A nice puzzle for you:

Person A washes a car in x hours. Person B washes the same car in y hours. Question: how long will it take A and B together to wash this car?

It's not that difficult, let's see who wins :)

xy/(x+y)

we have a winner :P

Next riddle:

At noon, the long and short pointer on a clock coincide. When, during the next 12 hours, do they coincide again?

01:05 pm

That is not entirely correct. You can say it more precise.

really? sounds right to me pim...

If you would count on with 1:05:

1:05
2:10
3:15
4:20
5:25
6:30
7:35
8:40
9:45
10:50
11:55

This is not right. You know they don't coincide at 11:55. They don't coincide at EXACTLY 1:05:00 PM.. You can give an expression in parts of an hour if you know what i mean.

ok i see what you mean. yeah it wouldn't be exactly 1:05 but i have to go to school,

laters

ah! I was walking downstairs when i realized the answer. Is it 12am?

wait thats not right, ok nevermind

noon is 12 AM right?? The question was when do they coincide in the 12 hours after 12 AM

no, noon is 12pm

ah okay, sorry, we don't use this am-pm system here.

nvm though, that answer was wrong, it has to be like 1.01 hours 6 minutes and some seconds but its way to early to do the math. have fun ryans

it is not that difficult :) You can write the answer in a fraction, in stead of splitting it up in seconds...

The minute hand and the hour hand will coincide again at 13 hours and 5 mins if you take noon as being 12:00 hours

The hour, minute and second hand will all coincide at exactly
13 hours, 5 mins and 5 seconds or 1 hour, 5 mins and 5 seconds after noon!:)

that's not correct ryans. I believe that there is no time that all 3 pointers coincide


edit: this is obviously wrong..

O.K. I am assuming that the steps of the minute hand are discrete and not continuous, i.e. the minute hand moves one unit clockwise each time the second hand goes past 12. Put up another problem, this one has too many different ways to percieve it. It is dependant on the clock you are using.

I assumed the steps of the minute and hour pointer to be continuous. I should have said that, sorry. Anyway, here is the answer i think is right.

Between 12 pm and 12 am there are 11 coincidings. These are spaced equally from each other because the pointers have constant speed. So 11 coincidings in 12 hours means the first coinciding of the minute and hour hand will occur at 1 1/11 hour, which is 1 hour, 5 minutes and 27 seconds.

Okay, now an easy one..how can you plant 4 trees in such a way that the distance between every 2 trees is equal

NO NO NO

Good effort pim, but the instant the second hand goes past the 12 at 1:05, the minute hand and the hour hand will no longer coincide, for they only coincide at exactly 01:05:00 pm. all 3 hands will only coincide at 2 places, 12 noon and 12 midnight.:)

Originally posted by Pim
Okay, now an easy one..how can you plant 4 trees in such a way that the distance between every 2 trees is equal Plant them in a square with 1 meter sides.
EDIT: wtf was I thinking?:)

you have to climb on a hill that has 3 sides, all are equilateral triangles.;)

Originally posted by ryans
NO NO NO

Good effort pim, but the instant the second hand goes past the 12 at 1:05, the minute hand and the hour hand will no longer coincide, for they only coincide at exactly 01:05:00 pm. all 3 hands will only coincide at 2 places, 12 noon and 12 midnight.:)


No that's not true if you assume continuous movement of both minute and hour hand...at 1:05:00 EXACTLY the hour hand will not be at the 1..but a little PAST it (1/12 of the distance between the 1 and the 2 to be precise), while the minute hand will be exactly at 1 at that time. So these two hands do not coincide then. 27 seconds later though, the minute hand will have moved a bit futher than the 1, and will coincide with the hour hand. Yes at 12 am and 12 pm they all coincide, didn't think of that :)

Another reason why i think they don't coincide at 1:05:00: if you go on with this you will get: 2:10:00, 3:15:00, ...., 11:55:00.
And you know by observation that they do not coincide at 11:55:00...

Originally posted by everneo
you have to climb on a hill that has 3 sides, all are equilateral triangles.;)

That's the solution :)

These are good problems pim, keep going

Ryans: did you read my last post about the clock?

OK, I guess this is math after all.

Three guys are stranded on an island. The only thing there is monkeys and coconuts. They can't catch the monkeys and not knowing when they might be rescued they decide they had best gather all the coconuts on the island to try and survive.

They put all the coconuts in a pile and agree that in the morning they will divide them evenly and should they run out each will just have to suffer the consequences.

That night the first guy wakes up and thinks, "I really don't know if I trust these guys, I'm going to make sure I get my fair share".

He divides the pile into three even piles but comes up with one coconut left over. Being a fair person he decides he couldn't live with himself if he took the extra coconut, so he throws it back in the jungle to the monkeys. Hides his 1/3 and restacks the remaining 2/3 into a single pile.

Later that night the second man wakes and has the same thoughts and goes through the same process ending up with one coconut left over and makes the same decision as the first.

He throws away the extra coconut, hides his pile and restacks the remaining coconuts.

Early the next morning the third man wakes and goes through the same routine. He throws away the extral coconut, hides his pile and restacks the remaining coconuts.

When the sun comes up all three men wake up and go and divide the pile into thirds. This time the pile divides evenly, none for the monkeys. Each takes his share of coconuts and puts them into hiding with his previous pile.

The question is "What are the fewest coconuts that can be on the island".?

Monkeys and coconuts
Depends on the how many coconuts each gets from the final division(how many when they get the even count=x)

total=((((x*3)*3+1)*3+1)*3+1)
total=x*81+13

Persol,

Nice try but you have sufficienct information to determine the total number of coconuts.

106 using vb... but now I'm wondering how I messed up going backwards

Persol,

Good job you have a correct answer. Actually, you can get multiple answers and I have not seen this happen before. If it is set up by algebra it will reduce to the "Fewest Coconuts" that are on the island.

But I didn't specify so you win.

But you have a good number also.

I have re-written the question to keep multiple answers from being correct.

Okay, here's one:

You have jeep sitting at the edge of a 500 mile expanse of desert that you need to cross. Your jeep gets 10mpg and has a 10 gal fuel tank. You can also carry one additional 10 gal drum of fuel.

When you start off, there are no supplies of fuel along your route, but you do have a supply of fuel drums at your starting point.

Questions:
How can you get across the 500 mile desert in your jeep.

What is the minimum number of fuel drums that it will take to do so.

Janus58,

It appears that your jeep can only carry 200 miles worth of gasoline regardless of how much fuel is available at the starting point. :confused:

Tom

Janus58,

I haven't worked out the number but the procedure is to use your tank to carry extra gas down route and have enought to return to get another load of fuel and keep extending your trip.

Back with an answer (I hope soon).

Originally posted by Prosoothus
Janus58,

It appears that your jeep can only carry 200 miles worth of gasoline regardless of how much fuel is available at the starting point. :confused:

Tom

MAcM hit on the method. You drive out part way into the desert and drop off fuel, repeat. This establishes a new refueling station. You use the same trick from the new fuel supply to establish another fueling station deeper into the desert. You keep doing this until youv'e crossed the desert.

The trick is to minimize the amount of total gas you need to start with.

you are the head chef of a new restaurant and you must cook a nine-minute egg for the restaurant reviewer. Just as you prepare to start cooking the egg, your watch stops. As luck would have it, you happen to have two hourglasses, one for seven minutes and one for four minutes. How can you time the nine minute egg?

