Nine identical balls. One is heavier. You have the scale and can measure only twice. How to find heavier ball?
113050
good question - took me a few minutes, so check if i'm right:\n\nA,B,C,D,E,F,G,H,I are the nine balls.\n\nweigh three v.s. three (A,B,C v.s. D,E,F)\n\nif scale is equal, then all six wieigh the same; now weigh G v.s. H\nif scale is still equal, then I is the heavier one.\nif scale tilts towards one side, say G, then the heavier is G. (same thing here for H)\n\nif first scale attempt is not equal, say the A,B,C side is heavier, then weigh A v.s. B (since one of the three here is the heavier one)\nif scale is equal, then C is heavier.\nif scale tilts towards one side, say A, then A is heavier. (same here for B)\n\nif first scale attempt is not equal and is heavier on the D,E,F side, simply repeat what i said in the previous paragraph, but with those balls.\n\nam i correct? :)
789266
Of course, it really depends on what lengths, masses and time increments you are trying to measure and why.  The key is to have a foundation reference point with each measurement.  But here are some ideas:\n\nLength:  A piece of string or a piece of paper is good tool.  If you need to know exactly how many feet something is, a common record album is 12 inches across ( 1 foot) or a standard floor tile which is 12 inches from one flat side to its' opposite.  Use this as your foundation length of your string.  For long distance you can guage the length of your footstep using the above method, then count your footsteps when walking constistently, or use the odometer in your car.  Not a precise technique but approximate.\n\nMass:  A bucket filled to the top with water, placed inside another bucket that can catch spilled water.  Let's say you want to find the mass of a bowling ball.  Simply place the bowling ball into the bucket filled to the top with water, and whatever water it displaces (and is caught in the second bucket), is its mass (example: 1 gallon).  Repeat these steps to find out the mass of other things like your dog.  It's not an exact science but it's a basis for comparison at home as long as you keep using the same buckets.\n\nTime:   The sun is always good for this since our clocks are based on exposure to it due to rotation of the Earth every 24 hours, and our revolution around it once a year.  It's how sundials work.  You use a shadow.  You can mark the position of a shadow with a pen or a scratch, then observe (and mark) how the shadow has changed at a later moment.  When sunlight is not available, candles are good to observe the passage of time in a similar way as they burn and melt, as long as you keep using the same kinds of candles.
356709
We think that the thing to do is weigh six balls against six balls, but if you think about it, this would give us no information the only different ball. As we already know that one side will be heavier than the other. \n\nSo that the following plan can be followed, let us number the balls from 1 to 12. For the first weighing let us put on the left pan balls 1,2,3,4 and on the right pan balls 5,6,7,8. \n\nThere are two possibilities. Either they balance, or they don't. If they balance, then the different balls is in the group 9,10,11,12. So for our second weighing we would put 1,2 in the left pan and 9,10 on the right. If these balance then the different ball is either 11 or 12. \n\nWeigh ball 1 against 11. If they balance, the different ball is number 12. If they do not balance, then 11 is the different ball. \n\nIf 1,2 vs 9,10 do not balance, then the different ball is either 9 or 10. Again, weigh 1 against 9. If they balance, the different ball is number 10, otherwise it is number 9. \n\nThat was the easy part. \n\nWhat if the first weighing 1,2,3,4 vs 5,6,7,8 does not balance? Then any one of these balls could be the different ball. Now, in order to proceed, we must keep track of which side is heavy for each of the following weighings. \n\nSuppose that 5,6,7,8 is the heavy side. We now weigh 1,5,6 against 2,7,8. If they balance, then the ball is either 3 or 4. Weigh 4 against 9, a known good ball. If they balance then the different ball is 3, otherwise it is 4. \n\nNow, if 1,5,6 vs 2,7,8 does not balance, and 2,7,8 is the heavy side, then either 7 or 8 is a different, heavy ball, or 1 is a different, light ball. \n\nFor the third weighing, weigh 7 against 8. Whichever side is heavy is the different ball. If they balance, then 1 is the different ball. Should the weighing of 1,5, 6 vs 2,7,8 show 1,5,6 to be the heavy side, then either 5 or 6 is a different heavy ball or 2 is a light different ball. Weigh 5 against 6. The heavier one is the different ball. If they balance, then 2 is a different light ball.
