harey
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Posts posted by harey
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The best way is to make them of equal volume ( suppose what I call a "3D square" is a prism?):
v_prism=a*a*d_prism
v_cylinder=pi*a/2*a/2*d_cylinder
=> d_cylinder=(4/pi)*d_prism
While it might work for small d, with growing d, it will became more and more perceptible that the cylinder should be a kind of barrel (or a pancake). The formulae get too complicated for my taste and my possibilities.
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The circle is inscribed to the square...
Going a bit further just for fun, which object can weigh the most, and which can weight the least?
That's easy. The circle is inscribed to the square.
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What you make us to believe is a cube might be three "squares" of unknown thickness. (Appologies to the purists.)
What you make us to believe is a sphere might be three cylinders of unknown height.
Therefore, the volume cannot be calculated.
Besides, I leave the question open whether "made of the same material" implies the same density.
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With d=2 and c=6, I can weight:
2=d
4=c-d=6-2
6=c=6
8=c+d=6+2=8
(It determinates odd numbers are between 1 and 9.)
To weight 10, I need a 3rd weight, I can go up to: b=10+c+d=18
10=18-6-2
12=18-6
14=18-6+2
16=18-2
18=18
20=18+2
22=18+6-2
24=18+6
26=18+6+2
To weight 28, I need a 4th weight, I can go up to: a=28+b+c+d=28+18+6+2=54
That's quite strange because with c,b,d, I can make any weight up to 26, so we can go with the marbles up to 81 (and usually problems are designed this way). So I continue.
28=54-18-6-2
30=54-18-6
32=54-18-6+2
34=54-18-2
36=54-18
38=54-18+2
40=54-18+6-2
42=54-18+6
44=54-18+6+2
46=54-6-2
48=54-6
50=54-6+2
52=54-2
54=54
56=54+2
58=54+6-2
60=54+6
4 weights, 3 postages, I did not earn much today
I checked and rechecked... A typo? Or a try to misslead us? -
I do not accept.
If I find out the full weight is i.e. 40, there can be 10 marbles weighting 4 each, or 1 marble weighting 40 and so on
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choose to order calibrated weights in marble units. Units are your choice. Then a calibrated weight of seven marble units would balance seven marbles. Would you accept the offer then?
I cannot purchase more than 4 weigths.
I do not need to weight 61 (nor any other odd weight) if I can weight 60.
a+b+c+d=60
a+b+c=58, => d=2
This should be solvable by brute force or even by the Monte Carlo method, but I hope to find something better.
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I think it is a pity that the solution is not posted when nobody finds it. So here we go:
Just integrate. As I safely integrate only x=k and x=e**x, I will leave the developement to someone else. (Anyway, there were "Best Answers" for less than that.)
The cannonball is 8 times faster than the bugs. Divide the band in more than 8 equal parts and mark the ends of these parts. Two adjacent marks will move away from each other by a speed that is smaller than that of the bugs. So the bugs will reach the first amrk, the second... and they will eventually meet (and if they contimue, one will reach the cannonball and the second one the fixation.
The speed of the cannonball is constant. The speed of the bugs is increasing.
Increasing speed is not enough. The bugs are accelerating and the acceleration is increasing.
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I do not accept.
If I find out the full weight is i.e. 40, there can be 10 marbles weighting 4 each, or 1 marble weighting 40 and so on
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Prayer to bona alma:
- correct the post 4 with the correction from the post 5
- delete in the post 4 remarks that are not necessary anymore
- delete the post 5
- delete this post
...and pray with me that bmad concedes it is the Best Answer
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Yes, and the beetle heading down will even reach the cannonball.
As I have already seen the problem, I will refrain from posting.
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Correction:
B sends:
- if(measurement<4) measurement else 7-measurement.
P.S. It would not be bad if the table in my previous post magically appeared in a fixed font.
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I made a lookup table for the processor.
1st line: measurements by A
1st column: measurements by B
x difference only one bit
. difference more than one bit
0 1 2 3 4 5 6 7
0 x x x . x . . .
1 x x . x . x . .
2 x . x x . . x .
3 . x x x . . . x
4 x . . . x x x .
5 . x . . x x . x
6 . . x . x . x x
7 . . . x . x x x
Observe the horizontal symmetry: B sends (7-measurement)
It makes me think of Grey code...
Sorry I cannot hide it. Maybe because of the font?
Edit: Definitely, I have a problem with the editor.
