harey
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Posts posted by harey
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You forget the lift by air depression on the wings.
When the space shuttle (which is basically a plane) takes off vertically, as long as the thrust is less than 10 [N/kg], it does not move. When the thrust exceeds this value, it moves up. [If you then cut the engines, it slows down, stops and falls like a stone.] That's what your equations describe.
When a plane flies horizontally at a constant speed, the gravitation is compensated by depression on wings - if you cut the engines, the plane glides. You still can keep it at constant horizontal speed, it slowly loses potential energy, but does not vertically fall like a stone. This implies that on constant speed/height, you need to furnish less energy than by a vertical takeoff and therefore less thrust.
Another approach: The energy of a cruising plane is constant, so the furnished energy must be equal to the energy lost. As the lost energy is partly transformed to lift on the wings...
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Spoiler
Monk M1 starts at 7:00 am at the bottom going up.
Monk M2 starts at 7:00 am at the top going down.
If there is just one path, they must meet.
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And what about this:
Spoiler1: 100 pay 10 with 100
2: 50 20 20 pay 9 with 20
3: 50 20 10 1 pay 8 with 10
4: 50 20 1 1 1 pay 9 with 20
5: 50 10 1 1 1 1 pay 8 with 10
6: 50 1 1 1 1 1 1 pay 10 with 50
7: 20 20 1 1 1 1 1 1 pay 9 with 20
8: 20 10 1 1 1 1 1 1 1 pay 8 with 10
9: 20 1 1 1 1 1 1 1 1 1 pay 18 with 20
1 1 1 1 1 1 1 1 1 1 1 -
Thinking it over and over again, I always finish with a problem I cannot solve.
The holes are numbered 1, 2, 3....
Three hunters check:
1 2 3
3 4 5
5 6 7
....On day n, they will have checked up to the hole 2 * n +1
The groundhog starts in the hole k and moves to the right.
On day n, he will be in the hole k+nQuestion 1: Will the hunters catch the groundhog (and if so, when)?
2 * n +1 grows faster than k + 1, I already have the answer. Nevertheless:
2 * n + 1 = k + n
n ⁼ k + 1
No matter how high the number of the starting hole the groundhog choses, it will be caught.Question 2: Is there a starting hole so that the groundhog is not discovered on day n?
k + n > 2 * n + 1
k > n + 1
No matter how long the hunters hunt, it always is possible that the groundhog started in a hole leading him outside the checked area. -
On 6/6/2020 at 12:13 PM, EventHorizon said:
Never mind how big the area you chose, I always can say "Bad luck, the groundhog is somewhere about 17 holes outside the defined area."
I agree that the size of the cleared area will -> inf. Just the size is expressed as a number. No matter how big it is, there always is a bigger number.
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I think I can do better:
Spoiler01: 100 pay 10 with 100
02: 50 20 20 pay 5 with 50
03: 20 20 20 20 5 pay 3 with 5
04: 20 20 20 20 1 1 pay 5 with 20
05: 20 20 20 10 5 1 1 pay 3 with 5
06: 20 20 20 10 1 1 1 1 pay 8 with 10
07: 20 20 20 1 1 1 1 1 1 pay 9 with 20
08: 20 20 10 1 1 1 1 1 1 1 pay 8 with 10
09: 20 20 1 1 1 1 1 1 1 1 1 pay 18 with 20
10: 20 1 1 1 1 1 1 1 1 1 1 1 pay 18 with 20But not sure it is the max.
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Please delete.
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Still confused.
SpoilerSay the numbers are 37 111 148.
On the end of the first round C thinks:
- I can have 111-37=74 or 111+37=148.
- If the guy with 111 sees 37 and 74, he announces 37+74=111 (as he knows he cannot have 74-37=37).
- As he did not announce 111, I only can have 148.
So why the second round?
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Spoiler
Label the square ABCD clockwise starting at the upper left corner.
Put Y next to the angle of 70 (I will use both X and Y as angles and as points).
BAY=180-90-70=20
DAX=90-20-45=25Rotate ABY over AY.
Rotate ADX over AX.ABY and ADX will exactly cover AXY - proof is left to the reader as exercise.
-> X=180-45--70=65
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1
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Nice beginning... (If this were the full solution, first round would be enough.)
Oh, I got it now!!! Great!!!