373613
try using a bathroom scale, measures gravity vs. mass quite accurately.
314372
I think I have the answer.  Make three groups of three balls.  Now, take any two groups and place one group on one side, and the other group on the other side.  If the balance indicates that the groups are equal, then set aside those balls.  Take the remaining group of three balls and place any one on one side, and another on the other side, leaving the third on the table.  If both balls are even, then the one on the table is the heaviest.  Now for the different scenarios.  Going back one step, if the scale indicates that one side is heavier, then that ball is heavier, and you've found the heaviest.  When placing the groups of three balls on the scale, basically the same procedure is used.  If one group is heavier, take that group and set aside all other groups.  Then place one ball on each side and find if the scale is balanced.  If it is, then the one on the table is the heaviest etc.
270269
Put them all on the scale at the same time and start by taking them off one at a time when the scale reading goes down by 9 grams then that is to box you are looking for.
547336
Divide into two equal groups of 4.\nOne will be heavier than the other.  Take that group off the scale.\nDivide the remaining four into two groups of two.\nOne will be heavier than the other.  Take that group off the scale.\nDivide the remaining group into two groups of one.\nThe lighter one is the answer.\n\nEDIT: This can be accomplished in just 2 weighings.\nSplit the 8 into two groups of 6 and 2.\nDivide the group with 6 into two groups of three.\nCase 1: If they weigh the same then the lighter one is in the group of two.  Weigh the two to see which one is lighter.\nCase2: If they don't weigh the same, take the lighter group of three and weigh one on each side.  If they are the same then you know which one is the lighter one.  If they are different then you still know which one is the lighter one.
105
The answer is 9 stones (or maybe more). \n\nAfter adding a look-alike to 9 stones, there will be 10. Name them ABCDEFGHIJ.\n\n"ABC<DEF" means "the stones ABC as a group weigh less than the stones DEF."\n\nWeighing 1. Weigh ABC vs. DEF.\n\nIf ABC<DEF:\n--Weighing 2. Weigh ABC vs. GHI.\n----If ABC<GHI, the look-alike must be light and among ABC.\n----Weighing 3. Weigh A vs. B.\n------If A<B, the look-alike is A.\n------If A=B, the look-alike is C.\n------If A>B, the look-alike is B.\n----If ABC=GHI, the look-alike must be heavy and among DEF.\n----Weighing 3. Weigh D vs. E.\n------If D<E, the look-alike is E.\n------If D=E, the look-alike is F.\n------If D>E, the look-alike is D.\n\nIf ABC=DEF, the look-alike must be among GHIJ.\n--Weighing 2. Weigh G vs. H.\n----If G=H, the look-alike candidates are I and J.\n----Otherwise, the look-alike candidates are G and H.\n----Weighing 3. Weigh one of the two candidates against A.\n------If they differ in weight, the look-alike is that candidate, otherwise it is the other candidate.\n\nIf ABC>DEF:\n--Weighing 2. Weigh ABC vs. GHI.\n----If ABC>GHI, the look-alike must be heavy and among ABC.\n----Weighing 3. Weigh A vs. B.\n------If A<B, the look-alike is B.\n------If A=B, the look-alike is C.\n------If A>B, the look-alike is A.\n----If ABC=GHI, the look-alike must be light and among DEF.\n----Weighing 3. Weigh D vs. E.\n------If D<E, the look-alike is D.\n------If D=E, the look-alike is F.\n------If D>E, the look-alike is E.