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Reasoning from a similar problem:
a) If no one answers immediately brunette, it means there is at most one blonde. (That one who sees 2 blondes answers immediately brunette.)
b) If the others see a blonde and a brunette, they will answer simultaneously brunette. (There are not 2 blondes.)
c) From b), everyone can deduce the others see two brunettes.
We have just to assume that all three think at the same speed.
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@Grimbal
You can continue:
- there are 50.5 boys born per 49.5 girls.
- women live longer. -
Really easy!!!
In a rectangle that's 2 by 200 units long, it's trivial to draw 400 non-overlapping unit-diameter circles.
But in the same rectangle, can you draw 401 circles?
And even more.
No one said that the 401 circles may not have a smaller diameter.
No one said that they may not ovelap.
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Labeling the pills A and B, you can have:
- 3A- 2A 1B
- 1A 2B
- 3B
I haven't yet seen a bottle of pills that has not the number of pills not written on it, so you can complete to 3A 3B. Dissolve in water and drink by thirds. Or mill them.
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If someone has 80 (and more), he will not (propose to) change. (If I have 120, as you cannot have 240, you must have 60).
If someone has 40-79, he will not change. (If I have 60: if you have 120, you will not agree to change; if you have 30, I am loosing).
If someone has 20-39, he will not change. (Same reason, you will refuse if you have 40-79.)
Recursivity comes to my mind. But what if I have i.e. 17?
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ThunderCloud
Hypothetically, had A in fact seen B=335 and C=334
All three stickers actually had the same number written on them.
We know that A sees B=334 and c=334 - A CANNOT see B=335 and C=334.
(Not to confound with: "As this information in not available to him, he ASSUMES he has 334 or 335.")
So my first answer was correct. In the 2nd try, I got lost and could not correct quickly enough.
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We know the distributed numbers (334 334 334) are therefore what A, B and C see.
A sees B=334 C=334, he thinks:
- I can have 334 or 335
- whether I pass or give up, B will know I have not seen B=335 C=334 -> B will announce 334
- C will know I did not see B=334 C=335 -> C will announce 334
-> I will not learn anything new, so I give up
B sees A=334 C=334, he thinks:
A did not see B=335 C=334 (he would announce 334) -> B=334
C sees A=334 C=334, he thinks:
A did not see B=334 C=335 (he would announce 334) -> C=334
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We know the distributed numbers (334 334 334) are therefore what A, B and C see.
A sees 334 334. He thinks:
1a) if I have 335, B will see 335 334 and will announce 3341b) if I have 334, B will see 334 334; not being able to choose between 334 335, he will pass
- > I will be able to know, so I pass
B sees 334 334
2a) not being able to choose between 334 335, he will pass (or even already give up, I am not very sure here)
C sees 334 334
3a) B did not see 334 335 (he would announce 344)
3b) -> B sees 334 334 -> I have 334
A knows he has 334 from 1b) or 3b) knows he has 334
C cannot know
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This did not happen:
1a) Number One sees 334 and 335 and announces 334
1b) Number Two and Three announce 1003-(sum of numbers they see)
1c) Number One announces 334
So this must have happened:
2a) Number One sees 334 and 334, gives up and commits suicide
2b) Number Two and Three announce 334
I know a better and harder version, if I find it, I'll post it. Does it matter if it is in French?
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I think it is still up for grabs, as I have not proven anything but simply given an single example.
Myself, I would prove it for N=0 and leave the proof for N>0 to the reader
If you try for some N, you quickly see the law. -
@bmad: I do not think my answer was the best one - after all, the chances were 50-50
phaze answered much better, he explained WHY.
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I like the Phil1882 idea of modulo and tried the following rule:
- sum the numbers you got
- add $ (your "position")
- modulo NIt works for some examples for N=4 (with no counterexample so far). The next steps should be an exhaustive computer simulation and a strong therie. Someone is willing to continue?
Two objects, Three views, Identical shapes
in New Logic/Math Puzzles
Posted
Yes, I see. There must be a lot of solutions, like taking a sphere and a cube and drilling holes into the cube until it becames lighter than the sphere - no formula, no misscalculation.
My preferred solution is still that of the three squares and three pankaces in the planes xy, xz and yz.
BTW, the Volume=3a2d-d3 - I have the same oversight in my post n. 8.
I just wonder how often we make false deductions in the real life in the style "three projections=circle => it must be a sphere".