Janus,

It seems to take twice as many trips every 50 miles. To get gas to the 400 mile marker I count 64 trips on the next cycle I must get gas to the 300 mile mark or another 32 trips. and to the 200 mile mark 16 trips. Then I don't need to drop off gas and retrun, I just keep filling up.

That should take 112 trips at 20 gallons each or 2,240 gallons for 500 miles - Must be driving an SUV @ 4.48 gallons/mile.:D

UpQuark,

Set both hour glasses. When the 4m runs out flip the 4m glass. There will be three minutes left on the 7 min glass. When the 7 min glass runs out you have 1 min left on the 4 m glass. Start boiling your egg. When the 4m glass run out, flip it twice for 4 m each = 9 minutes but it took 16 minutes to serve the inspector and your busted for slow service.:D

How many cubic feet of dirt in a hole 2 ft x 4 ft x 6ft?

Originally posted by MacM
Janus,

It seems to take twice as many trips every 50 miles. To get gas to the 400 mile marker I count 64 trips on the next cycle I must get gas to the 300 mile mark or another 32 trips. and to the 200 mile mark 16 trips. Then I don't need to drop off gas and retrun, I just keep filling up.

That should take 112 trips at 20 gallons each or 2,240 gallons for 500 miles - Must be driving an SUV @ 4.48 gallons/mile.:D

Nope. Your answer is too high. It can be done with less gas.

Originally posted by MacM
How many cubic feet of dirt in a hole 2 ft x 4 ft x 6ft?

Zero.
It is a hole.

Alternate method to the egg boiling....

set both running... when the 4 runs outs flip it over u have 3 in the 7 one.
when the 7 runs out flip it over u have 1 in the 4
when the 4 runs out flip it over u have 6 in the 7 when the 4 runs out again start timing you have 2 in the 7 when the 7 runs out again thats ur 9 minutes up this actually takes 19 mins to complete which is longer than the other solution but yeah still different solution

wow lateral thinking a hole :P haha bit of an oldy that one

one-raven,

You are quick or have you seen it before.;)

None is correct. If it had dirt in it it wouldn't be a hole.

Two identical twin brothers go into a bar. They order the exact same drink. The only difference is one brother drinks quickly, which the other drinks his much more slowly. The one who drinks slowly dies afterwards. the man who drank it quickly didnt die

Why?

It was poison. Both got poisoned, but you only mentioned one death You said that the 'only difference' was drinking speed, which could mean they both died.

ok ill edit that to specify that only the man who drank it slowly died, the man who drank it quickly was fine

There's poison in the ice -- the brother who drinks quickly doesn't allow the ice to melt.

correct Xlock, not particularly hard but a bit abstract well done

What rare characteristic do the following words all share?

Devil
Sinned
God
Warts

Originally posted by patty-rick
What rare characteristic do the following words all share?

Devil
Sinned
God
Warts

they all make words backwords?

lived
dennis
dog
straw

right... hmm well i obviously cant supply a challenge :P

Janus58,


How about 1,230 gallons?

Janus

Okay, here's one:

You have jeep sitting at the edge of a 500 mile expanse of desert that you need to cross. Your jeep gets 10mpg and has a 10 gal fuel tank. You can also carry one additional 10 gal drum of fuel.

When you start off, there are no supplies of fuel along your route, but you do have a supply of fuel drums at your starting point.

Questions:
How can you get across the 500 mile desert in your jeep.

What is the minimum number of fuel drums that it will take to do so.

It requires the use of 83 drums and therefore 830 Gallons of fuel.

I will not tell how I got this answer, but I'll let everyone hang and try to figure it out for themselves.

ryans,

I will not tell how I got this answer, but I'll let everyone hang and try to figure it out for themselves.

I'm going to wait and see if you are right first. Then I'll throw down my pencil.:D

Hmmm... I got 965

Possible spoiler?
(highlight to see)
I claculated backward. You need 20gal at 300m. Therefore 35g@250m... etc...
Every 50 miles, required:
965 485 245 125 65 35 20

(end of highlight)

i got 128 drums so a hell of alot of fuel, i dont see any other way ???

actually a few drums less than 128 like 123-125

My 965 is gallons of fuel... only 97 drums

How did you get your answers Algebra, computer, calculus? I am hanging to see what the correct answer is. I have checked my working and I cannot see any errors, but that is not to say that I did not make any.

Just to repeat, I got 83 drums.

i made mine the old fashion way , a nice diagram

I did it using a calculator, but now that I think about it (spoiler?) using 'sub-stations' every 1 mile instead of 50 may bring the number down(end spoiler)

i would think that would use more... as more back tracking

Yeah, just tried... it does use more. Common ryans, post the solution in white. I want to see the nswer so I can go to sleep:)

therefore, 50 miles is the longest distance between stations possible cause anyfurther and u cant get back

I think every 50 miles would be the optimum distance as it is exactly half the distance that can be travelled on one tank. If you say 1 mile, why not half a mile, why not a centimetre. 50 miles seems to make sense don't you think?

yeah and you use up half of your fuel -1/2 a tank each station u move along

Janus58, don't you think it is about time you put us out of our misery?:)

So far, everyone is still on the high side.

so below 90 tanks?

i have it!!!! 70 drums are needed!!!!

How did you get 70 patty-rick?

i have looked over and it doesnt seem to work any more

if you use 103 tanks at the start then u use 51.5 tanks to get 50 miles into the desert. so u have 51.5 tanks at 50 miles

then to get to 100 miles in you use 25.5 tanks and you have 26 left at 100 miles in.

to get to 150 tanks in you use 12.5 tanks so u have 13.5 tanks left at 150 miles.

you use 6.5 tanks to get that to 200 miles in so u have 7 tanks left at 200 miles.

you use 3.5 tanks to get to 250 miles so u have 3.5 tanks left at 250 miles

use 1.5 litres to get to 300 miles so u have 2 tanks at 300 miles which is enough to get you the last 200 miles.


rules are simple if you have an even/odd/or half number of barrels and you want to get to the next station 50 m away

if you have an even number of barrels you will be left with 1/2 the original number + 1/2 a barrel.

if you have an odd number of barrels you will left with exactly half the number you started with.

if you have 1/2 amount of barrels you will be left with half the original + 1/4 of a tank.

these rules are constant and i dont see how you could get another answer other than what i got

I can't see how this works. The answer I came up with was by progressively penetrating deeper into the desert until there were tanks at 100, 200 and 300 miles, then you can jump across the desert with the 2 tanks you have on board and the ones placed in the desert. The thing you have to find is the cost of placing the tanks in the desert in this configuration. This way I got 83 barrels.

I got 64.

I will post my math in a few minutes.

I hope I didn't miss something and look like a big fat fool! :)

im interested to see it

I DID miss something.

When I went through it I found the same answer as Ryans.

I started with 64 and worked my way up.

I am sure I am missing something here because all you need is two drums at 300 miles, and two drums at 100 miles to make it through.

I decided to post the way I was reasoning it out and maybe someone will see my flaw and find out how to reduce it.

I can't help but think it can be done in just over 64 barrels, but my mind doesn't seem to agree witm me right now.


Start with 80 barrels.

Trip 1 - Drive 50 miles out. Leave full drum. Return to start. (2 drums used. 1 left at 50 mile mark)

Trip 2 - Drive 50 miles out. Leave full drum. Return to start. (2 drums used. 2 left at 50 mile mark)

Trip 3 - Drive 50 miles out. Leave full drum. Return to start. (2 drums used. 3 left at 50 mile mark)

Trip 39 - Drive 50 miles out. Leave full drum. Return to start. (2 drums used. 39 left at 50 mile mark)

Trip 40 - Drive 50 miles out. Leave full drum. No need to return to start, so you still have 1/2 tank of gas left. (40 left at 50 mile mark. 1/2 tank in the jeep.)



