell my goods in complete kg's i.e only in whole numbers( 2.5 kg is out of scope). The problem is that I donot have
an electronic weighing machine and only have a traditional weighing machine. A friend of mine has told that I need only "4 different weights" to sell my goods if they are to be sold in the range of 1 to 40 kg. This is perfectly acceptable to me. Can you guess those 4 weights that I need to have so that I can weight anything between 1-40 kg? I am not allowed to use the weighing machine twice to weight a single request. i.e If someone wants to have 37 kg. of rice I will use those four wieghts (or less) to weigh 37 kg. of rice for him and not use the the weigh machine twise.
5)

Marbles

There are 8 marbles that weigh 1 ounce each, and 1 marble that weighs 1.1 ounces. The marbles are all uniform in size, appearance, and shape. You have a balance that contains 2 trays. You are only able to use the scale 2 times. How do you determine which marble is the heaviest using only the scale and marbles in 2 weighings

6) How do you six nines to equal exactlly 100(you can use addition, divistion, decimals... just no other digits.. ( Oh and you cant say that 99+ (99/99), because thats lame lol.... and you dont have to use all 9's)


7)
A smail is at the bottom of a 15 meter tree. He tries to reach the top. Each day, he manages to climb 4 meters, but during night, while he sleeps, he slips back 3 meters. How many days will he need to reach the top?

8)Mr and Mrs Smith have two children. You ring at their door, and one of their two children, a girl opens the door. What is the probability that the other child is also a girl? (Warning: medium difficulty :) )

Re: Maths problems (interesting ones)
 

4) I think it's 8,9,11,12 kg scales... I gotta go to work though, so I only tested it to liek 15 (and 30 once). So let me know if I got it. :P

5) Choose six marbles and measure 3v3.


If one pile weighs more than the other:
Take two of those three marbles and measure them. If they're equal, then the odd one out is the heavier one. If one weighs more than the other, than it's the heavy one.


If the 3v3 weighs the same:
Weigh the last two and the heavier one will show.


6) 9*9+9+9+9/9

That valid?


7) Twelve days?


8) 33%?

Re: Maths problems (interesting ones)
 

You mean a balance scale, right? The key is to recognize that you can put weights on both sides of the scale to either add to or subtract from the item being weighed. A "balanced ternary" number system fits this problem perfectly, so the weights are 1, 3, 9, and 27 kg.

Re: Maths problems (interesting ones)
 

6) (9*9) + 9 + 9 + (9/9) = 100

7) It will take him 12 days
8) Well, if you take it as an independent case, it's like a coin. It's either a boy or a girl; 50/50. If you take genetics into acount, then if I remember right, there's a slightly better chance of it being a girl.

Re: Maths problems (interesting ones)
 

fizman is totally right

Re: Maths problems (interesting ones)
 

Right on each one? What aboot the scale one? Are both me and Alan right?

And Dave, not quite on the money there... consider all the possible combinations:


Boy -> Girl
Boy -> Boy

Girl -> Girl
Girl -> Boy

We are asked what's the probability that they have two girls. We KNOW that they have 1 girl already, so that eliminates the Boy->Boy possibility.

Which leaves us with 3 possible outcomes.

Boy -> Girl
Girl -> Girl
Girl -> Boy

So what are the chances they had Girl -> Girl? 1/3 (or 33%)

Re: Maths problems (interesting ones)
 

Good catch. Never was very good with stats :o

Re: Maths problems (interesting ones)
 

oops
your wrong about the scale one
how do you make 1 kg?

Re: Maths problems (interesting ones)
 

I figure, you measure out 9 pounds, then measure out 8 pounds of it. Then you're left with 1 pound.

Re: Maths problems (interesting ones)
 

me 2. Those sort of problems always throw me. I say someone should go out and take a study!

Re: Maths problems (interesting ones)
 

Very fun puzzles...but here's one that has stumped pretty much everyone I know (similar to marble puzzle but harder)...

You have 12 identical-looking coins. One of the coins is a fake, yet you do not know which one it is nor if it weighs more or less than the other 11. Using 3 turns with a regular balance scale, determine which coin is the fake and if it weighs more or less than the other 11. (props to anyone who gets this in under a day)

Another one that I heard on the radio awhile back...(easier but fun)

There is a tree with exemplary fruit in the middle of a courtyard that a farmer has decided must be guarded from people trying to take the fruit. There are 7 circular fences around the tree with a guard at each. Wanting to get at the fruit so badly, you go up to the first guard and tell him that if he lets you through, when you return you shall give him half of all the fruit you have taken, but then the guard must return one piece to you. You continue to bribe every guard you reach until you have reached the tree, whereupon you take the fruit and upon leaving fulfill your deal with all of the gaurds. How many pieces of fruit did you take?

I'll try and come up with a few more or ask around for some...good luck!

Re: Maths problems (interesting ones)
 

I'll start thinking aboot the coin one, but right now with the fruit, I'm pretty sure you take just two pieces of fruit.

Each guard you give him half (1/2 of two fruit is 1 piece) then he gives you one piece back (back up to two). You never lose any fruit.

Re: Maths problems (interesting ones)
 

Look at it like this: there are two children, one who opens the door, and one who doesn't. Before she opens the door, the possibilities are:

Door Opener / Non Door opener

Boy / Girl
Boy / Boy

Girl / Girl
Girl / Boy

Once she opens the door, the first two possibilities are eliminated, leaving only the options that the shy child is either a Boy or a girl. In other words, the sex of the one child has nothing to do with the sex of the other.

