# The Naked Scientists Forum

### Author Topic: Try solving these easy problems by deductive thinking  (Read 45859 times)

#### Alan McDougall

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« Reply #25 on: 07/07/2008 19:43:19 »
"No it does not" If luck is not on your side your solution might require as much as 36 bouncing combinations before finding the odd ball.

I am aware the less massive ball will bounce back. In my youth I also played snooker.

Alan
« Last Edit: 07/07/2008 19:45:03 by Alan McDougall »

#### lightarrow

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« Reply #26 on: 07/07/2008 19:54:54 »
"No it does not" If luck is not on your side your solution might require as much as 36 bouncing combinations before finding the odd ball.
Actually, with only one shot, if you put all the balls in a row, quite distant one another, and you shoot the first (or the last).

#### Alan McDougall

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« Reply #27 on: 07/07/2008 22:46:15 »
Sorry it makes no sense!

#### lightarrow

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« Reply #28 on: 08/07/2008 13:59:41 »

#### Alan McDougall

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« Reply #29 on: 08/07/2008 16:03:38 »
It still makes no sense you are only allowed three weights/bounced/shots

Shots

If I understand you correctly, you want to bounce the balls against each other,hopefully by careful observation you would see the ball with less mass  bounce futher back than the one with the greater mass.

To do this you can choose and ball and bounce it against any other ball. If luck is not on your side it would need 11 bounces to slove the problem. Not three weighs on a pivot scale as in my 12 odd ball solution

#### Bored chemist

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« Reply #30 on: 08/07/2008 19:49:39 »
The whole point is that this way I need precisely zero weighings, I might need a dozen shots or so (I'm alousy player) but nobody said anything about them in the original question. In principle I could, as Lightarrow says, do it with one shot.I could do this in zero gravity with no weighings so that's less than three or do you somehow think 3 is not more than zero?

Let's make this as clear as I can.
I can solve the problem of finding the odd ball and whether it's light or heavy with zero weighings.

#### Alan McDougall

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« Reply #31 on: 09/07/2008 08:11:04 »
OH!!Man come on!! be real, your experiment solution would require a trip the the space atation. It is not practicle and completely irrelevant to my 12 all problem.

#### lightarrow

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« Reply #32 on: 09/07/2008 23:28:39 »
Alan, can you post the other problems you said to have? I find them interesting.

#### Alan McDougall

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« Reply #33 on: 10/07/2008 09:30:31 »
lightarrow,

Ok here is three more, I hope they are of interest!

Try these three problems and if they are found of interest I will pose more

1) Two brothers share a flock of x sheep. They take the sheep to the market and sell each sheep for \$x. At the end of the day they put the money from the sales on the table to divide it equally. All money is in \$10 bills, except for less than ten excess \$1 bills. One at a time they take out \$10 bills. The brother who draws first also draws last. The second brother complains about getting one less \$10 bill so the first brother offers him all the \$1 bills. The second brother still received a total less than the first brother so he asks the first brother to write him a check to balance the things out. How much was the check

2) Four dogs occupy the four corners of a square with side of length a. At the same time each dog starts walking at the same speed directly toward the dog on his left. Eventually all four dogs will converge at the center of the square. What path does each dog follow and what is the distance each dog walks until he reaches the center?

3) Five pumpkins are weighed two at a time in all ten sets of two. The weights are recorded as 16, 18,19, 20, 21, 22, 23, 24, 26, and 27 pounds. All individual weights are also integers. How much does each pumpkin weigh?

Some of the problems are from internet, most are mine collected over the years.  I love problems of deductive logical thinking.

We can go the route of the really difficult but that would be no fun I think

Regards

Alan

#### Bored chemist

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« Reply #34 on: 10/07/2008 18:47:48 »
"OH!!Man come on!! be real, your experiment solution would require a trip the the space atation. It is not practicle and completely irrelevant to my 12 all problem."
Nonsense.
At most, it would require a trip to the local pool hall. In fact anywhere with a reasonably flat table or floor would do.
The fact is that, if you happened to be on a space station, my method would still work, but your's wouldn't.
It's perfectly practical; what's impractical about playing snooker or pool?

Your original question was just a puzzle- any solution that meets the stated objective is valid. In terms of minimising the number of weighings, my solution beats yours.

#### Alan McDougall

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« Reply #35 on: 11/07/2008 00:02:39 »
Yes I wanted to say it could be done "your way" on earth. But your suggestion has nothing to do with the puzzle. To say anything that meets the objective is silly and outside the confines of the odd ball problem.