Starting from the 50 mile mark, continue outward in the same manner to 100 mile mark.
You are starting with 40 drums and 1/2 tank of gas.

Top off tank.
Now you have 39 1/2 drums and a full tank of gas.

Contuinue as above.
Each trip uses 2 tanks, except the first one because you already have gas.

Trip 1 - Drive 50 more miles out. Leave full drum. Return to start. (1 left at 100 mile mark, 38 1/2 at 50 miles)

Trip 2 - Drive 50 more miles out. Leave full drum. Return to start. (2 left at 100 mile mark, 36 1/2 at 50 miles)

Trip 3 - Drive 50 more miles out. Leave full drum. Return to start. (3 left at 100 mile mark, 34 1/2 at 50 miles)

Trip 19 - Drive 50 more miles out. Leave full drum. Return to start. (19 left at 100 mile mark, 2 1/2 at 50 miles)

Trip 20 - you are at 50 mile mark. No gas and 2 1/2 drums. Leave the 1/2 drum behind for the next poor unfortunate soul.

Now you have 20 drums at the 100 mile mark.


Trip 1 - Drive 50 more miles out. Leave full drum. Return to start. (1 left at 150 mile mark, 18 at 100 miles)

Trip 2 - Drive 50 more miles out. Leave full drum. Return to start. (2 left at 150 mile mark, 16 at 100 miles)

After trip 9 you have 2 drums left at the 100 mile mark. Fill your tank, and take one with you

Now you have 10 drums at the 150 mile mark 1/2 tank of gas.



Trip 1 - drive 50 more miles out. Top off your tank. Leave 1/2 drum. Return to start. (1/2 left at 200 mile mark, 8 at 150 miles)

Trip 2 - Fill tank. Drive 50 more miles out. Leave 1 drum. Return to start. (1 1/2 left at 200 mile mark, 6 at 150 miles)

Trip 3 - Fill tank. Drive 50 more miles out. Leave 1 drum. Return to start. (2 1/2 left at 200 mile mark, 4 at 150 miles)

Trip 4 - Fill tank. Drive 50 more miles out. Leave 1 drum. Return to start. (3 1/2 left at 200 mile mark, 2 at 150 miles)

Trip 5 - Fill tank. Drive 50 more miles out. Leave 1 drum. Return to start. (4 1/2 left at 200 mile mark, 0 at 150 miles)

No need to return after trip 5, so take that 1/2 drum and top off your tank.


You are now at 200 miles with a full tank and 4 drums.

Trip 1 - Drive 50 more miles. Drop off tank. Return to start. (1 left at 250 mile mark, 3 at 200 miles)
Trip 2 - Drive 50 more miles. Drop off tank. Return to start. (2 left at 250 mile mark, 1 at 200 miles, 1/2 tank in your jeep.)


Carry one drum (you have 1/2 tank of gas).

Get to 300 miles, top off your tank, leave 1/2 drum.
Return to start.
Fill your tank.
Drive to 300 mile mark.
Top off your tank.
Drive 100 more miles.
Run out of gas.
Die of exhaustion from driving back and forth for weeks on end with no food or water.
Be eaten by vultures 100 miles shy of your goal.

I wrote a program in GWBASIC to count the trips to stock weigh stations until I had enough fule to complete the trip without returning.

But apparently I made some bad or unecessary drops getting there, in the program.

You know garbage in garbage out.

Okay, Here's a hint that might help:


The fuel dumps won't be separated by equal distances.

If no one gets it by this evening, I'll post the answer.

Janus58,

The fuel dumps won't be separated by equal distances.


Opps. That blows my program. :bugeye:

patty-rick,
both of your solutions work. there is however one solution requiring only nine minutes, but, me not being specific gives you the correct answer.

UpQuark,

patty-rick,
both of your solutions work. there is however one solution requiring only nine minutes, but, me not being specific gives you the correct answer.


So what is it? Or are you waiting on others to try and find your 9 min solution?

Ok, got down to 42 drums
Starting Dist - Gas Needed
0- 418
47- 225
95- 124
147- 66
193- 43
251- 25
300- 20
500- 0

I got a formula for how much gas it takes to move X gas, and then calculated an overall mpg (which includes dropped gas). Then I just started at 300, and moved back the most efficient distance. 42 is a little high, as I rounded up.

turn over both hourglasses. when the 4 minutes run out, turn it back over. after 7 minutes, there is 1 minute of sand left in the 4 min glass. restart the 7 minute timer, after 1 minute elapses (the 4 minute runs out again) 8 minutes have passed, 1 minute has passed through the 7 minute hourglass. turn the 7 minute hourglass back over and 1 minute of sand will pass through arriving you at exactly 9 minutes provided you compensated for the time it took you to flip them.

UpQuark,

Wouldn't it be easier to convience him to just have them scrambled?

Good Job On Radiactive Waves.
There is the correct solution for the shortest amount of time to cook the nine minute egg.

Scrambled is alright, provided the reviewer supplies the cheese:D

Originally posted by Persol
Ok, got down to 42 drums
Starting Dist - Gas Needed
0- 418
47- 225
95- 124
147- 66
193- 43
251- 25
300- 20
500- 0

I got a formula for how much gas it takes to move X gas, and then calculated an overall mpg (which includes dropped gas). Then I just started at 300, and moved back the most efficient distance. 42 is a little high, as I rounded up.

BINGO!.

Here's the solution:

Jeep solution (http://home.teleport.com/~parvey/Jeep.html)

Before you read it , I think it should be fair to warn you that the solution has a problem, in terms of practicality, that I didn't notice until I wrote up the above explanation of the solution. It requires that you drop off more than 10 gals of fuel a trip most of the time. This is okay, in that you have the extra gas in the jeep's gas tank, The problem is, where are you going to put it ? If the drum holds only 10 gals, it's already full.

The only way I see around this is to assume that the drums can hold more than 10 gals, but you are still limited to carrying only 20 gallons total with the jeep (A weight load factor?). Given this extra consideration, the above solution will work.

Okay, here's another one. And I guarantee this one has no hidden bugs.

You have a solid sphere with a hole drilled all the way through it and passing through its center. The length of the hole, from lip to lip is 10cm.

What is the volume of the remaining solid portion of the sphere?

You have a solid sphere with a hole drilled all the way through it and passing through its center. The length of the hole, from lip to lip is 10cm.
When you say lip to lip... do you mean the length of the hole?

(|||) where |=10cm?

or
(|/|) where /=10 cm?

it requires that you drop off more than 10 gals of fuel a
trip most of the time. This is okay, in that you have the extra gas in the jeep's gas tank, The problem is, where are you going to put it ? If the drum holds only 10 gals, it's already full.

That's a bit unfair don't you think, you should have made an amendment to the problem so that we could have tackled it from these conditions.

Yeah, you had the first completely correct answer... but the most fuel you drop off is 10.4 gallons... which seems reasonable.

Strictly you can't, it wasn't specified in the problem, but don't worry about me folks, I am just bitchin'. Good problem though, stumped everyone for a while.:)

I suppose you want a general answer to this, as the question Persol raised makes a big difference.

You also either need the radius of the shere or the radius of the hole.

