G 000 100: Fill & Download for Free

Here's a nice solution that was shown to me long ago. (I would never have figured this one out by myself.)Here's the algorithm for solving the 12 balls in three weighingsproblem. It also happens to solve the 3 balls in two weighings or 39balls in 4 weighings, or 120 balls in 5 weighings or ..., problems.First, list all 3 digit combinations of 0, 1, and 2:000 100 200001 101 201002 102 202010 110 210011 111 211012 112 212020 120 220021 121 221022 122 222(For the 4 weighing problem, list all 4 digit combinations, and so on.)Next, cross out all sets with identical digits (000, 111, 222), and alsothrow out all whose first digit change (reading from left to right) isnot 01, 12, or 20. In other words, 112 is left in because the firstchange is from 1 to 2, but 212 is eliminated, since the first change isfrom 2 to 1. This leaves:(a) 001(b) 010(c) 011(d) 012(e) 112(f) 120(g) 121(h) 122(i) 200(j) 201(k) 202(l) 220where a, b, ..., l are the balls.Now, for the first weighing, weigh all the balls with a 0 in their firstposition against all the balls with a 2 in the first position, in otherwords, (a, b, c, d) are weighed against (i, j, k, l). If the balls withthe 0 are heavier, write down 0; if the balls balance, write 1,otherwise; write 2. Next, weigh all balls with a 0 in the secondposition against all balls with a 2 in the second position (a, i, j, k)against (f, g, h, l), and write down the digit corresponding to theresult of the weighing, and do it again for the third position in thethird weighing.As an example, suppose ball f (120) is heavy. In the first weighing, fis not involved, so write down 1. Next, (f, g, h, l) -- the "2" ballsare heavier than (a, i, j, k) so write down 2. Finally, in the weighing(b, f, i, l) versus (d, e, h, k), the "0" side (b, f, i, l) is heavier,so write down a 0. You have written "120" -- the code for ball f -- sof is the heavy ball.If you find that the combination you write down is not in the list, say211, then change the 2s to 0s, the 0s to 2s, and leave the 1s alone,giving 011. This is ball c, but since you had to flip the digits, ballc is lighter.With N weighings, there are 3^N combinations, of which we eliminate 3and cut the remainder in half, so for 4 weighings, one can distinguish(3*3*3*3-3)/2 = 39 balls, etc.In general, if you have n weighings, you can distinguishamong (3^n - 3)/2 balls and find the one that is either lightor heavy.The (3^n - 3)/2 is easy to explain. In each ball's "name"there are n slots to fill, each of which can be any of 0,1 or 2, so there are 3^n of those possibilities. But00...0, 11...1, and 22...2 are tossed out, so thereremain 3^n-3. This has to be divided by 2, since eachball really has two names -- one meaning "heavier" andthe othe meaning "lighter".Here is a list of the 39 names for lighter balls:1 00012 00103 00114 00125 01006 01017 01028 01109 011110 011211 012012 012113 012214 111215 112016 112117 112218 120019 120120 120221 121022 121123 121224 122025 122126 122227 222028 220029 220130 220231 200032 200133 200234 201035 201136 201237 202038 202139 2022Here are the weighings:1 2 3 4 5 6 7 8 9 10 11 12 13 <-> 27 28 29 30 31 32 33 34 35 36 37 38 391 2 3 4 31 32 33 34 35 36 37 38 39 <-> 18 19 20 21 22 23 24 25 26 27 28 29 301 5 6 7 18 19 20 28 29 30 31 32 33 <-> 11 12 13 15 16 17 24 25 26 27 37 38 392 5 8 11 15 18 21 24 27 28 31 34 37 <-> 4 7 10 13 14 17 20 23 26 30 33 36 39Suppose ball 21 is heavy. The results of the weighings willbe:equalright heavyequalleft heavyThis codes to 1210 which is ball 21.If ball 32 is light, here's what you'd get:left heavyright heavyright heavyequalThis codes for 0221 which does not appear in the list. Thusthe indicated ball is light, not heavy and has code 2001 whichis ball 32's code.

