It's puzzle time /sci/. I post a puzzle and you try to solve it. If you get it right, you get to pick the next puzzle (out of 3 listed). Let's start:
You walk one kilometer south, then one km east, then one km north. Now you are back where you started. Where are you? You are at the north pole. But are there other starting locations (on a spherical object like the Earth) where the same thing would happen?
2. Girlfriend dilemma
You mean: 1km/2*pi*k where k is a natural number.
But it's correct. Ferryboats:
Two ferryboats start at the same time on opposite sides of the river, traveling perpendicular to the shore. They both travel at constant speed, but one is faster than the other. They pass at a point 720m from the nearest shore. Each boats rests 10 minutes before returning back. On the return trip they meet 400m from the other shore.
How wide is the river?
1. Handshakes and Networks
2. Where are you going?
3. Desert Crossing
Let me pick it for you if you don't mind. Handshakes and Networks:
At a physics convention the number of physicists who shook hands an odd number of times, is even. Prove it (without using graph theory). Advanced version: Prove it using induction.
Next problems to pick:
1. The thick coin
2. Find the Oddball
3. Hole in Sphere
And here is The Thick coin:
How thick should a coin be to have a 1/3 chance of landing on the edge?
Iff you can solve this in less than 20 seconds in your head, you're smarter than his majesty pic related.
attribute a number to every physicist - how many hands he shaked
The sum of all those numbers is even - since you've counted every handshake twice.
If sum of some numbers is even, there must have been evenly many odd numbers among them.
Should one assume the coin loses all momentum as soon as it hits the ground and then it basically drops dead to its resting position?
Or some more complex bounce bounce bounce with a constant dampening factor?
You have a balance scale and 9 balls. One ball is heavier than the other 8. What is the minimum amount of weightings you need to find the odd ball?
The first time the slower boats meeet the faster one, it has traveled 720m and the boats together have traveled one river length. The second time it has traveled the length of the river + 400m, and the boats together have traveled 3 river lengths. So 3x720 = river + 400, subtract 400 and river is 1760.
I will pick for you. Probability of Matches:
A deck of 52 cards is laid out in a row, face-up. Beneath it another deck is laid out. What is the probability that there are exactly n number of matches (i.e. queen of hearts beneath queen of hearts).
This should keep you busy.
Let n be the number of physicists who shook hands an odd number of times.
- A handshake between 2 people both with an odd number of previous handshakes will subtract 2 from n.
- A handshake between 2 people both with an even number of handshakes will add 2 to n.
- A handshake between 2 people with an even and an odd number of handshakes respectively will not change n.
Therefore no handshake will change the parity of n. At the start noone has shaken hands so n is 0 (i.e. even).
This counts as proof by induction, right?
Their are 9 different possible states:
>Ball 1 is heavier
>Ball 2 is heavier
>Ball 9 is heavier
A weighing can give 3 different results
>Left side is heavier
>Right side is heavier
>The sides are in balance
For each weighing you have to distribute the remaining states evenly to the results. This way you divide the number of states by 3 every weighing.
3 groups of 3 balls (9 total)
Weigh only 2 groups against each other.
If those 2 groups are evenly weighted then the non-weighted group contains the heavier ball.
The rest is as you said, just adding the rest of the info left out. This of course assumes that all other balls are evenly weighted and not all random weights.
So one side of the scales has one normal ball plus the heavier ball, the other side has one normal ball. You now know the heavier ball is one of two balls, how do you know which one?
Nonono, the point is you weigh TWO balls from the three against each other, and leave one on the side. One ball on one side of the scale, one ball on the other side of the scale, and one not on the scale.
i think it is more intuitive to consider rotation around only one axis. This is how a coin toss is done, and it is more in line with angular momentum conservation.
I dont like his answer.
Because it's in 3 dimensions. Answer is 1/(2*sqrt(2)), see >>7790779
That von Neumann solved it in less than 20 seconds blows my mind. Very strong understanding of mathematics.