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It is not because of the gender. the Claytons indicates that Mrs Clayton is still alive and Jasper's mom is deceased. Therefore, Jasper is not a Clayton.
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On 12/7/2019 at 8:21 PM, flamebirde said:
The Clarks and Carters, staunch Republicans who are very good friends..." implies to me that both Carter parents and both Clark parents are still alive (since one would not say "the Clarks" or "The Carters" if only one Clark/Carter parent is still alive), and therefore Jasper is neither Clark nor Carter. Therefore, Jasper must be either a Clayton or a Cramer. I will assume that Jasper is a Cramer.
In the same way, you exclude that Jasper is a Clayton:
2. In deference to an influential family member, the Claytons agreed that if they ever had a daughter they would name her Janice.
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Spoiler
I see TWO lines EVERYWHERE (appearing as a single line if superposed).
Each step consists in turning one line by 45 degrees clockwise, (giving C).
It works even between the last element of a line and the first element of the next line.
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On 3/12/2019 at 5:53 PM, CaptainEd said:
my answer:3567 are True.
Argument:
The only distribution that has at least 4 Precious, and yields exactly 50% probability of finding a second P is (PP,PN,PN,NN).
There are 4 possible situations:
We picked P1 from PP
We picked P2 from PP
We picked P from PN
We picked P from other PNIf we swap boxes, then
* if in PP (half the cases), our chance of picking P is 1/2 * 2/3 = 1/3, as we might swap to NP, NP, or NN
* if in PN (half the cases), we will get PP in 1/3 cases, PN in 1/3 cases, and NN IN 1/3 cases, so 1/3 * (1 + .5 + 0)= 1/2So probability of picking P after swapping is (1/3 + 1/2)/2 = 5/12
Therefore 1 and 2 are false, 3 is true.
The rest of the statements are self-evident given this distribution.
There are 8 boxes, not 4.
And there is another possible distribution you have to exclude.
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The way how the marbles were selected is not known, so you cannot do better then Bonanova.
However, instead of grabbing the calculator:
Spoilere**x=limit(1+x/n)**n and take x=-1;
n=100 is large enough for a quick estimation: 1/e
The exact value is about 58% (Euler-Mascheroni constant), but do not ask me for details.
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3 hours ago, bonanova said:
I guess we can compute expectation value as well:
E = Sumk { outcomek } x { probability of outcomek }
One million simulations:
# holes % prob product
----------------------------------
1 hole 1.01% 0.0101 0.0101
2 holes 50.40% 0.5040 1.0079
3 holes 32.85% 0.3285 0.9856
4 holes 11.98% 0.1198 0.4790
5 holes 3.04% 0.030427 0.1521
6 holes 0.60% 0.006024 0.0361
7 holes 0.10% 0.000993 0.0070
8 holes 0.01% 0.000125 0.0010
9+ holes 0.00% 0.000021 0.0002
======
2.6792A big step forward. Now that we found the result, it remains to find the way to find the result.
SpoilerWhen I do simulations, a much lower number of iterations is sufficient (originally, we simulated it on Univac 1108 - it would have calculated for days). However, it is paying to try different rising n, it gives an idea (sometimes misleading, I agree) where the sums and the terms converge.
I can compute how many squirrels need to dig just two holes:
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
--+--------------------------------------------+
1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 |
3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 |
5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 |
6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 |
7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 |
8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 |
9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 |
--+--------------------------------------------+(n*n)/2: the part of the square under the diagonal (the diagonal becomes meaningless for large n).
Unfortunately, such a table is much harder to construct for the third dig.
For n=3, it is imaginable to place the number of dig nuts in 3D, but for 4 nuts, I surrender.
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@rocdocmac
Pas du tout. Mais c'est ma langue maternelle, je sais même écrire.
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On 3/25/2018 at 6:44 AM, bonanova said:
What I can't find is a corresponding analysis for the marbles problem (although plainglazed's approach is applicable to both problems) that will lead to that same sum, instead of yours, where both sums appear to be correct.
I see now: plainglazed's formula is the simplification of my formula I was so hard looking for. On paper, I got a kind of unreadable proof, so I fed them into my computer. Up to n=30, no difference with 6 decimals.
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2 hours ago, bonanova said:
There is survival in numbers.
Say the colony numbered N squirrels, and they each dug up 2 holes.
They would create N piles of nuts whose average size is 101 and whose standard deviation is about 40.
