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Forum » SpaceEngine » Off-topic Discussions » Challenge a Child's IQ (Be brainily brutal and get away with it)
Challenge a Child's IQ
midtskogenDate: Thursday, 05.09.2013, 20:37 | Message # 46
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Quote (Spyro)
I think I might have the first part right:

I'm afraid not. That wont work.





NIL DIFFICILE VOLENTI
 
Billy_MayesDate: Thursday, 05.09.2013, 20:40 | Message # 47
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Quote (Spyro)
Easy: What is the name of the game I largely enjoy involving evolution?
Medium: How much daily Calcium does an entire large bag of Goldfish have in it?
Hard: What did I name my art project? smile


1. Spore.

2. I have never eaten (or even seen) goldfish crackers, so no idea. wacko

3. I dont know. (Spyro?)





AMD Phenom II X4 955 3.2 GHz Quad-Core - AMD Radeon HD 6950 2GB VRAM - 4GB RAM - 1680x1050 75 Hz Samsung screen
 
WatsisnameDate: Thursday, 05.09.2013, 20:58 | Message # 48
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Spyro: Careful; what is midtskogen's problem asking for? The number of coins is not the same as the monetary value of those coins. smile

Quote (midtskogen)
Let's say I give you 1000 coins and 10 jars. Your task will be to distribute the coins into these jars and seal them, so that if I ask for any number of these coins, you shall be able to give me one or more jars that together contain the number of coins that I asked for. Can it be done, and how would you distribute the coins?







 
midtskogenDate: Thursday, 05.09.2013, 21:47 | Message # 49
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Whatsisname has the right answer!






NIL DIFFICILE VOLENTI
 
Gondor2222Date: Friday, 06.09.2013, 00:38 | Message # 50
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Quote (Watsisname)

(Solution)




Edited by Gondor2222 - Friday, 06.09.2013, 00:39
 
WatsisnameDate: Friday, 06.09.2013, 01:42 | Message # 51
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midtskogenDate: Friday, 06.09.2013, 06:57 | Message # 52
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Added (06.09.2013, 09:57)
---------------------------------------------
This thread reminds me of





NIL DIFFICILE VOLENTI
 
SpyroDate: Saturday, 07.09.2013, 13:24 | Message # 53
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I would have to say that the resistance on the infinite grid is 3 Ohms. There are only three segments to follow with no lesser resistance, and electricity always follows the path of least resistance.
3 Ohms.

Quote (Billy Mayes)
1. Spore.
Yes!

2. I have never eaten (or even seen) goldfish crackers, so no idea. wacko
... WHA

3. I dont know. (Spyro?)
Nope. It was Arbor McCarded-Boards.
I did that to you because this is what everyone else's questions feel like. :)


For anyone wondering what Arbor looked like, here he is:

Attachments: 2340607.png(10Kb)





I'm back, but am I here to stay? Hopefully!

Edited by Spyro - Saturday, 07.09.2013, 13:34
 
midtskogenDate: Saturday, 07.09.2013, 14:09 | Message # 54
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Quote (Spyro)
I would have to say that the resistance on the infinite grid is 3 Ohms. There are only three segments to follow with no lesser resistance, and electricity always follows the path of least resistance.

You forget to take into account that there are also infinitely many parallel paths: 5Ω, 7Ω, etc. Yes, for serial resistors you simply add, but when you have resistors in parallel the formula is 1/Rtot = 1/R1 + ... + 1/Rn. So it will be lower than 3Ω. Your argument only works for finding the upper limit.

There's a discussion of the problem here.





NIL DIFFICILE VOLENTI


Edited by midtskogen - Saturday, 07.09.2013, 14:19
 
SpyroDate: Monday, 09.09.2013, 21:27 | Message # 55
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Yea, anyone can see that I gave you that award for a reason. smile
But, I still got it RIGHT, right?
What was the difficulty of the question?

EDIT: When I asked myself that last question, 3 Ohms came into my head again. I also said that in Social Studies! Why is it in my head?!
Wait, I know! 3 Ohms!
OH COME ON
(Sorry, just amusing myself here tongue )





I'm back, but am I here to stay? Hopefully!

Edited by Spyro - Monday, 09.09.2013, 21:29
 
midtskogenDate: Tuesday, 10.09.2013, 08:42 | Message # 56
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Quote (Spyro)
But, I still got it RIGHT, right?

You get 3 Ω through the shortest path, but the current has an infinite number of extra paths in which to distribute its flow, so it must be less, potentially much less. But one node has four 1 Ω resistors connected to it, so equally obvious that it's less than 3 Ω, is that it's more than ¼ Ω.

The answer to the "nerd sniping" problem is 4/π-½ Ω. I.e. about 0.773 Ω. No, I didn't calculate that myself, and while it would have been fun to see if I had arrived at that before I googled it, what I really wish, besides the the time to solve such problems, is that I had the insight and mathematical intuition to see that the solution would involve π. But I generally don't. What I often find more intriguing than the solution itself is how numbers like π and e show up all the time, and how totally abstract concepts can create the tools needed to make a problem simple, kind of like a new mathematical space where different mathematical laws apply, yet problems in one such universe can be moved to another, solved, and taken back. As if when the laws of physics prevent us from doing something, we could say it doesn't matter, we just jump through a portal to a different universe where the physical laws allow it, and we take the result back to our normal universe. Engineers and physicists have much to envy mathematicians.

EDIT: Practical solution (see 7:15 for solution to the nerd sniper problem):





NIL DIFFICILE VOLENTI


Edited by midtskogen - Tuesday, 10.09.2013, 09:31
 
ElkarDate: Tuesday, 17.02.2015, 03:41 | Message # 57
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Challenge a child's IQ I dont know. biggrin




( ͡° ͜ʖ ͡°)( ͡° ͜ʖ ͡°)( ͡° ͜ʖ ͡°)

Edited by Elkar - Tuesday, 17.02.2015, 03:41
 
Forum » SpaceEngine » Off-topic Discussions » Challenge a Child's IQ (Be brainily brutal and get away with it)
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