Prove me wrong. :)


EDIT:

Even better, look at it like this:

FAMILIES WITH TWO CHILDREN:
25% have two boys - A girl will answer the door 0% of the time
50% have a boy and a girl - A girl will answer 50% of the time
25% have two girls - a girl will answer 100% of the time

If we take the product of the two percentages in each row, we get some more numbers to cloud the issue:

PROBABLILITY THAT A GIRL WILL ANSWER...
...and the family has two boys - 0%
...and there's a boy and a girl - 25%
...and the family has two girls - 25%

So, a girl will answer (0% + 25% + 25%) = 50% of the time.
Exactly half of those responses will come from families with two girls, and exactly half will come from families with a boy and a girl.

Or I could be completely off base - let me know.

Re: Maths problems (interesting ones)
 

so the odds are somewhere between 1/3 and 1/2?
Somebody start knocking!

Re: Maths problems (interesting ones)
 

diobsidian,

I've heard the first, so I won't answer.

As for the second, you forgot to disclaim that the person can't carry more then reasonably possible, or something like that. For example, FizMan's answer is the most efficient way to get 2, but if you want 3, all you have to do is start with 130. Or for 4, start with 258.

Another thing of interest is that if you were to start with 1, you'd cut it in half for the first guard, and then he'd have to somehow give you 1, and you'd have 1.5. After the next guard, you'd have 1.75. If there was an infinite number of guards, you would have 2, but it would probably be spoiled from having been cut open so long ;)

Re: Maths problems (interesting ones)
 

I think you're off Kris... I'm going to trust my highschool Finite math on this one...


We'll do it your way:


Door Opener / Non Door opener

Boy / Girl
Boy / Boy

Girl / Girl
Girl / Boy


Okay, a girl opens the door... so you say that eliminates the first two possibilities...

Why? Remember, two combinations of boy/girl exist, and you've eliminated one of them for the wrong reason.

Whoever opens the door does not determine what children exist in the family (with the exception of the boy/boy possibility)

If you want to keep your method of reasoning, we have to add more possibilities. That is, the order of children. I'll mention the first born child, the other is safely assumed to be the second born:


Door Opener / Non Door opener

Boy (1st born) / Girl
Boy (1st born)/ Boy

Boy / Girl (1st born)
Boy / Boy (1st born)

Girl (1st born)/ Girl
Girl (1st born)/ Boy

Girl / Girl (1st born)
Girl / Boy (1st born)


Eliminate the Boys answering the door, you're left with the last four. We can eliminate one of the extraneous Girl/Girl possibilities as well.

Girl (1st born)/ Boy

Girl / Girl (1st born)
Girl / Boy (1st born)

Voila, 1/3.

Re: Maths problems (interesting ones)
 

Okay. I, too, have heard those before, so I'll propose my own (well, as you might guess, it is by no means original with me). Suppose that there is a game where we flip a coin. If the first time the coin is flipped it lands on heads, I pay you $2 (2^1), if the first time it lands on heads is the second toss, I pay you $4 (2^2), if the first time it lands on heads is the third toss, I pay you $8 (2^3), ad infinitum. Using simple probability, tell me how much you would pay me to play this game and still be able to make a decent profit (but enough to tempt me to agree to the game).

Re: Maths problems (interesting ones)
 

You mean, for example, if I said, "Hey mtrawls, I'll pay you $20 to play <insert above mentioned game here> with me"?

Re: Maths problems (interesting ones)
 

Well, I tried. First I tried analytically, and had the same explanation as FizMan. However, when I tried to figure out why yours was wrong, I couldn't. So, I wrote a program to do it a bunch of times. In short (unless my algorithm is wrong), you're right, it is 1/2.

I've included the code below in case someone wants to review my work.

Re: Maths problems (interesting ones)
 

ok ok enough debate
the answer is 1/3




Right, 4 possiblities, as most of you got

boy/girl
girl/boy
boy/boy
girl/girl
from genetics
but if you find its a girl, you only eliminate one of them (boy/boy) not 2.
thus the probability - 1/3 (2 where the other is a boy and 1 where other is a girl) .

i quote an answer someone else wrote
You know that they have two children, that is four possibilities (Boy-Boy, Girl-Boy, Boy-Girl and Girl-Girl). Because you know that one of them is a girl, the first possibility is ruled out. So the probability that the second one is a girl is 1/3.


Another one :
Three cowboys are fighting together in a meadows. The first one, John, is a good shooter, he kills his target 2/3 of the time, and miss one third of the time. The second, Slim, is an average shooter, whose chances to kill and miss or even. The third, Bob, is a poor shooter, he only kills his target 1/3 of the time. They look at one another in silence... They still look... Sweat starts pearling one their faces... OK. Anyway. Their rule is the following. They shoot in rounds. One shot per round. They always shoot at the best shooter but themselves. They do as many rounds as neccesary so that either they all die or only one remains alive. Who has the most chances to survive? (Thanks to Matthias for this one)

oh and joe, dont answer this one lol seeings as you probably know where to find the answer lol


A grandfather clock chimes the appropriate number of times to indicate the hour, as well as chiming once at each quarter hour. If you were in another room and heard the clock chime just once, what would be the longest period of time you would have to wait in order to be certain of the correct time?


Let us go into heroic fantasy land. The story takes place in a antic land where dwarves are mining to extract gold. The king is visiting the mine when he ears voices around the corner: "I have a neat trick: we have to melt gold bars the size of this mold, that is exactly 100 grams. I remove just a tiny fraction to all my bars and they all weight 99 grams. Their scales are not precise enough, they don't see anything..." The king tries to run after then, but they ear him and manage to flee among the corridors. Annoyed, the king summons the gods, who grant him a gift from the future: a digital scale. The digital scales works as follows: you put whatever you want on the scale, (nothing happens then), you press a button and the exact weight of what is on the scale appears. Then, the scale vanishes. Note that before you have pressed the button, no weight appears and once you have pressed the button the weight won't change even if you add or remove things. Armed with this mighty scale, the king summons all his 100 workers. How will he find the cheater?