Another questions for the physicists (in case you aren't already familiar), 2D Random Walk:
Starting from origin a particle has a 1/4 chance to move either east, west, south, or north. What is the probability that the particle will return to origin after number of steps n.
You don't understand von Neumann's sheer mental ability.
>Two trains are on the same track a distance 100 km apart heading towards one another, each at a speed of 50 km/h. A fly starting out at the front of one train, flies towards the other at a speed of 75 km/h. Upon reaching the other train, the fly turns around and continues towards the first train. How many kilometers does the fly travel before getting squashed in the collision of the two trains?
>When posed with a variant of this question involving a fly and two bicycles, John von Neumann is reputed to have immediately answered with the correct result. When subsequently asked if he had heard the short-cut solution, he answered no, that his immediate answer had been a result of explicitly summing the series
>As far as I could tell, von Neumann was able on once reading a book or article to quote it back verbatim; moreover he could do it years later without hesitation. He could also translate it at no diminution in speed from its original language into English. On one occasion I tested his ability by asking him to tell me how the 'Tale of Two Cities' started. Whereupon, without pause, he immediately began to recite the first chapter and continued until asked to stop after about ten or fifteen minutes.
As a child he also used to multiply 8+ digit numbers in his head for fun.
0 if n is odd.
if n is even ...
we'll use the multinomial coefficient to calculate the nb of ways to place n distinct objects (step1, step2, ..., step n) into 4 boxes (north, south, east, west).
This will give you the number of ways to go back to origin after n steps
Sum i = 0 to n/2-1 of M(n/2-i,n/2-i, i,i)
If you divide by 4^n you get the probability.
I'm too lazy to try to simplify this...
During his graduate studies, and after working with von Neumann, Gillies became a fan of the book "One-upmanship" by Stephen Potter. John von Neumann was also a fan of this work, and was extremely successful at impressing others with his intelligence. An apocryphal math problem asks about a bumble bee flying back and forth between two approaching trains, and how far did it fly before colliding? When von Neumann gave the correct answer, the questioner asked if he used a standard time/rate-of-travel trick, and he replied, 'no, I summed the infinite series in my head' to impress the questioner. This method of impressing and astonishing others appealed to both Gillies and von Neumann.
Yes, the decks are shuffled.
Here's an easy one. The Flight around the World:
There is a group of planes on a small island. The tank of each is just large enough to take it halfway around the globe. Any amount of fuel can be transferred in flight. The island is the only fuel loss, and there is no time lost in refueling either on land or in the air.
What is the minimum number of plans to make a trip around the globe in a great circle, assuming the have the same ground speed and fuel consumption is constant?
Three. Send two flying in one direction, exchange fuel at one half pi, let one crash and the other continue. when that plane gets to pi send a third plane in the opposite direction and exchange fuel at 3/4 pi.
There, I gave it more thought...
There are a few steps involved
Nb of permutations for card 1 to match is 51!
Nb of permutations for card 1 and card 2 is 50!
Nb of permutations for card 1, 2 and 3 is 49!
Once you figure that out, you try to calculate the nb of permutations to have at least one card with a match (let's call that Z).
Have in mind that P(A u B) = P(A)+P(B)-P(A and B) ... use the generalized version with 52 events. and make sure you consider all intersections.
Z=sum (j=1 to 52) (-1)^(j-1) (j in 52) (52-j)!
52! - Z is the number of permutations with no match.
We also need to develop a similar expression for nb of permutations with no match for any nb of cards.
Prob (n matches) = (n in 52) * (nb of permutations with no match in 52-n cards) / 52!
Will continue to posts some riddles and games, if you are interested.
A man decides to shop on the town with his wife one evening.
They visit twelve stores.
At each store,
the man spends half of the money he had before he walked in to the store,
plus, 15 dollars. At the end of the trip,
the man has only 3 dollars left,
so he decides to return home with his wife.
At the start of the trip, exactly how much money did the man originally have?
(These are not mine by the way)
You're at a popular Italian Restaurant with your friends after a sports game.