Janus58,

Okay, here's another one. And I guarantee this one has no hidden bugs.

You have a solid sphere with a hole drilled all the way through it and passing through its center. The length of the hole, from lip to lip is 10cm.

What is the volume of the remaining solid portion of the sphere?


Unless I'm missing something here, I believe you must give us the diameter of the hole. I haven't bothered yet to compute the zones of one base but just assuming two different hole diameters I get different sphere diameters and net volumes that seem to indicate hole size will affect the answer.

Just in case not, I'm still calculating the two zones.


BACK: Assuming a hole diameter of 1 cm the solid volume equals 124.9945 (with some rounding and including the two spherical zones at the ends.

Assuming a hole diameter of 2 cm I get 107.103677


What is the hole diameter?

Originally posted by Persol
You have a solid sphere with a hole drilled all the way through it and passing through its center. The length of the hole, from lip to lip is 10cm.
When you say lip to lip... do you mean the length of the hole?

See what happens when you read quickly;)

Ok, the answer is 523cm^3.

Persol,

A calculation of 4,188 cm^3 = a sphere with a radius of 9.99937 without any hole?

10 cm lip to lip should be length of the hole. The radius is therefore something greater than 5 cm as a function of the hole diameter.

For anyone still thinking about coconuts, the fewest coconuts on the island are 25.

Lol... yeah, I realized that right after posting. I went back and changed it a minute before you posted:)

My basic reasoning here is that the volume is constant, so I calculated it with a hole of radius 0. As the hole grows, the sphere does too. To be any simple kind of answer, it has to be constant.

Persol,

Found my problem. TI-55. Boolean, gotta press = more often.

Originally posted by Persol
Lol... yeah, I realized that right after posting. I went back and changed it a minute before you posted:)

My basic reasoning here is that the volume is constant, so I calculated it with a hole of radius 0. As the hole grows, the sphere does too. To be any simple kind of answer, it has to be constant.

Correct.

Well, I screwed it up but it is really strange. I calculated Persol's first wrong answer backwards and got the right radius but running the TI-55 forwards I was getting incorrect answers for volume.. Wierd. I was looking to see if indeed the volume stayed the same but my results were saying they didn't. When actually they do.

Good problem though Janus58.

Disregarding the obvious that a half chicken doesn't exist, etc:

If a chicken and a half, lays and egg and a half, in a day and a half.

How many chickens does it take to lay a dozen eggs a day?

Originally posted by MacM
Disregarding the obvious that a half chicken doesn't exist, etc:

If a chicken and a half, lays and egg and a half, in a day and a half.

How many chickens does it take to lay a dozen eggs a day?
12

one-raven,

Nope.

12/1.5 = 8

8 x 1.5 = 12 but that is in 1.5 days.

Originally posted by MacM
one-raven,

Nope.

12/1.5 = 8

8 x 1.5 = 12 but that is in 1.5 days.

got me!

*tries to think of a mathematical riddle*

hmmmmmm

18 chickens

ryans,

Winner.

Question on mathematics.

As a magician (40 years ago) I did many tricks by using a unique feature of 9's.

Why is it that digits any multiple of 9 will add to nine?


i.e. - 9 x 71485 = 643,365

6
4
3
3
6
5+
---
27


2
7+
----
9

or 9 x 9 = 81

8
1+
----
9


And fractals of 9 also equal fractals of 9.


3 x 1 = 3

3 x 2 = 6

3 x 3 = 9

3 x 4 = 12 = 3

3 x 5 = 15 = 6

3 x 6 = 18 = 9

3 x 7 = 21 = 3

3 x 8 = 24 = 6

3 x 9 = 27 = 9

etc, etc., This 3, 6, 9 sequence continues forwver.

Originally posted by MacM
As a magician (40 years ago) I did many tricks by using a unique feature of 9's.
:confused: :eek:

everneo,

Cake in a hat, never ending ping pong balls from a card.

In "Cake in the Hat", I borrow a hat (deep like coyboy,etc) from the audience. My wife held it upside down and I crack an egg an put in it, pour a couple pints of milk in it, a cup of flour, etc. Stir it with my wand, heat it with a candle and wahla. I would dump out cup cakes or doughnuts.

The hat (unless I was nervous), suffers no damage.:D

PS: You don't want to play jpoker with me either. I can shuffle a desk of card until you say OK, pass the cards around for them to be cut by observers and then deal six full houses and an Ace high spade Royal Flush. Guess who gets the flush?

Guess who gets kicked out of the casino....:D

ryans,

Not exacatly, those suckers are sharper than me. I tried some stuff in Black Jack once and was up $120.00, They gave me strange looks and the next thing I new I was holding a cup on the sidewalk selling penciles.:D

Not actually but they did empty my billfold.

Originally posted by MacM
Well, I screwed it up but it is really strange. I calculated Persol's first wrong answer backwards and got the right radius but running the TI-55 forwards I was getting incorrect answers for volume. How did try calculating the volumes without using integrals? I don't know of any simple geometric way to do it, which is why I thought it must be constant.... riddles don't using integrals:)

persol,

Vt = 4*Pi*r^3/3

Volume Hole = Pi * r^2 * l

Volume of Zones of one base = 2 * Pi * h^2 * (r -h/3)

h = r - l

Ahhh. That works too... I tried simplifying symbolically and got something wrong in there.

Originally posted by MacM
Question on mathematics.

As a magician (40 years ago) I did many tricks by using a unique feature of 9's.

Why is it that digits any multiple of 9 will add to nine?


i.e. - 9 x 71485 = 643,365

6
4
3
3
6
5+
---
27


2
7+
----
9

or 9 x 9 = 81

8
1+
----
9



This has to do with the fact that 9 is the highest digit value in decimal. The same thing happens for 7 in octal, and F in hexidecimal.

First off, we'll show that any digit multiplied by 9 will result in a product who's digits add up to nine.

start with A*9, with A being any single digit number.

this is the same as A*(10-1) or A*10 - A

since A stands for a single digit we'll write A*10 as A0 which gives us:

A0
-A

Subtracting the 'ones column we have 0-A , borrowing the 'tens' column we get 10-A for our 'ones' digit of the product.

subtracting the 'tens' column, we have (A-1 - 0) since we borrowed. earlier. giving us the product digits of (A-1) and(10-A)

adding them gives (A-1+10-A) , the A's cancel out leaving 10-1 = 9

two digit numbers work like this:

AB*9

AB*(10-1)

AB0
-AB

if B>A this gives digits of (A),(B-1-A)&(10-B) added together:

A+B-1-A+10-B : A's and B's cancel out, Leaving 10-1 =9

B<A gives digits of (A-1),(10+(B-1)-A) & (10-B) added together:

A-1+10+B-1+10-B A's and B's still cancel out, leaving -1+10-1+10 or (9+9)

We've already shown that adding the digits of the sum of 9+9 or 2*9 will equal nine.

three digits:

ABC * 9

ABC *(10-1)

ABC0
-ABC

..... and so on

Its easy to see that no matter how many digits we have:

The Individual digits(A B C's etc) cancel out

The one's column produces a 10 which results in a -1 somewhere down the line.

Every 10 produced by borrowing creates a -1 , meaning after you cancel out the individual digits, you are always left with at least one (10-1) pair each which reduces to a 9.

result: the sum of the digits of any product of 9 equals 9 (in decimal)

Janus58

Good to see a semi-rigoous algebraic proof.