Re: Maths problems (interesting ones)
 

Yes, it IS 1/3. REMEMBER what I told you? The door does NOT, I repeat, had NO bearing on the possible combinations of children. IF you CHOOSE to go that route of using the door for logic, you ARE EXCLUDING possibilities.


And it's too early in the morning for me to figure out your code right now, but I'm pretty sure it's flawed too since again, you're using the door for reference and therefore probably excluding possibilities.


Grandfather Clock Question:
1 hour and a half. Five possibilities:

Either it hit a quarter of an hour (on some hour OTHER than 12 o'clock) OR it hit a quarter of an hour (15, 30, or 45 minutes) on 12 o'clock OR it hit 1 o'clock

In the first possibility, all you have to wait is 45 minutes, in which case you'll hear it chime more than once.

In the second possibility (15 minutes) you'd have to wait 1:45 for the clock to chime more than once.

In the third possibility (30 minutes) you'd have to wait 1:30 for the clock to chime more than once.

In the fourth possibility (45 minutes) you'd have to wait 1:15 for the clock to chime more than once.

In the fifth possibility (1 o'clock) you'd have to wait 1 hour for the clock to chime more than once.


As you wait, if the clock still chimes once at an HOUR AND A HALF, you've elminated all the possibilities but one (number 2) and you know for a fact that it's CURRENTLY 1:45 and when you heard the first chime, it was 12:15.



Cowboy question:
I'm going to go with a 54.78% chance that Bob (the poorest shooter) will win. I hope I'm right; good ol' notepad:

I kind of stopped after that when I realized that the possibilites won't top the first one (in which Slim and Bob manage to shoot down John. And John shoots down Slim)

Re: Maths problems (interesting ones)
 

yes right for grandfather , however i havn't worked out the cowboy one yet as i am meant to be working

Grandfather clock answer

You would have to wait 90 minutes between 12:15 and 1:45. Once you had heard seven single chimes, you would know that the next chime would be two chimes for 2 o'clock.

Re: Maths problems (interesting ones)
 

The details are simple but hard to explain without graphics. Label the twelve coins A through I. Weigh ABC against DEF, then ABC against GHI. The results will tell you which group of three is different and whether it is heavy or light. Then you weigh any two of the different group's coins against each other to find out which is the odd one; if they match, the third one is the different one.

if ABC < DEF and ABC < GHI, one of ABC is light
if ABC < DEF and ABC = GHI, one of DEF is heavy
if ABC = DEF and ABC < GHI, one of GHI is heavy
if ABC = DEF and ABC > GHI, one of GHI is light
if ABC > DEF and ABC = GHI, one of DEF is light
if ABC > DEF and ABC > GHI, one of ABC is heavy

The other three combinations are not possible if exactly one coin is different.

Once it's narrowed down to three coins, just compare two of them. If they don't balance, you already know whether the fake is heavy or light; just choose that one. If they do balance, the other must be the fake.

If I take two, then I can give one (half of my two) to each guard and according to the agreement he'll give it back to me, leaving me with the two I took.

Re: Maths problems (interesting ones)
 

I hate to break it to you after all your work, but there are 12 coins, and lettering them A through I is only 9. :ahh:

Yep you got that one! :D

Keep working on the 12 coin problem - its a real toughie!

Re: Maths problems (interesting ones)
 

You're missing the fact that there are two different ways in which the door may be answered by a two-girl family. Your analysis is looking at the probability of a girl answering, which doesn't matter for this problem, because a girl answering is given.

Re: Maths problems (interesting ones)
 

I understand that the answers you've looked up or been given in finite math class are what you've demonstrated so far. And I understand your answer, and see why it works. What I don't see is why mine was wrong. Your proof above was a bunch of arbitrary hand waving. Why do we need to "eliminate one of the extraneous Girl/Girl possibilites as well" ? If you don't eliminate that - the probability is (surprise surprise) 1/2.

Think for yourselves and come up with the answer! Don't tell me that your high school math teacher told you this was the answer, and therefore it's so. There is more than one way to look at any problem - the challenge here is to determine, at a fundamental level, what is different between the way your math teacher looks at the problem and the way I've outlined it above. I don't see anything fundamentally different, but just because I don't doesn't mean it isn't there.

My proof and Joe's simulation both show the answer to be 1/2. I've also included an Excel simulation that does the same - why are we wrong?

Re: Maths problems (interesting ones)
 

Whoops. :o

I still have a 3/4 chance of finding the coin and identifying it as heavy or light in three weighings if I simply ignore the last three, and it will only take one extra weighing in that 1/4 case, but that obviously doesn't meet the requirements of the problem, does it?

I know how to do the harder problem in theory, choosing appropriate groupings of coins to get the equivalent of three equations with three unknowns, but I lack the time to figure out the details today.

Re: Maths problems (interesting ones)
 

Actually, I'm the only one catching it - it's twice as likely that a girl will initally answer from a G/G family than from a B/G family, and there are twice as many B/G families. Therefore, the other sibling has a 50/50 shot of being girly.

Re: Maths problems (interesting ones)
 

No, you're counting families too soon. Only half of of the boy/girl families had a girl answer the door, so the others don't meet the criterion for this problem.

None of the boy/boy families are included. Half of the boy/girl families are included. All of the girl/girl families are included. But once a girl answers the door, you have to completely forget all the families that didn't get included. You can't count them, because they don't exist in the set of familes that did have a girl answer the door.

Re: Maths problems (interesting ones)
 

Okay let's try this again...

