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File: 1477688055359-0.png (58.61 KB, 300x72, handshaking problem.png)

File: 1477688055359-1.png (14.45 KB, 300x125, basic coin problem with twist.png)

File: 1477688055359-2.png (174.48 KB, 300x213, Super Hard squares problem.png)


Lets have a puzzle thread. I am not very good at maths, so they might be a bit trivial, but hopefully some of you can get some enjoymeny out of these :-)


File: 1477688182106.png (59.92 KB, 200x104, p problem - unsolved.png)

Please note that for the one about Coin A and Coin B, the question should be "how many rotations" not "how many revolutions", and also note that the answer is NOT 3.

It has a little twist to it, its not quite that simple.


dA = (dB)/3


I don't get it, what's the twist? Is it in the the way the question is made instead of the actual mathematics? "3 times smaller" might be slightly ambiguous.

Anyway, if 3 is not the answer, then please post the answer (spoiler it if you can).

Also, the first challenge (handshakes) is pretty badly worded.


I just got the dumb `twist'.
This is more a word problem than math.


coin 'a has a perimiter of 2*pi*r and coin 'b has a perimeter of 2*pi*3r= 6*pi*r 3*'a
So it takes 3 revolutions and goes back to the starting position
I feel that I might be wrong, though the reasoning looks correct ... halp?


wait actually it'd take 6 revolutions for the coin to get to it's original position, because it takes 2 revolutions for a coin the same size.


For question 1 if there were 2 people on the planet we could let the handshakes be 1.For question 2, when the circle turns upon it's internal axis this is called a rotation.When on the axis of another it is called a revolution.The answers I've seen so far claim to answer for a revolution, but if it was a revolution the answer would be one.3 is obviously not the answer for rotations (OP said so) also how is six derived from all that? #204


I'm guessing you tried it with two equal sized coins to get that result but it doesn't quite scale up like that. Spoiler for answer.

It's 4 rotations. The coin will do (r1/r2)+1 rotations. The r1/r2 is like you calculated, the extra one comes from the fact that the coin rotates around its own axis as well. https://en.wikipedia.org/wiki/Coin_rotation_paradox


I think you misunderstood the first question - it should be solved for any number of people on earth with any amount of total handshakes.


>It is important to understand the difference between rotations and revolutions. When an object turns around an internal axis (like the Earth turns around its axis) it is called a rotation. When an object circles an external axis (like the Earth circles the sun) it is called a revolution.


File: 1477798047615.png (1.55 MB, 200x113, 20161030_122330.jpg)

I think I got it?

Is it 4?


I doubt this answer for the first puzzle is best, but I tried:
1. Let the person who has shaken hands odd times "odder" and the person who do even times "evener"
2. If odder and odder have shaken hands, then they became evener. This don't change the parity of odders.
3. The case of evener and evener also don't change the parity, and also the case of evener and odder don't change either.
4. Every case of handshakes doesn't change the parity of odders.
5. the parity of odders on the initial state is even, therefore, the statement holds. Q.E.D.


The squares problem is famous.
For the first one it's helpful to consider first the easier case of 2 integers instead of four. If you remember that |zw|=|z||w| for complex numbers, this leads to the formula (ac-bd)^2+(ad+bc)^2=(a^2+b^2)(c^2+d^2) (which can now be checked by expanding everything too)
A similar formula exists for sum of four squares, this time using quaternions instead.
ii) can also be proved using quaternions and abstract algebra (with somewhat more difficulty) this is done in the book "topics in algebra" by herstein.


Here's a pretty rigorous mathy solution for the first, that proceeds by contradiction.

Notice that the total number of handshakes people have participated in is an even number. If two people engage in a handshake, each will have done one handshake; their total number of handshakes is therefore two. Another way to look at it is that all handshakes must be done in pairs (doing a handshake with yourself is not allowed*).

So, write 2H for the total number of handshakes, for some natural** number H. Assume for the sake of contradiction that an odd number of people participated in an odd number of handshakes. We can then write

2H = [sum(2k_i) for i=1 to N] + [sum(2h_j + 1) for j=1 to 2M + 1]

where 2k_i is the (even) number of handshakes person i did and 2h_j + 1 is the (odd) number of handshakes person j did (for natural numbers k_i, h_j). In other words, we partition the total amount of handshakes into two groups: those done by people who did an even amount, and those who did an odd amount. The above can be rewritten into

2H = 2[sum(k_i) for i=1 to N] + [sum(2h_j) for j=1 to 2M + 1] + 2M + 1

= 2[(sum(k_i) for i=1 to N) + (sum(h_j) for j=1 to 2M + 1) + M] + 1

Let K=[(sum(k_i) for i=1 to N) + (sum(h_j) for j=1 to 2M + 1) + M]. As all k_i's h_j's, as well as M, are natural numbers, K must be a natural number. But there is no natural number K for which it holds that 2H=2K+1 (this corresponds to finding an even natural number which is also odd), so we have our contradiction. That is, the number of people who did an odd amount of handshakes cannot be odd and must therefore be even.

* if it was, a trivial counter example would be one person who did a handshake with himself. This is an odd number of people (1) who did an odd number of handshakes (1) -- It is also pretty sad imo.

** Natural numbers are assumed to contain 0. I.e., the set {0,1,2,3,..}



Yeah, I'd say so.

Me and a buddy discussed (or, he convinced me) this in our university Friday bar and came to the same conclusion.


see the hint >>211 its a trick question