Of course you probably could have written a thesis on this, stating that A,B,C... are elements of some field yada yada yada, but this is proof enough for me.

Good stuff:)

Janus58,

The same thing happens for 7 in octal, and F in hexidecimal.


Interesting thanks. One of the tricks is to add any mixed seven digit number (1639425) across as an addition column.

1
6
3
9
4
2
5+
---
30

Subtract the answer from the original number.

1639425
.........-30
------------
1639395 ; reverse it and place it below and add.
5939361
-------------
7578756; added = 45 (A multiple of 9)

Now without having looked at the numbers you told them to perform, have them circle any digit and add the remaining in the final added answer:

.............(8)
i.e. 757....756 = 37. You pause and think. The next higher multiple of 9 over 37 is 45:

"You circled an 8". (Note you can't tell the difference in them circling a "0" or a "9" so you simply say "0" or "9" if that happens.


Another one: Write down the number 1,089 on a small piece of paper and put it in some hard to reach place. An envelope, the guests shirt pocket, etc.

Have them write a mixed three digit number (no picket fences, i.e. 111 or 222, etc.)

i.e.: 792

Have them reverse the number. If it is larger in value place it above the original or if smaller place it below.


792
297 - Carry all "0"'s i.e. 190 becomes 091, etc. and subtract
-------
495 Now reverse this answer and add to the number


495
594 +
-------
1,089 Have them open the hidden paper.

Originally posted by MacM
Janus58,




Interesting thanks. One of the tricks is to add any mixed seven digit number (1639425) across as an addition column.

1
6
3
9
4
2
5+
---
30

Subtract the answer from the original number.

1639425
.........-30
------------
1639395 ; reverse it and place it below and add.
5939361
-------------
7578756; added = 45 (A multiple of 9)

Now without having looked at the numbers you told them to perform, have them circle any digit and add the remaining in the final added answer:

.............(8)
i.e. 757....756 = 37. You pause and think. The next higher multiple of 9 over 37 is 45:

"You circled an 8". (Note you can't tell the difference in them circling a "0" or a "9" so you simply say "0" or "9" if that happens.


Another one: Write down the number 1,089 on a small piece of paper and put it in some hard to reach place. An envelope, the guests shirt pocket, etc.

Have them write a mixed three digit number (no picket fences, i.e. 111 or 222, etc.)

i.e.: 792

Have them reverse the number. If it is larger in value place it aboe the original or if smaller place it below.


792
297 - Carry all "0"'s i.e. 190 becomes 091, etc. and subtract
-------
495 Now reverse this answer and add to the number


495
594 +
-------
1,089 Have them open the hidden paper.

As an somewhat out of practice Amateur magician myself, I'm fairly familiar with such tricks, though I didn't really go in for the Mental Magic type tricks myself.

ryans,

Another one that will get you in trouble in the casino.

I can shuffle a deck of cards to your satisfaction. I'll fan the cards and look at the color sequence. Then hand you the deck and have you deal the cards face down into two seperate piles based on your instinct of "Is it red or black".

When you finish I will let you turn over one stack while I turn over the other.

Guess what they will be in perfect color in each stack!. The dealing is totally at your will. that is on impulse 1 here, 3 there, 2 here, 5 there, 1,1, 4, etc.

It really is impressive. Maybe we'll run into each other sometime. I'd love to show you that one. (Not how it is done of course).:D

Janus58,

What type of things did you do? Cards, general silks, rings, illusions (i.e - levitation)?

I also did a limited amount of hypnosis. Not on stage but at private parties, etc. My wife and I work on stage for about 3 years while in Europe.

OUr car broke down once and we had to ride the Blue Goose to the show wearing an extended tail tux. That was fun.

PS: For those not familiar with the term "Blue Goose". It refers to a bus line in France's public transportation.

Originally posted by MacM
Janus58,

What type of things did you do? Cards, general silks, rings, illusions (i.e - levitation)?

I also did a limited amount of hypnosis. Not on stage but at private parties, etc. My wife and I work on stage for about 3 years while in Europe.

OUr car broke down once and we had to ride the Blue Goose to the show wearing an extended tail tux. That was fun.

PS: For those not familiar with the term "Blue Goose". It refers to a bus line in France's public transportation.

A mixed collection of close up magic including the things you've already mentioned, plus stuff that could be adapted to a small stage/audience. The largest illusion I did was when I built a Sword box. I only did a few small stage shows,(for the some of the nearby Granges). And once entertained the kids at Camp Easter Seal. That was arranged through a friend who was working there for the summer.

Janus58,

Kids are great - ahhhh. I was doing my ping pong balls from a card routine for a Boy Scout Troup once, placing the balls on a rack as I produced them and had about 18 on the rack when they couldn't stand it any more.

They come running up on the stage, knocked over the rack and I had ping pong balls going everywhere. All you can do is laugh.

I got started as a kid because we had a neighbor back on the farm that was a pro and he gave me an "Abbot's" catalog. I used to read the catalog like a comic book. Fascinated by all the stuff you could do. Never could afford the larger illusions though.

Do you do the "Red & Black" pile cited above? If not I can send you a PM on how to do it.

What's the next number in the following sequence?

1
11
21
1211
111221

When does a 4 occur for the first time?

This is not that difficult :)

Working on it.

Got a hint?
:)

It's very hard to give a hint without giving the solution :) Try not to think too far.

Originally posted by Pim
A nice puzzle for you:

Person A washes a car in x hours. Person B washes the same car in y hours. Question: how long will it take A and B together to wash this car?

It's not that difficult, let's see who wins :)

I am going to go back to this. Although the guy won, he did not give a reason for it. So, being my natural self, I had to work it out myself.

Let us setup three equations. The first two describe how long is takes persons A and B to wash a car.

T_A = x*c_1
T_B = y*c_2

Because they are washing the same car, c_1 + c_2 = 1 (one car)

Assume they wash different parts are the car and finish at the exact same time.

T_A = T_B
x*c_1 = y*c_2

c_2 = 1 - c_1

x*c_1 = y*(1 - c_1)

Solving for c_1, we get
c_1 = y/(x + y)

Because they are finishing at the same time,
T_A = T_B = (x*y)/(x+y)

James Sibley

Originally posted by Pim
What's the next number in the following sequence?

1
11
21
1211
111221

When does a 4 occur for the first time?

This is not that difficult :)

1
11 (one 1)
21 (two 1)
1211 (one 2 one 1)
111221 (one 1 one 2 one 1)

Each next number describes the previous set of numbers.

The pattern continues:

312211
13112221
1113213211
31131211121221
13211311123112112211
11121221123112132112212221

I hope everyone caught on to the pattern. Notice we have yet to see a 4 in there.
Assume that after a finite number of iterations, we find a 4. The number 4 describes how part of the last sequence was assembled.
Let us work with this exampe and work backwards.

41 (this means in the last seqeuce, there were four 1s
1111 (this is part of the last sequence.)
11 (because the last seuqnce says one 1 one 1...)

Ok, we have worked backwards. Let us work forward now.

11
21
1211
We have a contradiction. 1211 != 41

The other example would be something like:
14
4n where n is some number. This case has already been proved.

James Sibley

James: Yes that is the correct answer to the sequence, a 4 will never occur :)

The car wash riddle can be solved more easily I think.

Person A washes the car in x hours, that is 1/x car per hour. Person B washes in the same way 1/y car per hour. Together they wash 1/x + 1/y car per hour. This means it will take them
1/(1/x+1/y) = x*y/(x+y) hours to wash the car :)

Originally posted by Pim
James: Yes that is the correct answer to the sequence, a 4 will never occur :)

The car wash riddle can be solved more easily I think.