Four possible combinations of children. Right? That's simple math too, you have two inputs, four outputs.

A girl answers the door... This is GIVEN.

We've now determined that it is impossible for the family to have a boy/boy combination.

Since a girl answering the door is GIVEN, it leaves us with three possibilities.

Two of those possibilites have boys. One possibility has a second girl.


I don't know how else to go aboot this, but I for one am going to trust my OAC finite before I trust logic that I feel is flawed.

Re: Maths problems (interesting ones)
 

I think it comes down to, how are the kids selected to answer the door. I don't beleive the problem is clear enough, and that's why you have two different arguments for the answer.

When Kris reads the problem, he sees no method for picking which kid goes, and so assume random. That throws out the possibility of 2 boys, and also 1 of the 1 boy 1 girl possibilities. You get 1/2

When Denman and FizMan read the problem, they see that if a girl is in the family, she is sent out first. Thus, you can only eliminate the 2 boys possibility, and get 1/3.

Is one or the other the correct reading? Make your arguments to convince me.

Re: Maths problems (interesting ones)
 

I think we can all agree on the four possible combinations of children, right?

So imagine four houses, each having one different combination of children.

The question specifically states that a girl answers the door. I read that as a given. Leaving three houses left; two with a boy, and one with a girl. If you randomnly chose either house than it'd be a 33% chance.


IF you guys were going by the probability that a girl answered the door in the first place... then consider this. We can naturally exclude the boy/boy combination from calculations. This leaves FOUR girls and TWO boys in our last three possibilities.

So what are the chances that a girl will answer the door? 66%

If you choose randomnly between the houses, House A (has girl/girl) will yield a 100% girl answering door chance. Whereas Houses B,C (girl/boy x 2) have a 50% chance each. But the probability of having either possible combination is 33% (three possibilities)

1/3 x 100% + 1/3 x 50% + 1/3 x 50%
33% + 16.5% + 16.5% = 66% (2/3)

So you have a 66% chance that a girl will open the door, this is where your calculations should start. I agree with the idea that at this point, it can be a 50/50 chance for the second child being a boy or a girl (just like flipping a coin really), but since we already have the 66% chance of a girl opening the door in the first place, we multiply it by the 50% and we end up with a 33% chance of a girl/girl possibility through this method.

Re: Maths problems (interesting ones)
 

If you know stats then think of it as permutations / combinations
i will write a full answer later, but its only 9am and i am tired lol

Re: Maths problems (interesting ones)
 

There is a five-digit number having digits 1-9, no two digits being identical. Two are prime numbers, two are square numbers, and one is neither. The third digit is twice the fifth, the fourth is six greater than the second and the last is three less than the first.
This is really hard actually, i havn't worked it out yet lol

Re: Maths problems (interesting ones)
 

73894

Here's my notepad logic. Then I just fiddled around with numbers for five minutes at the end until I got something that worked.

Re: Maths problems (interesting ones)
 

its right!
the webpage i got it from had several riddles and this was the most interesting one. To getto thesolutions page you had to solve that one you just did lol, and i checked and it works!
Well done :)
i just didn't have the patience

Re: Maths problems (interesting ones)
 

here is my favorite math problem. it's pretty cool. :D

i tried uploading it again in response to the post below

Re: Maths problems (interesting ones)
 

The document had a figure and the question, but there was no shaded region...try uploading again maybe...

Re: Maths problems (interesting ones)
 

This page really explains where the two different answers come from:

http://mathforum.org/library/drmath/view/52186.html

It's interesting and worth a read if you're into this kind of thing.

Re: Maths problems (interesting ones)
 

Exactly. And for those interested in finding the answer without doing the math (shock! pure blasphemy, I daresay), try searching for the St. Petersburg Paradox from our good friend Bernoulli (ah, but which one?)

Re: Maths problems (interesting ones)
 

Neat problem.


<SPOILER ALERT>
(use pythagoras and you get the answer to be pi)

Re: Maths problems (interesting ones)
 

sory i been a bit busy lately.
some new problems coming up later

Re: Maths problems (interesting ones)
 

Ok, I know this is a little late, but here goes...

The whole 1/2 1/3 debate was bugging me, so I e-mailed her about this with links to a couple of different arguments (in addition to the thread it self). Here's a copy:

She is a profesor at a local college.

Re: Maths problems (interesting ones)
 

I just wrote my own computer program.


IF you choose a random child to open the door then it's an outcome of 50%

IF you choose a girl to answer the door (I still say the question provides that as a given), then the outcome is 33%


But I don't know... I honestly don't agree with your teacher's logic... yes I agree that the gender of one doesn't influence the gender of the second... but she seems to ignore the fact that there are twice as many families with a boy and a girl than there are with two girls. The gender of the second kid isn't generated when you meet them (i.e. flip the coin); it's already been made... and fact is twice as many "tails" get flipped than "heads."


Also, why does opening the door and getting a girl eliminate 500 doors? Shouldn't it only eliminate 250 doors (b/b) as even in the b/g & g/b families, either one could have answered the door? And wouldn't that leave 250 doors with GG and 500 with some combination of GB?

Re: Maths problems (interesting ones)
 

and that, boys and girls, is the sound of my head exploding.

Re: Maths problems (interesting ones)
 

Of course there is. GG, GB, BG, BB. But that assumes you can identify A and B. But in this instance, you can only identify them by which one opened the door, so you have to account for both instances. A or B may open the door!

A opens:
+GG
+GB
-BG
-BB
B opens:
+GG
-GB
+BG
-BB
It's still 1/2, 50-50.

Also, That list is generated from the combinations of 2 mutually exclusive chances, a table:The bottom line is: THE RESULT OF ONE DOES NOT AFFECT THE RESULT OF THE OTHER.