800535
jinde, hi!\n\nwhat an amazing question!  i remember reading about this somewhere, so it took me so long to research it...\n\nsee, when we talk about weight, it varies from spot to spot in the universe.  you may weigh 60kgs on Earth.  but on the moon, you'd be 10kgs.  on Jupiter, you'd be 600000kgs!!!\n\nwhat we need to discuss is the mass of the object, because that would be the only thing that makes sense on the Galactic scale.  are you ready for the answer?  it's gonna be mind-boggling!\n\nThe Earth's mass is approximately 6,000,000,000,000,000,000,000,000 (6E+24) kilograms!!\n\ni know what you're next question will be?\n\n"How did they measure that?"  i have the answer for you!\n\nthe measurement of the planet's weight is derived from the gravitational attraction that the Earth has for objects near it.\n\nit turns out that any two masses have a gravitational attraction for one another. if you put two iron weights near each other, they will attract one another gravitationally. the attraction is extremely slight, but if your instruments are sensitive enough you can measure the gravitational attraction that they have on one another. from that measurement, you could determine the mass of the two objects. the same is true for two marbles too! but the attraction is even slighter because the amount of gravitational force depends on mass of the objects...\n\nNewton showed that, for spherical objects, you can make the simplifying assumption that all of the object's mass is concentrated at the center of the sphere... and the following equation expresses the gravitational attraction that two spherical objects have on one another :\nF = G * M1 * M2 / R2\n\n    * R is the distance separating the two objects.\n    * G is a constant that is 6.67259x10-11m3/s2 kg.\n    * M1 and M2 are the two masses that are attracting each other.\n    * F is the force of attraction between them. \n\nassume that Earth is one of the masses (M1) and a 1-kg sphere is the other (M2).  the force between them is 9.8 kg*m/s2 -- we can calculate this force by dropping the 1-kg sphere and measuring the acceleration that the Earth's gravitational field applies to it (9.8 m/s2)...\n\nthe radius of the Earth is 6,400,000 meters (6,999,125 yards)... if you plug all of these values in and solve for M1, you find that the mass of the Earth is 6,000,000,000,000,000,000,000,000 kilograms (6E+24 kilograms / 1.3E+25 pounds)...\n\nhope this made sense... and if it did, i hope it was useful to you...\n\ndo write me if you need more clarification...\n\nenjoy!
899962
Let x equal the cost of each ball.........\n\n5x </= to 20\n\nNow try your hand at solving it.
997488
The explanation is really simple, really!\n\nOkay, take a look at the mystic ball chart right now. Take a look at the numbers 0, 9, 18, 27, 36, 45, 54, 63, 72, and 81. They are all the same symbol, right? And they are all multiples of 9! Now, before doing anything, just click on the mystic ball. You'll notice that the symbol that pops up is the same symbol that the numbers mentioned above show. Click on the blue link to try it all again. You should get another sheet with different symbols. Now, look at the very same numbers that you looked at before. They all share the same symbol, but now it is a different one, right?\n\nThis is the trick. No matter what number you choose from 0 to 99, the final answer will always be a multiple of 9. Try it out yourself! Go through each number and figure out what the final number of each number would be. Figure out what the final symbol would be, too. You will see that no matter what number you choose, the same symbol will keep popping up.\n\nIf you want the equation, it would be 10a + b - (a+b). 'a' represents the first digit of the number you chose, and 'b' represents the second digit of the number you chose. '10a + b' represents the number as a whole (for example, let's choose the number 32. a = 3, b = 2. 10(3) + 2 = 32. Make sense?). the '- (a+b)' part represents the time when you add the digits together and subtract it from the original number to get your final answer. After reducing the equation, you get 9a, which is why the answers all come out to be a multiple of 9.\n\nHopefully this makes sense!
449872
The both weigh the same
1087044
Zero out all scales. You start will the scale with the largest ballast (big heavy thing). Move it larger by notches until the scale tips the other way, then move it back one notch. Repeat this on each successively smaller scale. Then your weight is the sum of each scale.\nSo 150 + 10 + 5 + 3 would be 168.