Regards

Alan

#### lightarrow

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« Reply #36 on: 13/07/2008 12:55:24 »
2) Four dogs occupy the four corners of a square with side of length a. At the same time each dog starts walking at the same speed directly toward the dog on his left. Eventually all four dogs will converge at the center of the square. What path does each dog follow and what is the distance each dog walks until he reaches the center?
The distance walked is a. The path is an exponential spiral with respect to the square's center (a/2;a/2):

r(t) = -a/2 exp[-θ(t)]*{cos[θ(t)] + sin[θ(t)];cos[θ(t)] - sin[θ(t)]} + (a/2;a/2)

|r(t) - (a/2;a/2)| = a/sqrt(2) exp[-θ(t)]

About the distance walked my reasoning is probably not simple:
after a time dt the dog (I considered the one at the left down corner in the origin of coordinates) walks for vdt up and vdt at right so he is now at a distance sqrt[(a-vdt)^2 + (vdt)^2] ~ a - vdt so it's now in a corner of a new square whith side of lenght: a - vdt; so the differential variation of the square's side lenght is da = (a - vdt) - a = -vdt and this equation is valid in every moment. Integrating from 0 to t:

∫da = ∫-vdt --> a(t) - a = -vt --> a(t) = a - vt

When they meet, at the instant T, they are in a square of side lenght 0, so:

0 = a - vT --> T = a/v

Since the speed v is constant, the total lenght L is simply v*T:

L = v*a/v = a.

About the path followed, my reasoning was even less simple, because I solved it analitically (complicated computations), so I assume there must be a simpler way.

#### Bored chemist

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« Reply #37 on: 13/07/2008 17:34:43 »
"To say anything that meets the objective is silly and outside the confines of the odd ball problem."
Yet it still solves the problem of spotting the oddball, and it does so with fewer weighings.
« Last Edit: 13/07/2008 17:43:40 by Bored chemist »

#### Alan McDougall

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« Reply #38 on: 14/07/2008 09:34:51 »
Lightarrow,

Lightarrow,

I like you approach to this not so simple little problem

Hard Solution
It is intuitive that at all times the dogs will form the corners of a square, ever decreasing in size and rotating about the center.
Let C be the center of the square at point (0,0).

Let the dogs be at initial points (.5,.5), (.5,-.5), (-.5,-.5), and (-.5,.5).

Let dog 1 bet at (.5,-.5) who is chasing dog 2 at (.5,.5). Consider the line from C to dog 1.

Next consider the line from dog 1 to dog 2. The angle between these lines is 135 degrees.

Theorem: Let C be a fixed point, D a point on a curve, r the distance from C to D, x the polar angle formed by r, and y the angle formed by the angle between CD and the tangent line to the curve at point D. Then tan(y)=r/(dr/dx).

So the line of sight from the center to the initial point of dog 1 would be be at a 270 degree angle. The line of sight from dog 1 to dog 2 would be a vertical line of 90 degrees. The angle between these lines is 270 degrees. The tangent of 270 is -1. So we have r/(dr/dx) = -1. Thus r(x)=c*e-x.

At the moment the dogs start walking x equals 0. At this moment the distance from the center to any of the dogs is 2-1/2. So r(x)=2-1/2*e-x.

The arc length is given by the formula:

Integral from 0 to infinity of ((f(x))2+(f'(x))2)1/2 =

Integral from 0 to infinity of ((2-1/2*e-x)2 + (-2-1/2*e-x)2)1/2 =

Integral from 0 to infinity of (2-1*e-2x + 2-1*e-2x)1/2 =

Integral from 0 to infinity of e-x =
-e-x from 0 to infinity =
=0 - (-e0) =
0 -(-1) = 1 The path the dogs take is called a logarithmic spiral.

It is an interesting paradox that the dogs will make an infinite number of circles, yet the total distance is constant. Every revolution the size of the square will to e-2*pi/21/2 = 0.03055663 times it's size before the revolution.

Easy Solution
David Wilson suggested the more simple solution which follows...
Suppose dog A is pursuing dog B who is pursuing dog C. During the entire pursuit, the dogs remain at the corners of a square, and angle ABC is a constant 90 degrees. That is, B's path towards C is always perpendicular to A's path towards B, and B has zero velocity in the direction along A's path. Since B neither approaches nor recedes from A during the walk, A simply covers the intial separation of 1.

Antonio Molins went further to state that as dog A is heading direcctly towards dog B, dog B is moving towards from dog A at rate equal to cosine of any of the interior angles of the shape made by the dogs.

He also provided the following graphics.
For example consider a pentagon. Remember that for a n-gon of n sides each angle will be 180*(1-(2/n)).

So a pentagon will have angles of 108°. Let's assume each dog walks at a rate of 1 unit per minute, where the initial shape has sides of 1 unit. As pointed out by David Wilson we can forget about the spiral paths but instead think of the dogs walking in straight lines.