Re: Maths problems (interesting ones)
 

I don't understand what your + and - signs are denoting.

Re: Maths problems (interesting ones)
 

the + seems to represent a possible combination and the - shows that it does not work. That post aims to prove that there are 2 ways that the Girl-Girl combination works (because either one could open the door and then the other child would be unknown), which would then make the odds 50-50.

Re: Maths problems (interesting ones)
 

Right, i remembered this thread from a while ago, so i decided to post some new problems up
Number 1

given that the rectangle is 2x4, find the radius

Number 2
Divide $261 (in whole $ increments) into a number of bags so that I can ask for any amount between $1 and $261, and you can give me the proper amount by giving me a certain number of these bags without opening them. What is the minimum number of bags you will require?

When i get around to it i'll add a couple more

Re: Maths problems (interesting ones)
 

See attached drawing. (The original diagram looked something like the top left picture.)

Re: Maths problems (interesting ones)
 

AHHHHH my brain hurts ... i need the solution!

Re: Maths problems (interesting ones)
 

You don't need to use the scale at all.

First, if you need a small graduated cylinder. 11 coins should be exactly the same, so they all have the same density. The fake, although not necessarily weighing the same, will have a different density than the others. Simply put in X mL of water in the cylinder and calculate the displacement of the water for each coin dropped, and the fake coin will be different.

Re: Maths problems (interesting ones)
 

Well yea i think that is definitly an option if i had to do this in reality... but we're not in "reality", we're in Steve howlands reality... and he descided to give us merley a stupid scale with a bunch of stupid coins... AHHH i need to know the answer dangednabbit! lol

Re: Maths problems (interesting ones)
 

ok, here got it from a book some place and i figured out the solution (actually there are two but only two)

--so here it goes: a 9-digit number (123456789; the digits only represent the position for further labeling),
--now each position has to be divideable (english, gramar, I sorry.) by the #of the position (like 1/1=n; 12/2=n; 123/3=n; 1234/4=n and so on; n representing a natural number)
--and finaly, the number consist of the digits 1 through 9, but they can be used only once

So get the solution, and I'm sorry for my terrible english

Re: Maths problems (interesting ones)
 

Right,

For the coins one you can't do it unless you know whether its heavier or lighter.
if you know whether it is heavier or lighter:
put 6 on either side of the scales and weigh
One side will be lighter than the other so you take the side witth the heaveier or lighter one and weigh 3 on 3. One side will be heavier/lighter. Then weigh 1 on 1 of and leave one off. If it is even then the 3rd is fake. If it is not even then the heavier/lighter side is fake

Coin Problem
 

I got the solutions for the coins! However I need 4 steps. Maybe I can get it with 3 (I need some time on that)

I'll post the solution with 4 later today (School now).
I know it doesn't count but at least it's something and maybe then someone else can do it then with three...

Re: Maths problems (interesting ones)
 

Hi

for the coin question:

label coins a-l
measure 4 (a-d) coins against another 4(e-h)
if the same then the spare 4 (i-l) contains the odd one out, measure 1(i)v1(j) of these
if i and j are the same take one of the spares (k) and measure against one of those you know to be true (i or j), if the same, the unmeasured one (l) is the fake, if different kmust be fake
if the i and j are different you know any of the remaining coins are real so measure i against any of the others to know if it is fake, if it isnt j is.

if a-d are different to e-h (remember which is heavier)
take two from each set a,b,e,f and measure the against i-l if they are different you can work out whether the fake is heavier or lighter; which means you know whether a or b are fake or e or f are fake, you can then eliminate which one in the same method as i or j above

if they are the same take c and g and measure against i and j
if they are different, you know whether the fake is heavier or lighter so can work out whether it is c or g that is the fake
if they are the same mesure one of them against a real coin and find which one is fake in the same way as i and j above




now for another coin question:
there are a thousand coins lined up in a row, all with heads up. If somone went along and turned over every coin, and when they finished that someone else came along and turned over every second, and when that was done someone turned over every third, and so on up to a thousand how many coins would be facing heads up when it was finished

oljomo

Re: Maths problems (interesting ones)
 

Hey you brits are smart! congratulations.
I probably won't have to post the 2 solutions with 4 ways I found?!
So now it's your turn!

Re: Maths problems (interesting ones)
 

Hi

Another problem:

ian and jenny both have a whole amount of pounds (im british)
Ian says to Jenny "if you give me £3 i will have n times as much as you"
Jenny says to Ian "if you give me £n i will have 3 times as much as you"
Given that n is an integer find the possible values for n

oljomo

Re: Maths problems (interesting ones)
 

OK.. Here goes...

If ian has £x and Jenny has £y, then 3+x=n(y-3) and n+y=3(x-n)
use equation 2 and y=3x-3n
use equation 1 and 3+x=3xn-4n^2-3n
solving equation 1 for 0, 0=3xn-4n^2-3n-x-3
(you could work the other way with equation 2 and get to this same point)
pluging in some numbers for x (3,4,5,6,7,8,9,10) you get n=1 or 2 for x=5 and n=3 or .75 for x=6 and the rest are very odd or imaganary.

Therefore n=1,2 or 3

Re: Maths problems (interesting ones)
 

Hi

solution is

381654729


jdiwnab, you have not got all the answers.

oljomo

Re: Maths problems (interesting ones)
 

Contenuing on with the numbers from the above question: and then 7 (if there are more, it is not in the range of 3-147)

Re: Maths problems (interesting ones)
 

Lets wrap up the unsolved puzzles.