Person A washes the car in x hours, that is 1/x car per hour. Person B washes in the same way 1/y car per hour. Together they wash 1/x + 1/y car per hour. This means it will take them
1/(1/x+1/y) = x*y/(x+y) hours to wash the car :)

lol. Yes. Much much easier. I miss the obvious. :p

Thank you,
James Sibley

Pim,

What is your reason why 4 will never occur?

James Sibley

Because a 4 would only occur if a number would be 4 times in the previous element of the sequence, which is not possible because this can always be simplified.

Example: 11 gives 21 and not 1111.

But you said it very good in your post :)

Originally posted by Pim
Because a 4 would only occur if a number would be 4 times in the previous element of the sequence, which is not possible because this can always be simplified.

Example: 11 gives 21 and not 1111.

But you said it very good in your post :)

That was the reason I was trying to get to. I have a hard time explaining things. I may have left out one or two key points in my explaination but with some thinking, I am sure anyone can fill them in.

:)

James Sibley

How about this...

x^2 = x + x + x + x + ... (an x number of x's)
d/dx [x^2] = d/dx [x + x + x + x + ...]
2x = 1 + 1 + 1 + 1 + ...
2x = x
2 = 1

Hehe... obviously something is wrong with this. Can anyone find it?

:)

James Sibley

What

x^2 does not equal x+x+x+x+x....That's what's wrong with it

x^2=x.x

Originally posted by ryans
What

x^2 does not equal x+x+x+x+x....That's what's wrong with it

x^2=x.x

Good job :)

You solved it :p

James Sibley

(Not math)

Father and son are in a car accident. They are both severely hurt and need medical attention. They are rushed to the hospital. When the son is on the operating table, the doctor comes in and says: "I can not operate on this boy, this boy is my son."

Who is the doctor?


(as allways..should be easy but so many miss it ;-)

the boys mother.

The doctor is his mother.

offcource :-)

I have tried this one on a lot of my collegues and I think it is all the "boy" and "son" talk that throws them of. "grandfather" is a common answer hehe.. but It's also the fact that a lot of people think that the doctor automaticly is a man :-D

I heard that one at uni during a psychology class.

Still driving a car through the desert...

I have an elegant solution of 45 barrels using equally spaced stations of 50 miles, never transferring over 10 gallons, and arriving with an empty tank.

The algorithm is simple:
Assume 1 station every 50 miles.
On each trip into the desert, fill up your tank from each station you come to, until you reach an empty station. Drop off the spare drum at this point and turn around, filling up again from each station (you will always transfer 1/2 a tank).
Repeat until you drop off a drum at the 300 mile mark.
Repeat again until drop off a drum at the 200 mile mark.
Repeat again until you drop off a drum at the 100 mile mark.
Drive through the desert, reaching your goal with an empty tank!

There's a nice binary pattern. It takes 1 trip to get a barrel to the first station, 2 trips to the second, 4 for the 3rd, etc. Each new station reached means consuming all fuel at previous stations.

Placing the 300 mile drum costs 32 drums.
Placing the 200 mile drum costs 8 drums.
Placing the 100 mile barrel costs 2 drums.

42 drums, plus the three placed, gives 45. A little over the 42 solution, but neater and adheres strictly to the rules.

No no no....this car thing is easy. You stop by the convenience store 10 miles away and pick up a whole bunch of rope, maybe trade them some of your unlimited fuel or something, then you go back to your fuel dump, tie up a whole bunch of barrels and drag them through the desert.

Bloody pragmatist! :D

How can you place a tank at 100 miles in 2 tanks, this is impossible.

you have 1 full tank of fuel and carry another full tank. I tank gets you a hundred miles.

Drop 1 tank off at 50 miles.

50 miles out, 50 miles back, 1 tank used, 1 dropped off

Fill up your tank.

Go to 50 miles. Use 1/2 a tank, fill up half with tank at 50 miles
1/2 tank left at 50 miles.

With full tank go to 100 miles and drop off tank and go back to 50.

I tank at 100 miles. Empty tank filled with half tank left at 50 to get you home.

Thus in total you have used 3 tanks to drop off one tank at 100.

Now you may think wow, 1 tank extra, big deal, but the number of tanks used to place tanks further away increases exponentially, and so it magnifies. Being totally within the rules I used this algorithm to show that you used something like 83 tanks. Nobody found less by working totally within the rules

As I am usually right, I believe there is a mistake in your algorithm for finding 45 tanks working totally within the rules, so you better go back and check!

This has used 4 tanks.:)

That indeed seems correct, however I'd say it would be more entertaining to harness the vultures circling around you. First, pretend to be dead...when they come down, take some of that convenience store rope, tie it around their necks (we know that the number of vultures in a desert is infinite. When you kill one, 5 more appear from nowhere. The laws of conservation don't hold for vultures), kill one, eat it for dinner. Repeat until you have captured enough to fly you and your jeep out of the desert.

ryans, on this occasion you are correct.

This is a nice one:) How does this work:

http://mr-31238.mr.valuehost.co.uk/assets/Flash/psychic.swf

I like that one. Very clever presentation of maths magic :)

Pete: you discovered how it works? It's not that hard! :)

The number you get finally is always a multiple of 9. 9 and its multiples have the same symbol. that is the symbol is already decided and that of 9.

Yes, that's what i figured out too everneo :)

Time for a new riddle:

In an alley two ladders are placed crosswise as in the image. The lengths of the ladders are respectively 2 and 3 meters. They cross at 1 meter above the ground. What is the width of the alley?

http://puzzle.dse.nl/complex/ladder.gif


edit: I renamed the thread, because it has nothing to do with car washing anymore :p

edit 2: Ah, i can not seem to change the thread's title..:bugeye:

Pim,


2.309 meters. (Excluding the affect of the thickness of the ladders).




Knowing to believe only half of
what you hear is a sign of
intelligence. Knowing which
half to believe will make you a
genius

MacM: that's obviously not the right answer since one of the ladders is 2 meters long, so the width of the alley has to be smaller than 2.

Indeed the thickness of the ladders can be neglected, should have said that, thanx :)

Pim,

I agree, I should have read closer. I looked at the 2m lable as segment and just did trig from there. Will have to go back an think on this one.:D






Knowing to believe only half of
what you hear is a sign of
intelligence. Knowing which
half to believe will make you a
genius

Pim,
if i say the width is 1.231m approx., you should not be surprised.;)

everneo: that is indeed the correct answer. Would you care to explain how you obtained it? :)

the link to the answer is also available in that page. sorry i could not overcome the temptation..;)

*ducks and runs away*

*deleted*

I don't understand your last post?

me too. i'm going to delete my 'last post'.

Well, as everneo pointed out the answer is on the same page as the picture..(thnx for that). There is also an explanation of how it is derived. It is in Dutch though, good luck :p

Well, i hope people will try to solve it without looking there :)

you are right. let me edit it again.:D

The answer is 1.2311857 meters.

How did I do it?

First, I set up the distance forumla for both lines.
Then, I made a linear equation for both.
I knew f(x) = g(x) = 1, so I solved for 'x'.
I set both x's equal to eachother, and solved for 'w,' the width.

As far as anyone wanting to see the work, there is not much to it. Like I said in a previous thread, I visualize most stuff...