Everyone would experience a 10 000 g force for [math]1 \times 10^{-9}[/math] seconds. This is more than the g force experienced by pistons in Formula One engines, which can rev up to 15 000 rpm.The highest sustained g force that a person has survived under experimental conditions, according to Wikipedia, is 25 g for 1.1 seconds with a peak of 46.2 g. According to this study, a person can only survive 100 g for 0.003 seconds of upwards acceleration (“eyeballs down”) or 0.01 seconds of horizontal acceleration (“eyeballs in”) before they sustain severe injury. These are similar to the g forces experienced in lethal car accidents.But, we are only talking about one nanosecond.How far will your body collapse in one nanosecond? Let’s use our simple kinematics from high school.[math]\text{distance} = (\text{initial velocity})(\text{time}) + \dfrac{1}{2}(\text{acceleration})(\text{time})^2\tag*{}[/math]which simplifies to:[math]d = \frac{1}{2}at^2\tag*{}[/math][math]= \frac{1}{2}(9.806 \text{ m/s}^{2})(10\text{ }000)(10^{-9} \text{ s})^2\tag*{}[/math][math]= 4.9\times10^{-14} \text{ meters}\tag*{}[/math]Not too bad. That’s only a few times larger than a large atomic nucleus and thousands of time smaller than an atom.So how easy is it to shake off the energy of your entire body suddenly collapsing by the width of a few atomic nuclei with a force of 10 000 g?Let’s figure out the velocity that things will be moving after one nanosecond.[math]\text{velocity} = \text{initial velocity} + (\text{acceleration})(\text{time})\tag*{}[/math][math]v = at\tag*{}[/math][math]= (9.806 \text{ m/s}^2)(10\text{ }000)(10^{-9} \text{ s})\tag*{}[/math][math]= 9.8\times10^{-5} \text{ m/s}\tag*{}[/math]or about 0.1 millimeter per second. Not very fast at all.Now we need to know how much energy that translates into. For a 100 kg human,[math]\text{Kinetic Energy} = \frac{1}{2}(\text{mass})(\text{velocity})^2\tag*{}[/math][math]T = \frac{1}{2}mv^2\tag*{}[/math][math]= \frac{1}{2}(100 \text{ kg})(9.8\times 10^{-5} \text{ m/s})^2\tag*{}[/math][math]= 0.48\text{ microjoules}\tag*{}[/math]So, a human body subject to a one nanosecond increase in gravity to 10 000 g would have to absorb about half a microjoule of energy. That’s about half of the energy of the proton collision in the Large Hadron Collider at CERN which created the Higgs boson “God particle”. Perhaps that sounds impressive, but put another way it is less than the impact you would feel from a single mosquito flying into your body.So if the human race can absorb mosquito impacts, we should be okay with a one nanosecond aberration of 10 000 g, as long as it instantaneously goes back to normal.It doesn't look like you’re going to get a trademarked Everyone Dies answer this time.Note: everyone put up similar calculations while I was writing this. Also thanks to Kumar Saurav for helping with my formatting. I learned a couple new tags!

Short answer: ~ 50 000 000 000 000 000 000 atoms.50 quintillion (short scale) in the US, Canada or UK.50 trillion (long scale) in the civilized part of the world... ;)50 hexillion if you like greek (not a standard).50 million million million atoms, some would say.50 x 10^18 is another way to write it, although:5 x 10^19 atoms would be the standard expression.Long answer: It varies.There's roughly in the neighbourhood of 1 - 100 quintillion/trillion atoms. Quite possibly more, quite possibly even less.That's 1 - 100 million million million atoms.Sand is made up of different materials, making it difficult to do a generalized calculation.However, the most frequently mentioned material is silicon dioxide, or SiO2.The atomic weight of Si (Silicon) is 28 units.The atomic weight of O(Oxygen) is 16 units.That puts the atomic weight of SiO2 at 28 + 2 x 16 = 60 units.Avogadro tells us that we have 6.023 x 10^23 SiO2 units per gram, so there would be 6.023 x 10^23 / 60 ≈ 1 x 10^22 SiO2 units in a gram of pure SiO2, and with SiO2 composed of 3 atoms, that puts us at 3 x 10^22 atoms per gram of SiO2.Now, a grain of sand does not weigh 1 gram. It weighs less. So we continue:Sand is defined as having a size between 0.0625 mm and 2 mm.Say a grain of sand is shaped like a cube, with each side exactly 1 mm in length. It then has a volume of 0.001 cm^3. This is not fine sand.1 cm^3 of SiO2 weighs about 2.65 gOur little cube of "sand" then weighs 0.00265 gTo find the number of atoms in this cube we then multiply this weight with the number of atoms in a gram of SiO2.3 x 10^22 x 0.00265 = 7.95 x 10^19 atomsSo, our little cube holds 79.5 quintillion/trillion atoms.Given the size and composition variations, even the range estimation given above of 1 - 100 quintillion/trillion is going to be inaccurate, but I hope this answer still gives you an idea of the numbers we are dealing with.