11)
Here are your chances of winning amounts of money:

50%.......0$
25%.......2$
12.5%....4$
6.25%....8$
3.125%..16$
etcetera

The average winnings of the game, is therefore: 0*(0.5)+2*(0.25)+4*(0.125)+8*(0.0625)+16*(0.03125) ...
This infinite series yields, amazingly, infinity. Therefore, if my math is correct, I could pick an arbitrarily large number of games, pay you whatever you wanted, (lets say 3 million games for 10 million dollars), and I would come out ahead.
Yet, am not certain in this result, so I would stick to paying a dollar per game.

Short answer: Any amount?

Another thing: if you were to stick to your payment scheme, you would pay me a dollar even if I loose the first round. However, this does not effect the results.

14)
This problem requires that you can take portions of each person's supply. So take 1 bar of the first person's supplies, 2 bars of the next person, 3 bars of the next, and so on until you have some of every worker's suplies on the scale. When the scale is read, you can then determine which person cheated based on how much off the scale was to the ideal amount. In this example, with 100 workers and starting with 100 bars, 505000 grams - Measured grams = number of person who cheated.

17)
Have bags that are filled with 1$, 2$, 4$, 8$, 16$, 32$, 64$, 128$, $6
That is 9 bags.

This is what the binary system is all about; the minimum number of on/off digits to get any value. With 8 bags, you can get any number from 0 to 255. With 9, you can go from 0 to 511.

19)
31 will be be tails
969 will be heads

All those numbers that have an even number of factors will be heads at the end. But those that have an odd number of factors will end up tails. The only way a number can have an odd number of factors is for a factor to repeat itself, like 16 has 1, 2, 4, 8, and 16. Because the factor repeats itself, the number is a square of an interger. There are 31 squares of intergers between 1 and 1000 (31 squared being 961 and 32 squared being 1024).

And now for some new ones!

21)
1000! (1000 factorial) has how many zeros at the end?

22)
Which is bigger and by how much? 1000 to the 1001 power or 1001 to the 1000 power? Once you get the solution, can you prove that one divided by the other is close to a famous irrational number?

Sorry for the long post.

Re: Maths problems (interesting ones)
 

I think you made a mistake right there.-- You only have $261, so you have to divide it some how different (even though I think that is impossible).
Or maybe I understood the problem wrong.

I think that would be 700 zeros at the end. (1000/2) + (1000/5) = 700

Great job on the other ones!

Re: Maths problems (interesting ones)
 

Neither have you! There is a second solution to it... Very similiar!

Re: Maths problems (interesting ones)
 

Hi

for the first one, it should have a 0 for every multiple of 10, 100 and also one for every multiple of five that can be coupled with an even number, (with so many thats going to be all of them), another hundred, if you don't count the multiples of ten. This should mean that there are 200 zeros at the end (i think)

For the second answer surely the question is contradictory, if its an irrational number by definition it cant be represented as a fraction (or at least i thought)
so the division cannot represent an irrational number.

1000^1001=10^3003
(1000+1)^1000=10^3000+1000*10^2997+500500*10^2994+ ..........+500500*10^6+1000*10^3+1 (by binomial distributions (only doing the first and last three you can see that the numbers get smaller as it goes on
=10^3000+10^3000+5*10^2994+........+5*10^11+10^6+1
which must be less than 10^3003

Therefore 1000^1001 is bigger

And for the money splitting up one unapiedrs point is easily overcome if you take the 256 bag and replace it with a 1, 2 and 2 bag, so you have 2 $1 bags and 3 $" bags in total) then you use $261, and can give out any amount.

And i cannot find the other solution (im sure ive tried all the ones that might work ad they fall down at the seven), so can you tell me the other answer

oljomo

Re: Maths problems (interesting ones)
 

your right unapiedra, you only have 261 to start with thus you can't do that method.
Its nearly right though

nearly
but you can get it less than that.

Re: Maths problems (interesting ones)
 

hi

silly me you can get away with just a single $5 bag

oljomo

Re: Maths problems (interesting ones)
 

well done :p

Re: Maths problems (interesting ones)
 

Sorry about the irrational number statement, I meant "close to" an irrational number. It is fixed it now.

Also, I fixed my answer to the money bag problem, so it only takes $261.

I am not sure yet what the answer for the factorial is, but I will have it soon. As for the exponent problem, I think you may need to use logs to get the "how many times bigger" part.

Also, I wrote a javascript program to find out the average winnings of the coin game (11). The answer seems to be between $5 and $15, but even with 5000 games played, it varies a lot. To run the program yourself, save the txt as a .html file and open it in your web browser.

Re: Maths problems (interesting ones)
 

Hi

for the log question

1000^1001/1001^1000
=e^ln(1000^1001)/e^ln(1001^1000)
=e^ln(1000^1001)-ln(1001^1000)
=e^1000ln(1000)-1000ln(1001)+ln(1000)
=e^1000ln(1000/1001)+ln(1000)
=368.0633043

therefore one is 368 times bigger than the other
(the reciprocal
is 0.0027169, which is near to e\1000

oljomo

Re: Maths problems (interesting ones)
 

(#58 in this thread)

I wrote a program (see attachement) for a brute force algorithm and I only got one solution: 381654729. With a very high likelihood ;-) there is no second solution.

Carsten

Re: Maths problems (interesting ones)
 

Hi

Thats what i got, i narrowed it down to about 12 possible results, and then tried all of these and that was the only one that i got too work.

oljomo

Re: Maths problems (interesting ones)
 

Ok well I know that there has been debate on the Door issue.