With all the work, you land to this:

1/sqrt(9-w^2) + 1/sqrt(4-w^2) - 1 = 0

James Sibley

Back to the desert! The Car Problem(tm):

Geez, this has taken me forever. Using calculus and separating into variables, dg/dx (change in gas per unit distance) = - (amount of gas needing to be transfered) / 20, the transfer size, * the MPG. The rate is defined to be negative. I am transfering the whole load as I go, shuttling it back and forth a meter at a time, so the number of trips I take is almost entirely dependent upon how much gas I have, not how far I am traveling. Then


200/G dg = - dx

200 ln g = - x + C, a constant

using the point (g, x) = (20, 300), having ... is that where I screwed up. Well, the latest place. (I was off by a factor of 10, couldn't figure out why, etc. etc., and now I see I did it in another place too.) 20, 300, because once you get there you can just take off and can stop worrying about this back-and-forth crap. Anyway, you get

200 (ln 20) + 300 = C

Solving for g from the previous equation,

g = e^([C - x] / 200)

x = 0 should give the original amount of gas before setting out with my tons of fuel tanks.

g = e^( [200 (ln 20) + 300] / 200 - 0 )
= e^( ln 200 + 300/200 ), where the ratio 300/200 is derived from the "final" location of 300 miles and the product of 20 gallons * mpg.

g = 20 e^1.5

Which is, presumably, incorrect. (It's about 89 gallons.) Can anyone tell me what I did wrong?

If you say that you can only leave 10 gallons at a time, your fuel load is effectively 10 gallons (the unused 9.99 gallons just stay in your tank), and you get g = 20 e^(3/1) = 401.

Just to doublecheck -- this was when I was still getting "reasonable" values of about 430 and 460 gallons for the 20/10 tank size calculations, I tried writing a program for it. :)

I based it off of Persol's sequence, so I'd get 20, 35, 65... gallons needed at 300, 250, 200... miles respectively, working backwards. Here is the relevant code, after literally hours of finding at isolating bugs:


Const curMileage As Currency = 10
Const curLoad As Currency = 20
Dim curMyLocation As Currency
Dim curStepsize As Currency
Dim curMyGas As Currency

curStepsize = Val(txtInputStep.Text)
curMyGas = 20
curMyLocation = 300
Do Until curMyLocation <= 0
curMyGas = curMyGas + DeltaGas(curMyGas, curStepsize)
curMyLocation = curMyLocation - curStepsize
Loop


Private Function DeltaGas(curMyGas As Currency, curStepsize As Currency)
'calculate gas needed to iterate the step

Dim curDelta As Currency

'stepsize / mileage = gas per trip
'tip: next time, dimension #of trips as a separate variable!

If Round(curMyGas / (curLoad - curStepsize / curMileage)) < curMyGas / (curLoad - curStepsize / curMileage) Then
curDelta = curStepsize / curMileage * (2 * (Round(curMyGas / (curLoad _
- 2 * curStepsize / curMileage) + 1)) - 1) '2 * single-trip = roundtrip for TotalLoad/perload,
'then subtract one because don't go back.

ElseIf Round(curMyGas / (curLoad - curStepsize / curMileage)) > curMyGas / (curLoad - curStepsize / curMileage) Then
curDelta = curStepsize / curMileage * (2 * (Round(curMyGas / (curLoad _
- 2 * curStepsize / curMileage))) - 1) 'rounding is STUPID in VB! have to do two cases since I don't know the language very well.

Else
curDelta = curStepsize / curMileage * (2 * curMyGas / (curLoad _
- 2 * curStepsize / curMileage) - 1)
End If
DeltaGas = curDelta
End Function


It's in Visual Basic -- sorry, hehe, but it's the only programming language I am learning. btw, I managed to get Persol's 965 value somewhere along the way, but that was without rounding up to the nearest whole number of one-way trips... grrr. Something like that. It's why I had to code the stupid rounding bits in the code.

Anyway, some sample values are:
Stepsize (distance between stations) | gallons required at start:
0.5 | 422
1 | 425
5 | 462
10 | 508
20 | 735
25 | 566.67 (has to do with the non-linear nature of it. how many refueling trips to make, how many total stations, etc. One more station can make a significant difference.)
30 | 740
33 | 1019
40 | 1532
49 | 2348
50 | 1240
59 | 4174
60 | 2150
75 | 4385
90 | 190910
98 | 123,774,510
99 |1,990,297,029.8

Kind of surprising, since the peak of distance * fuel deposited does peak at 50 miles. I will not guarantee these values are accurate, but until someone finds a hole in my logic or coding ineptitude I will view them as precedent. Too bad my calculus is faulty... gr.

peace. :m:

It takes 86 drums to cross the desert.

3 drums to place a drum at 100 miles.
15 drums to place a drum at 200 miles.
63 drums to place a drum at 300 miles.

3 + 15 + 63 + 3(drums at each station) + 2 (drums from starting point) = 86 drums of fuel.

This follows the method that ryans posted earlier.

The problem with the formula to get 42 drums of fuel is the distance unit. The maximum distance you can travel to allow you to shuttle fuel between stations is 100 miles, not the 200 miles that was used.

Phoenix.

You are right; the original equation should have been dG = (2 * dx) (G/20) [<=the number of trips taken, no units] * 1 gallon / 10 mi.

Unit analysis is miles * gallons / miles.

And that is assuming that the amount of gas spent in shuttling fuel around is negligible when compared to how much there is in all, since in reality less trips would have to be made.

That still gives an end equation of 20 ln (e^3), or the 401 gallons from the "conservative" estimate. This is still different from the one from the program, lower than the answer on the website, and much lower than anything anyone else was able to come up with!

so what am I doing wrong? :)

This is another problem i have been thinking about for a while.

Let a coordinate system be given. A man is standing in point (0,0). His dog is standing in (0,2). The man starts walking along the positive x-axis with constant speed. The dog starts walking in the direction of the man with the same speed. The dog hence walks along a curved line. The question now is if an explicit expression can be determined for this curved line, that is, a function y in terms of x.
What do you think?

No response to my method... lol. Oh well. I am curious how everyone else did it. I would like to see other methods.

As for the man/dog problem... I am thinking about that one. I got something close...

something like f(x) = sqrt[ (r/r_1)^2 - 1 ]*x + 2

I just have to get rid of r/r_1 and convert it to something involving x...

JS

4DHyperCubix: You mean your solution to the ladder problem? I think it's a very elegant way to do it like that :) I used geometrical properties of the triangles in the picture to compute the width. Don't feel like posting that here though :p

What kind of geometrical properties? Knowing that will be good enough for me :)

How far along have you got with that man/dog problem?

Originally posted by 4DHyperCubix
What kind of geometrical properties? Knowing that will be good enough for me :)

How far along have you got with that man/dog problem?

Well, i don't know the english word for it..i think it's something like congruent triangles? Does that makes sense? ("Gelijkvormig" in Dutch)

I haven't been able to come any further with the dog-man problem. I tried to determine the equation for the derivative of the curve the dog travels, but that is not that easy since the dog moves too...

Gentlemen,

I guess I am just thick even after coffee. Can sombody explain to me why we are talking about a curve in the dog/man problem?

It seems to me that the line created is a linear or straight line since both have the same speed. It matters not what direction and for that matter that they have the same speed.

Plotting coordinates of motion of two uniformly moving bodies seems to result in a straight line.:o

What am I missing here?