In the case of the penagon for every one unit dog A draw towards dog B's initial position, dog B will draw cos(180°)=-0.309 units towards dog A.

The question is, how long will it take for them to meet? Let the answer to that question be t. In t seconds dog A will cover t units, and dog B will cover .309*t units. The total distance traveled by A equals the distance traveled by B plus 1 (the initial separation). So solving for t: t=0.309+1 --> t=1/(1-.309) --> t=1.4472.

What do you think?

Alan

#### Alan McDougall

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« Reply #39 on: 15/07/2008 06:01:00 »
OK,

A horse is tied to a stake with a rope 5 metres in length, 10 metres in front of him is a bunch of nice hay. The hay is firmly tied down and can not be moved.

The horse somehow gets to the hay and begins to eat. How does he do it?

Regards
Alan

#### Alan McDougall

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« Reply #40 on: 15/07/2008 06:15:28 »
Now try this

Please don't muse over the request act immediately you read what you must do

Before anything get a pencil and paper ready. please don't scroll down yet

OK scroll down now:.................? far down please!!

Try to think of two objects or shapes "one inside the other" under the circumstances I suggested and "IMMEDIATELY DRAW THEM"
as soon as the image pops into your mind.

Note don't reveal what you drew I will attempt to predict that later, after a few replies, such as OK Alan I have my drawing!!

#### BenV

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« Reply #41 on: 15/07/2008 09:50:38 »
OK,

A horse is tied to a stake with a rope 5 metres in length, 10 metres in front of him is a bunch of nice hay. The hay is firmly tied down and can not be moved.

The horse somehow gets to the hay and begins to eat. How does he do it?

Regards
Alan

Forgive me if I'm being too logical about this one, but horses are pretty good at chewing through rope.  If was a chain, I'd have to think about it again...

Oh, and I've thought of a shape within a shape.

#### lyner

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« Reply #42 on: 15/07/2008 13:23:06 »
Perhaps someone takes pity on him?

#### Alan McDougall

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« Reply #43 on: 15/07/2008 13:39:58 »
No no guys with repect there is a more logical answer. The horse must get to the hay all on his own, no outside help permitted

Regards

Alan

#### BenV

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« Reply #44 on: 15/07/2008 15:00:48 »
So "he chews through the rope" doesn't score any points?

#### Bored chemist

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« Reply #45 on: 15/07/2008 20:18:56 »

I don't know how to hide an answer, but I doubt this is the answer being looked for.
The puzzle states that the hay is tied down. It does not state that the stake is driven into the ground so maybe the horse just walks over to the hay dragging the stake.(Though my first thought was that he chews through, or breaks the rope.)
« Last Edit: 15/07/2008 20:24:06 by Bored chemist »

#### LeeE

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« Reply #46 on: 15/07/2008 20:28:50 »
OK,

A horse is tied to a stake with a rope 5 metres in length, 10 metres in front of him is a bunch of nice hay. The hay is firmly tied down and can not be moved.

The horse somehow gets to the hay and begins to eat. How does he do it?

Regards
Alan

The horse is standing at the extreme length of the rope, five meters from the stake.  The horse is facing the stake, and five meters further away in the same direction is the hay.  The horse walks the ten meters across the 'circle' and eats the hay.

#### Alan McDougall

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« Reply #47 on: 16/07/2008 00:23:25 »
BoardChemist,

Quote
I don't know how to hide an answer, but I doubt this is the answer being looked for
.

"" to the mind guess"" of prediction test in  "blue letters", This is post separate from the horse hay one.

Quote
The puzzle states that the hay is tied down. It does not state that the stake is driven into the ground so maybe the horse just walks over to the hay dragging the stake.(Though my first thought
was that he chews through, or breaks the rope.)

Yessssssss!! The above is the correct answer a little silly I know but many people most often just presume the skate is firmly hamered into the ground

Regards

Alan

#### BenV

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« Reply #48 on: 16/07/2008 17:57:06 »
I'm fairly sure I know the answer to the pumpkins one, just haven't yet given myself the time to write it out! Back with you soon on that...  and knowing horses, a horse would have chewed through the rope anyway, even if the stake wasn't in the ground.  They just like to be a pain.

#### Pumblechook

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« Reply #49 on: 16/07/2008 21:26:24 »
Lost track of all that..

1)Weigh 4 against 4..   Either balance or not.

2)Swap 1 from one pan to the other...  Replace 3 balls with 3 un-weighed balls.

Lock at change of state between first and second weighing.  In all cases you can find the odd one..  Might have to indentify one as a control.

It come down to one of two balls, lighter or heavier... Weigh one against a control.

OR one of the three where you now know you are looking for a lighter or heavier.  Weigh any two together.

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« Reply #49 on: 16/07/2008 21:26:24 »