What you are looking at are too completly different events. There are two children ( any thing can be substituted for children typically coins as they are a favorite in most statistics books. Ok now forget about the door you have to people each has a 50% chance of being male and a 50% chance of being female. Thus there is a 25 % chance of both being male 50% chance of one being male and ne being female and a 25% chance of both being female ( the favorable outcome) Thus the answer would be 25 % if the question asked what are the odds of both the door opener and their sibling being female. How ever because gender is completely independent you know that their is a 50% chance that the first is feamle and a 50% chance that the second is female thus .50*.50=.25 (see earlier explination) once you know the gender of the first child the problem becomes 1.00*.5=.5 or 50%.

To disprove the answer being 33% you have to view it as the two children being the door opener and the non opener (as said earlier) rather than being the elder and the younger with the door opener being an additional variable

Example:
opener/non opener
Girl/Girl
Girl/boy
Boy/Girl
Boy/Boy
The last two are invalid because the Opener must be female.

Example2:
There are three combinations of children
2 Girls
1 of Each
2 Boys
Since one must be female the final solution is invalid thus it must be one of two solutions or 50%

What it appears is happening is that those who are arguing 33% dont truly understand probability and are instead repeting the answer to a simmilar problem that is part of Algebra 2 as well as Bio 1 and Genetics
There are 2 children what are the odds that they are both female if both are not male thus there are 3 possible results elder/younger
male/female
female/male
female/female
only one is favorable thus there is a 1/3 or 33% chance

This is the law of independent statistics or something to that effect (I dont remember the name)

This is also what says If you play powerball (Lottery game in which 6 numbers out of 60 are pulled, jackpot is won if all 6 match) the odds are 60*59*58*57*56*55 or 1 in 632,385,600 how ever if you lose the first 632,385,599 times your odds of winning the last time are still 632,385,600

Re: Maths problems (interesting ones)
 

Hi

There is an easier way, tha finds all the answers for certain

take the equations from the question
P+3=n(j-3)
j+n=3(P-n)
rearrange the first, substitute it into the second and you end up with
j+n=3nj-9n-9-3n
j-3nj=-13n-9
j(3n-1)=13n+9
3j(3n-1)=39n+27
3j(3n-1)=13(3n-1)+40
(3j-13)(3n-1)=40
as j and n have to be integers this allows you to work out all solutions which are 1,2,3,7

oljomo

Re: Maths problems (interesting ones)
 

The first and last ones need to be counted twice, because there are two ways to assign the opener and non opener. That is why the answer comes out in thirds.
But only half of the families with the second combination meet the requirements to be included in the count. Again, the answer comes out in thirds.
I believe you're thinking about conditional vs. unconditional probabilities. However, the problem as stated is definitely about conditional probabilities, asking "Given that x is true, what is the probability that y is true?"

Re: Maths problems (interesting ones)
 

Oops, of course you are right there is only one solution! I did this one a long, long time ago, so please understand me ;-) (I am getting old...)

The solution is 381654729 as stated before!

Re: Maths problems (interesting ones)
 

No one has gotton 21 (1000 factorial) correct, but lets get some new questions going. These are not that complex, but we need some new material. This girl/boy debate has gotten a bit out of hand (The answer is 1/2, but I do not wish to debate it).

23) What number comes next: 77, 49, 36, 18...

24) What is the maximum number of bishops you can put on a chess board so than none of them can take another in one move.

Re: Maths problems (interesting ones)
 

24) 14. Line up each of the end rows with bishops on all but one corner of each row. This is assuming all bishops can take all other bishops. If we go white can't take white, then correct answer is 64: put bishops on all squares of the board.

Re: Maths problems (interesting ones)
 

Ok let me try again; 2 and 5 are the prims of 10. The factorial of 1000 can be divided by 2 (1000/2)=500 times. It can also be divided by 5 200 times. therefore 1000! can be by 10 divided 200 times.
Knowing this 1000! has 200 zeros on the end.
Tell me if I am wrong.

Re: Maths problems (interesting ones)
 

I think you're wrong. :)

Every 2nd number is divisible by 2, every 2nd one of those has a second factor of two, every 2nd one of those a third, and so forth. The total number of factors of two, then, is 500 + 250 + 125 + 62 + 31 + 15 + 7 + 3 + 1 = 994.

Similarly, the number of factors of 5 is 200 + 40 + 8 + 1 = 249.

If you had as many factors of 5 as factors of 2, you would have 994 zeroes. But there are only 249, so the answer (unless my analysis is way off) is that 1000! has 249 trailing zeroes.

Re: Maths problems (interesting ones)
 

I believe that one's 8.

Re: Maths problems (interesting ones)
 

You made your point. At least the answer is based on my idea... :)

Re: Maths problems (interesting ones)
 

Yes. Thank you for the hint. :)

Re: Maths problems (interesting ones)
 

Yup, all three are right:

21) 249
23) 8
24) 14

I am starting to get low on problems. But, here is one more:

25)
'SEND
+MORE
MONEY

What numbers do each of the letters stand for? (' is a place holder)

Re: Maths problems (interesting ones)
 

why 8?

Re: Maths problems (interesting ones)
 

23) What number comes next: 77, 49, 36, 18...

7x7=49
4x9=36
3x6=18
1x8=8 ;)

Re: Maths problems (interesting ones)
 

Here's an interesting one I picked up a while back - the solution is pretty neat:

You have been commissioned by the town to place four colored lamp posts in a circular area of a park. The lamps are red, green, blue, and yellow. You must place them inside of the circular area, which will later be fenced in. The town tells you to place the lamps in an orientation so that a person walking all the way around the circular fence looking inward can see every possible arrangement of colors ordered from left to right in their field of view.

Is there any orientation of lamp posts that makes this possible? If not, what is the maximum number of arrangements the person would be able to see? Prove it any way you can.

Top view of circular area attached:

Re: Maths problems (interesting ones)
 

26)

You can only get 14 combinations.

