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.
Spyro: Careful; what is midtskogen's problem asking for? The number of coins is not the same as the monetary value of those coins.
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?
Since log21000 < 10, it is possible to divide the 1000 coins amongst the 10 jars through binary division. I.e., distribute as follows: Jar 1: 1 coin Jar 2: 2 coins Jar 3: 4 coins Jar 4: 8 coins Jar 9: 29-1 = 256 coins.
At this point we have used up 1+2+4+8+...+256 = 511 coins, so there are 1000-511 = 489 coins remaining. We can place these in the 10th and final jar. Then, if you were to ask for all the coins, you would receive all the jars. For 999 coins, you would receive all the jars except the first. For 998 coins, all the jars except the second, and so on.
Aside: Being base 2, this is quite like how computers handle numbers.
It's more complicated than binary though, because more than half of the numbers have to be represented using a digit that doesn't represent a power of 2. In the base we've created with the jars, 111 111 111+1=1 000 010 111
I think you're missing the idea. Although not all decimal numbers are powers of two, all decimal numbers have a binary representation, or the sum of powers of two. The nice thing about it is that you only need one or none of each of those powers of two to represent the number. This is the same principle that computers work with.
I think Gondor2222's point is that while jar 1-9 can represent binary digits, the 10th jar doesn't work as a 10th binary digit for the reason he made, i.e. addition wont work. In binary arithmetics 111 111 111 + 1 = 1 000 000 000, but in the jar system that's 511 + 1 = 489. Had I used 1023 coins, it would work, but I felt that it (210-1) would hint too much about the solution.
All numbers have one binary representation and only one. But in the case of the jars the numbers 489 - 511 have two: 489 is either 1 000 000 000 (10th jar only) or 111 101 001 (256 + 128 + 64 + 32 + 8 + 1).
Added (06.09.2013, 09:57) --------------------------------------------- This thread reminds me of
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:
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.
Yea, anyone can see that I gave you that award for a reason. 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 )
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):
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