So if they share among themselves, they will all get through the winter.
If each eats only what s/he finds, maybe only 2/3 of them will survive.Nice try, but they do not share and they survive all.
Hint:
SpoilerVery few will find 100 nuts and do not dig a second hole.
Some will find find 100 nuts in two holes and do not dig a third hole.
How many will dig a third hole? And a fourth? -
3 hours ago, bonanova said:
I think this puzzle is exactly the same as the traffic jam puzzle I posted recently, although that's not obvious at first glance.
@plainglazed found a solution in which he formed clusters of cars by recursively locating the slowest of a group of cars, assuming on average it was in the center of the remaining cars. This corresponds to "grabbing" on average one-half of the remaining marbles from the bag. Picture the marbles in a line and, grabbing a random percentage of them starting from one end. This has to end up having the same number of marble grabs and car clusters. It leads to a logarithmic answer, but to the wrong base -- it should be the natural loge, not log2, which gives too large an answer. Instead of decreasing the number by 1/2 each grab, the remaining number is decreased by 1/e each grab. Same must go for locating the slowest of the remaining cars.
@plasmid found a solution that leads to Sum { 1/k }, which as you point out is ln { n } + gamma, and is here confirmed by simulation. What I can't find is a corresponding analysis for the marbles problem (although plainglazed's approach is applicable to both problems) that will lead to that same sum, instead of yours, where both sums appear to be correct.
This has been fun to think about.
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Though they are similar, I see at least one huge difference. In the traffic jam puzzle, you cut ANY part. In the marble problem, you cut the REMAINING part. Enough for different formulae.
Remains me of http://brainden.com/forum/topic/18168-squirrel/
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A large colony of squirrels dug holes during the summer and in each hole, they put between 1 and 100 nuts (each quantity has the same probability).
If each squirrel has to eat 100 nuts during the winter, how many holes must he find (in average)?
Each hole contains 50.5 in average, so 2 should be enough, right?
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4 hours ago, bonanova said:
That agrees with simulations I ran, which gave, approximately,
4, 10, 30, 84 and 244 (geometric series in e) as the number of marbles that require integer numbers 2, 3, 4, 5 and 6 grabs.
So if you multiply the number of marbles by e, it takes one more grab.exp { 1.4 2.4 3.4 4.4 5.4 } =~ { 4 10 30 84 224 }.
That's interesting.
... but not so surprising if you consider that the formula for (1+1/2+1/3+1/4+ +1/n) is ln(n)+0.5772156649. Adding exp to exp usually gives an exp, too.
Still, I wonder whether there is a formula for 1 + (1/(n-1)+2/(n-2)+ +(n-1)/1) / n .
Nice solve.
Thanks.
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On 2/1/2018 at 8:56 PM, rocdocmac said:
1 If the “needle” (stylus) is poised 12.5 mm directly above the beginning of the first song and 2 mm away from the edge of the record, how far does the “needle” travel to the point where the last song just ended?
SpoilerI do not get what the 12.5 mm are there for. Do you mean after the begin of the first song? I just ignore it and assume the first r=290 mm.
The usual name for these records in French is "33 tours" (to distinguish it from the previous standard of 78 rpm). (In fact, the speed is 33 1/3 rpm.)
In 18 minutes, the disk will make 600 rotations.
197 * 600 =~ 120,000 -> without any doubt, the last r=17 cm, average r=23 cm.
23 [cm] * 600 * 3.14 = 433 [meters].
I recall these LPs, the last r=17 cm seems a little bit high to me. They had so about 20 minutes playtime on each side, but I would guess the inner r nearer to 7 cm then.
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Spoiler
By definition, p(to be correct at random)=(number of correct answers)/(number of answers)
As the (number of correct answers) is not known, p cannot be calculated.
Even if the possible answers were {15%, 25%, 35%, 45%}, the problem would remain meaningless. One could believe 25% to be the correct answer. Why should it be correct? Don't tell me
It is correct because it is correct.
Chance of getting his seat.
in New Logic/Math Puzzles
Posted
Assume that passengers 2-98 ask (if necessary) the forgetful passenger to leave his seat and THIS passenger then choses a seat at random.
When the last passenger boards, 98 are correctly seated. The forgetful passenger occupies either his own seat or that of the 100th passenger.
So p=50%.