MacM: since the man is moving, the direction of the dog changes every moment, and thus the dog walks along a curve..If it would be a straight line, the dog would walk in the same direction all the time. which does not make sense because the man is not at the same point all the time, he is moving

Man&Dog problem:

I get a set of parametric differential equations that I have no idea how to solve:

(dy/dt)² + (dx/dt)² = 1
dy/dx = y/(t-x)
Initial condition: (x,y,t)=(0,2,0)

Later I will try posting what I know about the problem.

As for right now, maybe knowing that the tangent line to every point on the curve passes through the point that the man is on at the time the dog is at the given point. Maybe that makes sense?

James Sibley

Pim,

Got it now thanks.

After searching a bit on the internet i found that the man-dog problem is a known problem, there was also a solution given which i never would have found. I will not post it here yet though :)

The dog cannot continuosly adjust his position function of the movements of the man,he can do that only in a discrete manner let's say at very small intervals dt.For example initially the dog walks toward the point [0,0].At dt he observe that the man is in the point [vdt,0] so that he walks now toward it and so on.If dt is chosen as being small enough [let's say dt=0.1 seconds] then an interpolation approach [using Lagrange's method] is possible.y(x) is seeked in the form of a polynome,the number of points chosen must be high enough for a good approximation.The angle between dog's direction of movement and the 0y axis varies with time but it's tangent can be calculated as much as the step of the time dt is chosen [let's say dt=0.1 sec].As a result of this procedure the coordinates (x_i,y_i) of the dog can be computed as a function of v,the number of nodes 'i' being in function of the intervals of time dt chosen.Finally using Lagrange interpolation an approximation of the funtion is obtained in the form of a polynomial.
The solution is viable however it's not easy to find a general formula for y(x) when the general variable 'v' is used,the number of calculations is great enough,but for particular values of v this solution is feasible.

Okay...this is what i found on the internet, it is somewhat different, now there is a leash between the dog and the man, but i think that doesn't change the problem?

http://mathworld.wolfram.com/Tractrix.html

Metacristi: Let's assume the dog CAN change his direction continuously :)

Basically if dt is chosen small enough this is equivalent to a continuous changement of direction.
But in the article it is said that Leibniz found the curve otherwise...There is no clear explanation but I assume that he had not known the equations previously [otherwise he could have found the curve by simply plotting some experimental data in the derivative of the known equations].I will try further...

I have a small logical problem for you all to solve. No math involved I'm afraid.

3 man are standing in line before a wall so that the first man only sees the wall and the third one the 2 men in front of him and the wall.

Now each is put on a hat which they themselves can not see, chosen out of 3 white hats and 2 black ones.

When the first man asks the third man what color of hat he thinks he's wearing the third man replies : " I don't know"

Then the first man asks the second man to tell what color of hat he thinks he has on and he replies after thinking a while also: " I don't know"

"Oh" says the first man; " That means that I'm wearing a white hat"

He was correct but can anyone explain me the logic he used to come to this conclusion.

Originally posted by Pim
After searching a bit on the internet i found that the man-dog problem is a known problem, there was also a solution given which i never would have found. I will not post it here yet though :)

Mind at least posting the link? lol. I cannot find it myself. I just keep finding pages about dogs, even though I use keywords like "calculus" and "equation." Oh well.

James Sibley

Good riddle. Here is the logic.

When the man at the rear responds that he doesn't know what colour hat he is wearing, this permits three possible scenarios:

I will give scenarios in the format (front | middle | rear)

i) (B | W | ?)
ii) (W | B | ?)
iii) (W | W | ?)

When he asks the gentleman in the middle, who also responds he doesn't know, this eliminates possibility number (i). This is because we know that there are three white hats, and two black hats. If the man in the middle looked forward and saw a black hat, he would know that scenarios (ii) and (iii) aren't possible. Having eliminated (ii) and (iii), scenario (i) would tell the gentleman in the middle that he is wearing a white hat.

However, because scenario (i) was discounted, both scenarios (ii) and (iii) have the gentleman in the front wearing a white hat, thus the man at the front having heard the conversation and knowing the number of hats provided knows he is wearing a white hat.

Thanks!

In the spirit of the man/dog problem, and the man/dog problem (with a leash - see the link provided earlier - http://mathworld.wolfram.com/Tractrix.html), here is another one.

Consider three dogs, each at the corners of an equilateral triangle. Simultaneously, each dog begins to run towards the dog on his/her left (to help people out, the path of the three dogs will look like three curves spiraling in towards the centroid of the triangle).

The question is, what is the path length the dog travels before it reaches the centroid?

P.S. I forgot the answer to this one, but will see if I can recall it...

Enjoy!

Originally posted by sdeliver645
Good riddle. Here is the logic.
*snip*

Wow. Kick ass. I got the three senerios then I stopped. I had to get ready for work :(

Good work though! :)

James Sibley

Hi Pim,
Having a fixed distance between man and dog is a different problem, as in that case the dog's speed will vary (starting at zero, and approaching the man's speed asymptotically).

Pete: i think it is the same problem. Since there is a leash between the two, i think they must have the same speed. The man also starts at zero speed. Think about it..in one second walking the dog travels the same distance as the man because they can be seen as "one" because of the leash, so their speeds are the same. I think adding a leash does not change the problem.

I don't think that a fixed distance implies that the dog is travelling at the same rate as the man. This would be the case of the orientation of the leash remained unchanged, but that is not the case.

You can think of the motion (dog's velocity) as having a component due to translation, and one due to rotation (changing the angle at which the leash is oriented).

Having said that, I don't have a solution.

Yes, maybe you're right...i have to think this over for a while..having a leash does not imply the same speed always. For example, the dog can run in a "circle""around the man, and thus have a much higher speed, but doesn't the dogs speed has to be at least the man's speed?

A: In the context of this question (tractrix), yes, the dog is always travelling AT LEAST as fast as the man. This can be seen IF the following assumption is true:

Assumption: That the dog's speed decreases monotonically.

This seems reasonable, doesn't it? One could always check this by cheating and looking at the solution (on the Wolfram site - see link previously given by someone else). Once the parametric eqn's of motion are known, one can determine the instantaneous velocity and then the speed of the dog.

If this is true, then the dog's speed decreases to that of the man's asymptotically. In the limit that the man is pulling the dog for an infinite amount of time, the dog's trajectory will become coincident with that of the man's (coincident with that of, say, the x-axis, if the man is moving along the x-axis). When this occurs, the dog and the man will be both moving at a fixed speed.

In the leash situation, the dog's speed is always less than that of the man. In the first instant when the man is walking at right angles to the leash, the dog's speed is zero.

If the dog were to travel at the same speed as the man and maintain the same distance between them, the dog must walk parallel to the x-axis.

In the original problem with the dog and man moving at the same speed, the dog is always getting closer to the man.

(x+y)*2

:-))

Originally posted by Pete
In the original problem with the dog and man moving at the same speed, the dog is always getting closer to the man. I agree with Pete, but his last sentence raises a quandry in my head. If they're travelling at the same speed, does the distance between the two asymptotically approach zero, or does it approach a nonzero number?

I think I partially answered my own question. The x component of the dog's velocity can never be >= the man's velocity. Since the x component of the dog's position will obviously lag that of the man in the beginning, the dog will never be able to make up the difference. The question is, what's the distance between them at infinity?

Originally posted by Pim
After searching a bit on the internet i found that the man-dog problem is a known problem, there was also a solution given which i never would have found. I will not post it here yet though :)

Pim, you just reminded me of Fermat right there.