there is a pile of 12 identical bricks.  one of the bricks weighs heavier OR lighter than the other 11.  you have a libra balance scale to compare groups of bricks BUT you can only perform 3 comparison weight measurements.  there IS a solution and it is complex.  the solution algorithm MUST be able to determine the odd brick everytime and in every case. it is one of the simplest stated problems and requires no knowledge of advanced math or logic but does require ONE remarkable insight to solve.  i solved the problem without any help but it took me several days.  to date i dont know anyone who has been able to solve this problem that i know.  perhaps some of you out there will be able solve this very interesting puzzle.
130237
Spoiler warning!\n\nSolution is given in the source links below.
222262
Sounds like a Math/Science type question to me. Why are you posting it here?\n\nBesides, I have the the slightest idea.
235327
Let the weight of the brick be x.\n\nx = 3/4 + (3/4)x\nx - (3/4)x = 3/4\n(1/4)x = 3/4\nx = (3/4)/(1/4)\nx = (3/4)*4\nx = 3 pounds
247543
Yes, you place a pile of 6 coins on either side of the scales, which ever set of six is lighter you split into 2 sets of 3 coins and do the same, then with the lightest set of 3 coins, you weigh 2 of them, if they weigh the same, the coin you didn't weigh is the counterfeit.  Or which ever of the 2 coins was the lightest.
625535
I may be barking completely up the wrong tree here. Are you comparing one ratio with another?\n\nIn the first, there are 45 parts, in the second there are 19.\n\nMost math problems have a glimmer of sense about them, in that when you hit upon the right way to solve them or what they are really asking, the calculation is quite easy.  I was expecting the number of total parts in each expression to be an easy multiple of each other, which they are not.\n\nIf you are expected to apply the first ratio to the 18 side of the second, that looks a little more reasonable (45 and 18 are both multiples of 9.)\n\nDoes that help without seeming to? (Also I understand you to mean 'factor' and 'easy')
444364
If I have interpreted you correctly, each of the 49 packages must contain either BBB, BBS, BSS, or SSS.\nThen there are:\n22,100 combinations.\n\nThis is the same as the number of solutions of the equation:\nx1 + x2 + x3 + x4 = 49\nin non-negative integers.\n(i.e., x1 is the number of BBB packages, x2 is the number of BBS, x3 the number of BSS, and x4 the number of SSS)\n\nThis is C(49+4-1, 4-1) = C(52,3) = 52*51*50/(3*2*1) = 22,100.\n\n*************************\nADDITIONAL DETAILS:\nOk, I saw that you reposted this question.  However, you are still missing something or have the numbers or the question incorrect.\n\n285 = 3*5*19 .  This number is not a result formable by the combination function: C(a,b) = a!/[(a-b)!b!]\n\nThe closest would be 286 = C(13,3).  This would be the number of ways that 3 items could be chosen from 13 items.\n\nAlso, as you have it written on the other post, my answer above is still correct.  There are 22,100 ways that 4 different types of items could be distributed into 49 packages.  This assumes that the packages are indistinguishable from each other.  If they weren't, i.e., if BBB in package #1 and BBS in package #2 is different from BBS in #1 and BBB in #2, then the number of combinations is MUCH larger.  (It would be 4^49 ~ 3.2 * 10^29)
474700
It helps if you know something about Pythagorean triples (integer solutions to the Pythagorean equation for right-angled triangles).\n\nIf you do, you will know that there are no primitive triples in which the difference between the legs is 3, The smallest such differences are 1, 7 and 17 (always prime, but not all primes will serve).\n\nTherefore this must be a non-primitive triple which is a multiple by three of a primitive triple with the legs differing by 1. There is a series of such triples, starting with (3, 4, 5) which can be generated from one another.\n\nSo we divide 87 by 3 and get 29 and recognise that this must be the triple (20, 21, 29), the second in the series.\n\nSo plainly the answer is length = 63 cms and width = 60 cms (i.e. 3 x 21 and 3 x 20)\n\nTHE ALGEBRA \n\nL = B + 3\nL^2 + B^2 = 87^2\n2 B^2 + 6B + 9 = 7569\nB^2 + 3B - 3780 = 0\n(B - 60) (B + 63) = 0\n\nAs rectangles in the real world have sides of positive value, ignore B = -63 and take B = 60, yielding L = 63,\n\nBut none of that is necessary if you know your triples!
164606
I think practice is the only way. I use several different methods. But usually, I start with the obvious ones. Look within the boxes first to fit in the easy ones. Do that once again in case you missed any, then look at each row and column. Sometimes, if there looks like a few possibilities, you can work out if there is a number which can only go in one square which you might not see straight away. Sometimes, there will be squares where only 2 numbers are possible. If there are 2 squares in the same row/column/area which have the same 2 possiblities and then there is only one square remaining then you know the last square must be the other number. Some people don't think of that.
89131
They used log tables.\nSimple problems like this could also be solved by using Newton's method.