With 4 lights there are 24 combinations:
RGBY RGYB RBGY RBYG RYGB RYBG GRBY GRYB GBRY GBYR GYRB GYBR BRGY BRYG BGRY BGYR BYRG BYGR YRGB YRBG YGRB YGBR YBRG YBGR

The problem can be simplified as such: Draw four points on a plane and then draw a line to connect every two points. Then draw a circle enclosing the four points. The lines will divide the outside of the circle into a number of regions. Since each region represents a different order of lights, the goal is to get the maximum number of regions outside of the circle.

When you put 4 points that do not make any parellel lines or have 3 points on a single line, you get 18 regions. 4 of them are inside of the points, so you can have up to 14 outside the circle. 2 of them will start far outside of the circle, so only 12 will be on the perimeter. It is impossible to get all 24.

Here is a new one that is neat:

27)

A camel works for a banana plantation that is 1,000 miles away from a market. This year, the harvest contains 3,000 bananas that need to be moved to the market. The camel can carry 1,000 bananas at a time, however, the camel eats one banana for every mile it walks. What is the maximum number of bananas from the harvest that the camel deliver to the market?

Re: Maths problems (interesting ones)
 

Bingo. I didn't even think about the two outside the circle...that's interesting.

Re: Maths problems (interesting ones)
 

Problem 21: 1000!

You are right, 249 trailing zeroes. Here is the concret value, calculated by a self written little java program.

1000! = 40238726007709377354370243392300398571937486421071 46325437999104299385123986290205920442084869694048 00479988610197196058631666872994808558901323829669 94459099742450408707375991882362772718873251977950 59509952761208749754624970436014182780946464962910 56393887437886487337119181045825783647849977012476 63288983595573543251318532395846307555740911426241 74743493475534286465766116677973966688202912073791 43853719588249808126867838374559731746136085379534 52422158659320192809087829730843139284440328123155 86110369768013573042161687476096758713483120254785 89320767169132448426236131412508780208000261683151 02734182797770478463586817016436502415369139828126 48102130927612448963599287051149649754199093422215 66832572080821333186116811553615836546984046708975 60290095053761647584772842188967964624494516076535 34081989013854424879849599533191017233555566021394 50399736280750137837615307127761926849034352625200 01588853514733161170210396817592151090778801939317 81141945452572238655414610628921879602238389714760 88506276862967146674697562911234082439208160153780 88989396451826324367161676217916890977991190375403 12746222899880051954444142820121873617459926429565 81746628302955570299024324153181617210465832036786 90611726015878352075151628422554026517048330422614 39742869330616908979684825901254583271682264580665 26769958652682272807075781391858178889652208164348 34482599326604336766017699961283186078838615027946 59551311565520360939881806121385586003014356945272 24206344631797460594682573103790084024432438465657 24501440282188525247093519062092902313649327349756 55139587205596542287497740114133469627154228458623 77387538230483865688976461927383814900140767310446 64025989949022222176590433990188601856652648506179 97023561938970178600408118897299183110211712298459 01641921068884387121855646124960798722908519296819 37238864261483965738229112312502418664935314397013 74285319266498753372189406942814341185201580141233 44828015051399694290153483077644569099073152433278 28826986460278986432113908350621709500259738986355 42771967428222487575867657523442202075736305694988 25087968928162753848863396909959826280956121450994 87170124451646126037902930912088908694202851064018 21543994571568059418727489980942547421735824010636 77404595741785160829230135358081840096996372524230 56085590370062427124341690900415369010593398383577 79394109700277534720000000000000000000000000000000 00000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000000000000 000000000000000000

The upper calculation cost 2,5 seconds. And the calculation for 10000!, which has 2499 trailing zeroes, runs about 2,5 hours on a normal PC.

Best regards
Carsten

Re: Maths problems (interesting ones)
 

Prove that theta is a good approximation for sin(theta) when theta is small. Prove that 1+x is a good approximation for e^x when x is really small. They are both the same problem and involve some advance calculus. The first problem is really important since it drastically simplifies the pendulum calculations if theta is really small.

Re: Maths problems (interesting ones)
 

Starting with the second problem:

Maclaurin series e^x to get
e^x = 1 + x + x^2 / 2! + ... + x^n / n!

Take the limit of that series as x->0 and you'll notice that it equals 1, aka e^0 = 1.

Supposing that x is not zero, but infinitely close to zero, you have 1 + x + (small number) / 2 + (smaller numbers) / (bigger numbers).
The smaller number over a number approaching infinity is zero.
Therefore, you have 1+x for small values of x.

Similarly.... sin (theta) = theta for small values of theta. Do a Maclaurin... and you'll see as x gets increasing small, all other terms of the series except x become 'nothing.'


Not the best proofs by any shot... but they explain the problems. :)

Re: Maths problems (interesting ones)
 

My own quote
That took me all of five seconds to figure out the answer. They are the same thing except the maclaurin series is a taylor series when a=0.

Re: Maths problems (interesting ones)
 

I think this thread can use a little revival.

The unsolved puzzles are:

25)
27)
Any new puzzles would be appreciated. Thank you.

Re: Maths problems (interesting ones)
 

27) 500 bananas

28) Here's one I thought of at breakfast on Sunday, after seeing a watch upside down:
The minute hand of an analog clock is pointing directly at 12, while the hour hand is halfway between 1 and 2. This is not a valid state for the clock to be in, probably because the clockworks have rotated relative to the face (the part with the numbers). How far do I have to rotate the face of the clock, and in which direction, to make the time valid, and what time is it, to the nearest second?

For bonus points, name all the possible times that this arrangement of hands could be, if I rotate the face all the way around.

Also, please note that I'm not an experienced creator of math problems, so the answer isn't pretty.