A waitress takes your ticket to the chef. You've just ordered seven pizzas,
for the guys and gals.
The restaurant offers two choices of pizza, one choice is considered regular, but the other choice is premium, using a different recipe, which makes the pizza weigh slightly more.
All of the pizzas appear the exact same, and you and four of your friends order the usual pizza.
But two of your friends however order premium pizzas of a different recipe.
The chef has baked five pizzas of the same weight, following the same recipe.
The chef has also baked two pizzas of the heavier weight,
which weigh the same to each other,
following the different recipe.
Again, all of the pizzas appear the same.
after they come out of the oven,
he obviously can't tell which are which.
The chef decides to weigh each pizza.
In order to weigh the pizzas,
he needs to place them on a two-pan balance,
and weigh them against each other on both pans.
At most, he can fit three pizzas on each pan.
(You can weigh each at a time,
two at a time on each pan,
or three at a time on each pan,
or any combo of that.)
How many weighing actions are necessary
to tell which are the heavier pizzas?
This problem has more than one solution.
Answer is $135138
You can recursively add and multiply, but I wanna model is as so:
3+sum[for x 1 to y](33*(2^x))
For y number of stores, cuz it looks cooler that way...but wait! You can optimize it for use in computing:
Where I is initial value
M is multiplicative constant
A is additive constant
S is number of stores to visit
There are a number of ways to do this, as you said. Here's one:
Set one pizza aside and call it A.
Weigh the rest with one half the pizzas on each side. If they are balanced, we know A cannot be a premium, and that each side of the scales has exactly one premium pizza.
With each of the sets of 3 on the scales, set one pizza aside, and see if the other 2 balance. If they do, the one set aside for that group is premium, otherwise the heavier side is premium. This requires one weighing for each of the sets, resulting in a grand total of 3 weighings.
If the scales are UNBALANCED at the first weighing, we must first determine whether A is premium. Suppose side B is heavier than side C. Side C must not contain any premiums, so if you weigh one pizza from side C with pizza A, you can determine whether A is premium.
Then you weigh the pizzas from side B the same way that you split each side in the earlier scenario, e.g. set one aside, weight the other 2. If A is premium, you deduce the other premium the same way as before. If A is NOT premium, then:
If the scales are balanced, both are premium. If they are unbalanced, the lighter one is not premium, and the other 2 are.
This scenario also results in 3 weighings.
So...3 is my guess.
Theres a power tower next to a river, and a house on the other side of said river. (pic related)
We need to connect the house to the power spending the less amount of money.
The costs for the wire are :
$1 a meter for wire over the shore or the bridge.
$2 a meter for wire over the water.
Whats the cheapest way to do it, and if over the water, whats the angle that the wire makes with the shore?
Consider a triangle
Hyp is the length of the wire if it goes above water.
A is length of base and B the height.
If we can find a triangle for which
2*hyp < A + B then it is cost effective to go above water
with A = h sinx and B = h cosx
we cover all possible triangles
Is it possible to find an angle x that satisfies
2h < h(sinx + cosx)
2 < sinx + cosx
Unfortunately, it is not possible because the max of sinx + cosx is 2^0.5=1.41... so it is never cost-effective to go above water.
Cheapest way to do it is 75$.
Is the following accepted as proof?
-Let a,b be sides of a right angled triangle and c be the hypotenuse.
(a-b)^2 + 2c^2 > 0
3c^2 - 2ab > 0 (pythagoras)
3c^2 > 2ab
4c^2 > a^2 + 2ab + b^2 (pythagoras)
2c > a + b
c > (a + b)/2
Prove that a sphere with 5 points on its surface must have a closed hemisphere with 4 points in it.
It's a pretty well-known question, so if you know the solution, don't spoil it for others.
draw a circumference around the sphere through any two of the points. This divides the sphere into two hemispheres.
Since there are three points left and only two hemispheres to place them in a basic application of the pidgeonhole principle tells you one must have at least two points, combined with the two on the circumference makes four.