522588
Can you please check you haven't made a mistake somewhere? I have just calculated (with 99% certainty) that there are no solutions to those equations with those restrictions.\n\nEdit - OK, I assumed you made exactly one mistake. By removing each possible word and trying to solve it, I discovered that removing 'SIGH' was the only option that gave an answer.\nt 1\nh 8\na 3\nn 7\nk 6\ns 5\ng 9\ni 4\nv 2\nwhere SIGH = 26, not 25. Then THANKSGIVING = 65.\n\nThat was actually a cooler puzzle that the original, assuming you made one mistake :)
307030
That would be simple assuming the piles of gold were made up of similar pieces, like gold bars or coins of a supposed known weight and the scale was a simle weighing device instead of a balance type scale.  Take 1 piece from the first pile, 2 from the second, 3 from the third, and 4 from the fourth.  Weigh all the pieces.  The pieces from the good pile would weigh 10% more than the bad ones.  Assume each bar is supposed to weigh 1 ounce/gram whatever.  There are 10 bars on the scale (1+2+3+4).  If all of them were of the correct weight, then they would weigh 10.0 grams or whatever, but we know that at least 6 and at most 9 will weigh less.  If the pile weighs .6 ounces less than 10.0, then the real pile would be #4, because 4 weighed correctly and 6 did not.  The same logic applies if the weight is .7, then the third pile would be correct, .8 less means the second pile is real, and .9 less means the first pile is real.\n\nIf it was a balance type scale or if the piles of gold were gold dust, something else would need to be figured out.
47677
you can use these numbers 1, 3 , 9 , 27\nactually these numbers are 3^0, 3^1, 3^2, 3^3 \nand here you are using 3 sounting system \nSo three counting system can use 0 , 1, 2 of each weight\nbut since we don't have except one weight of each unit, mathematics allow to use weights -1 , 0, 1 instead of 0, 1 ,2 so you can either put the weight unit with the weighed object (-1) or on the other side (1) or don't put it at all (0)\nBest Regards
89827
1+1=1\n\nOne pile of rocks and another pile of rocks can be combined to make what is just one, albeit larger, single pile of rocks.
77425
I will assume that the one golf ball is heavier than the other twelve.\n\n1st weighing:\n\nSplit the golf balls into two groups of six, leaving one out.  Weigh the two groups.\n\nIf the groups balance, the 13th ball is the heavier of them all and the puzzle is solved.\n\nIf one group tips the scale, the heavier golf ball is in that group.\n\n2nd weighing:\n\nSplit the heavier group of six into two groups of three golf balls.  The heavier golf ball is in whichever group of three tips the scale.\n\n3rd weighing:\n\nSplit the heavier group of three into two groups of one and leave one out.  Weigh one golf ball against the other.\n\nIf they balance, the heavier golf ball is the one you did not weigh out of this group.\n\nOtherwise, the golf ball that tips the scale is the heavier golf ball.
315436
It's not square *roots*... it is the series of PERFECT SQUARES!!! How many times are people going to ask this? And isn't it obvious? This is part of some Google DaVinci Code contest, isn't it?\n\nWhy don't you just search for the same question because it has been asked about 31 times now!!!
121441
The cubes, when stacked, form a brick with sides divisible by 3. \nTwo dimensions of the tin are divisible by 3, while 8, the height, is not. This means we will only be able to put two stacks (height of 6) into the box with 2 inches of air on top. The volume of the stack is 6*12*9 = 6*4*(3^3), so we will only be able to fit 24 cubes inside, unless we cut up the remainder (which will, incidentally, fit inside).
17241
Jan 17 Note: If this is hard to understand, after sleeping on it, I may have an easier construction.\nIf you need further explanation, add another note to your question.\n\n-----------\n\nI think I have a solution.\nI'm making the assumptions that the Babylonians didn't have an acurate value for pi and used the value three. If I do that, every thing falls out.\n\nDraw a cirlce. Mark off 2 arcs. Make one the top portion of a circle say from 10 o'clock to 2 o'clock, the other the matching arc from 4 o'clock to 8 o'clock. Note: These arcs are just to get a representation of the problem and are not an acurate length.\n\nNext draw the two chords for the arcs, that is, the straight line from 10 to 2 and the other from 8 to 4.\nNow draw in the diameter of the circle from 12 o'clock to 6 o'clock.\nThe two chords cut off pieces on the ends of the diameter. Label these as length 2 (2 is the perpendicular) "Double 2 and get 4" That's the length of the two end pieces of the diameter.\n"Take 4 from 20 and get 16" This is where using 3 as a value for pi comes in. If pi 3, then a circle of circumference 60 will have a diameter of 20, so let the length of the diameter be 20. Taking 4 from 20 gives you the length of the diameter that is between the 2 chords, 16.\n\n"Square 20 get 400, square 16 get 256, take 256 from 400 get 144. Find the square root of 144. 12, the square root, is the chord."\n\nWe're obviously doing some Pythagorean theorem here so where is the triangle? \nDraw the diameter of the circle from 8 o'clock to 2 o'clock, that's the hypoteneuse as has length 20.\nThe distance between the two cords is 16. That's also the distance from 8 o'clock to 10 o'clock. Draw that line and label it 16. (Remember the drawing isn't accurate just enough to show us how things are layed out).\nThat leave the 3rd side of the triangle as the chord between 10 o'clock and 2 o'clock. \n\n20^2 - 16^2 = 12^2  so 12 is the length of the chord.\n\nHow does that sound?\n\nIn modern English, the statement of the problem would be something like this:\n\nLet AB be an arc of a circle of circumference 60. Let X be the chord connecting A and B.\nLet Y be the perpendicular bisector of the chord. If the distance from X to the circle along Y is 2, find the length of X. (You may assume that pi = 3 for the purposes of solving this problem.)