293931
This problem was fun and took me a few minutes to realise the solution.  First I will restate the problem just for clarification.\n\nWe have 12 coins, one of which is fake (lighter or heavier).  We can only make 3 weight comparisons to discover which of the coins is the fake one.  My solution will tell you which coin is fake and if it is heavy or light in exactly 3 comparisons.\n\nI number the coins 1 - 12 and make the following 3 comparisons on a scale\n\nleft    vs   right\n-------------------\n1 2 3 10  vs  4 5 6 11\n1 2 3 11  vs  7 8 9 10\n1 4 7 10  vs  2 5 8 12\n\nnow I know which coin is false.  How do I do this? well I created a table which has 12 unique outcomes of 3 weighing comparisons on which side of the scale went down. for example left left left indicates that all 3 comparisons had the left side dropped.  Here is my table:\n\nfake coin  | correspondin weight comparisons\n--------------------------------------------\n1          |  left left left (heavy) or right right right (light)\n2          |  left left right (heavy) or right right left (light)\n3          |  left left balanced (heavy) or right right balanced (light)\n4          |  right balanced left (heavy) or left balanced right (light)\n5          |  right balanced right (heavy) or left balanced left (light)\n6          |  right balanced balanced (heavy) or left balanced balanced (light)\n7          |  balanced right left (heavy) or balanced left right (light)\n8          |  balanced right right (heavy) or balanced left left (light)\n9          |  balanced right balanced (heavy) or balanced left balanced (light)\n10         |  left right left (heavy) or right left right (light)\n11         |  right left balanced (heavy) or left right balanced (light)\n12         |  balanced balanced right (heavy) or balanced balanced left (light)\n\nnow I can find exactly which coin was the fake one by comparing the results of my 3 wieghs against my table. For example if I get two balances and then a right, I know the 12 coin is the fake and is heavy. Note that if I get balanced balanced left the the 12 coin is light.\n\nNotice that I first created this table of possible outcomes and then generated from it which coins would be compared against which. \n\nHeavy or Light:  I created this table in which no result on any row is the reverse of any result on any other row (as the reverse indicates whether it is light or heavy)\n\nThis is the simplest solution to this problem.  I hope this made sense.  Was really fun.
885944
Using a 1-pan Weighing Scale:-\n\nAnswer: 6 weights\n\nthe weights are 1,2,4,8,16 and 32\n\n1=1\n2=2\n3=2+1\n4=4\n5=4+1\n6=4+2\n7=4+2+1\n8=8\n9=8+1\n10=8+2\n11=8+2+1\n12=8+4\n13=8+4+1\n14=8+4+2\n15=8+4+2+1\n16=16\netc..\n\nGeneralized Formula:-\n\nLeast value of (2^n - 1) such that 2^n -1 >= given weight.. n is the minimum number of weights required\n\nHere, given weight = 40\nTherefore, minimum 2^n-1>=40 is for n=6\ni.e 2^6-1=63>40\n\nNote:- Using these 6 weights, we can weigh different weights from 1 upto 63\n\n\n\nUsing a 2-pan Weighing Scale:-\n\nAnswer: 5 weights\n\nthe weights are 1,3,7,15 and 31\n\n1=1\n2=3-1\n3=3\n4=3+1\n5=7+1-3\n6=7-1\n7=7\n8=7+1\n9=7+3-1\n10=7+3\netc..\n\n\nGeneralised Formula:-\n\nSummation of (2^i - 1) where i varies from 0 to n >=Given weight\nand n is the minimum number of weights required\n\nHere, given weight = 40\n\n1+3+7+15+31= 57>40\n\nSo minimum number of required weights = 5\n\nNote:- Using these 5 weights, we can weigh different weights starting from 1 upto 57
1038919
Write down the symbols of several of the other numbers-symbol combinations first then click the silver ball. You will see that the symbols changed. Next on the new page do the math that you were told to do on the first page for several different 2 digit Numbers. You will see that on the page of symbols it is impossible to do the math that they ask you to do without coming up with the same symbol as the symbol that will later be in the Chrystal ball. The math leads to, as the above answerers have said, a multiple of 9. \n\nIt is just that the human mind doesn't really remember what all of the symbols were, so it doesn't recognize that every time you reset the page it has been restacked to give a different symbol, thus giving the illusion of always being right. Also, people seldom will try doing the same number twice, but if they did they would quickly see through the trick because if you picked the same number twice in a row you would realize that the same number lead to a different symbol.\n\nBy the way good job to those above answerers for the math involved I prefered to just explain the trickery.
350288
You question is incomplete....or at least requires more information, as it is, it is too general.\n\nWhat are the conditions are are talking about here?\nAre both balls pushed with some constant force?  As they rolled down an incline?  Are they already in motion?\n\nWhat type of surface are they on?  A smooth (frictionless) surface?  A rough surface where friction needs to be taken into account?  Are both balls solid, hollow, one each?\n\nThey are many many variables which need to be taken into account, you cannot just say "big" and "small" and get an accurate answer.\n\nAnd by "greater distance" do you mean translational displacement, or angular displacement?