Tricky puzzle: From which number can you take away half and leave nothing?
352166
8\ntake away o from top and you're left with o on bottom - half - nothing\nor i guess 10 could work, too (or any double digit ending in 0) take away the first number and you're left with 0
166778
29786\n+  850\n+  850\n_____\n31486\n\nThe way to solve this is to first note that the units digit of Y + N + N is Y. Therefore N must be 5 or 0. If N is 5 then you'd need to carry the one so that now the units digit of 1 + T + E + E is T. But this is impossible because E + E is an even number. Therefore N is 0. Using similar logic, we find E must be 5 and we do carry a one to the next column. Let's look at what we have now:\n\n___1\nFORTY\n+__T50\n+__T50\n_____\nSIXTY\n\nNow, recognize that T50 and RTY are less than 1000 and FO thousand plus two thousand (and change) can only equal SI thousand if O is 9 and I is 1. This gives:\n\n___1\nF9RTY\n+_T50\n+_T50\n_____\nS1XTY\n\nF and S must be consecutive and R and X are not both odd or both even (since we carried the one). If R is 8,=> T is 7,=> X is 3, but there remain no consecutive numbers available for F and S, so R is not 8, yet 1 + R + T + T must be at least 22, so T is therefore 8. R = 6 also fails to allow consecutive values for F and S, so R = 7.\n\n___1\nF978Y\n+_850\n+_850\n_____\nS1X8Y\n\nFrom here it should be easy to see X is 4 and F and S are 2 and 3 respectively leaving Y to be 6 and solving the puzzle.
23859
I run through the numbers from 1-9, looking at each to see where the constraints ensure that there is only one spot left to place an additional number.  To do this, realize that each row, each column and each block need exactly one of each number, so starting with 1, I look for combos with limited possibilities.\n\nIf the blocks 1,2 have a 1 in them, then 3 should be easy to find, look at blocks 6,9 to see if they have a 1, and mentally cross out the rows and columns that those 1's eliminate and see whats left.  Similary, if 2 and 6 have 1's in them try looking at blocks 5 and 3 for enough eliminated squares to ensure there is only one remaining spot for 1 in those blocks. Once you've tried all the possibilities for 1, move on to 2, and keep going until you've been through the whole list of numbers once without finding any new entries.\n\nFinally, a trick I really like, is when a block has only one row or one column with space remaining. You know that the remaining numbers in this mostly complete block will be in that row or column, and so you can use that information and apply it to other blocks even though you could not exactly place the numbers in the mostly complete block.\n\nHope that helps!
133517
the solutions is 7,8,3,2 or 7,6,5,2. The problem works with both sets of numbers.
945730
Just bypass the whole word thing. Letters contain letters. That's not mathematically contradicting.
172875
Obviously you are doing homework and don't want to figure it out yourself.  It is easy, just write them out.  Obviously you can disregard the B, because there is no choice (both dominant).  Now do all the combinations with a "A" and then double your answer to get the "a".\nACD\nACd\nAcD\nAcd\nThat is four combinations.  Double it for "a" and you get \n\nEIGHT\n\nTyping out the question is harder once you learn the rules.
674697
For five consecutive, there are 34 possibilities because you could start with 1, 2, 3,..., or 34\nYou can't start with 35 because then you would have\n35, 36, 37, 38 which is only four numbers.\nI'm not sure about the other two.\nNote: The answer to the other two are not 35 and 36.
885691
Part (a) of problem 1 is pretty easy.  There are 10 marbles total, and 6 of them are red.  On your first draw, you have a 6/10, or 0.6 (60%) probability of drawing a red sock.  In other words, there are 6 ways of drawing a red sock out of a total of 10 possibilities.\n\nFor part (b), you need to know that the probability of two independent events happening is given by the product of the probabilities of each of the individual events.  In this case, the two independent events are that the first marble drawn is red AND the that the second marble is also red.  We already calculated the probability for drawing a red marble on the first draw in part (a).  After the first draw, the box contains 9 total marbles, 5 of which are red, so the probability that the second marble is red is 5/9.  The probability that both are red is given by the product, 6/10 * 5/9 = 30/90 = 0.3 (30%).\n\nThe reasoning for part (b) is the same as for part (a).  The probability that the first marble drawn is white is 4/10, and the probability that the second marble drawn is also white is 3/9.  The probability that both are white is given by the product, 4/10 * 3/9 = 12/90 = 0.133... (13.333...%).\n\nFor part c, there are two ways one can get a red and a white marble: either the first marble is red and the second white, OR the first marble is white and the second red.  When there is more than one way to achieve the desired result, the probability of getting that result is given by the SUM of the probabilities of each of the different ways to achieve that result.  In this case, the probability of drawing red-white is 6/10 * 4/9 = 24/90, while the probability of drawing white-red is 4/10 * 6/90 = 24/90.  The total probability of getting a red and a white marble is then 24/90 + 24/90 = 48/90 = 0.53333....\n\nUsing the same reasoning, you should now be able to solve parts (a) and (b) of the second problem.  You should find for part (a) that the probability is equal to 0.42424242...., and that the probability for part (b) is 0.09090909....\n\nFor part (c), you would use a similar logic as used for part (c) of the first problem.  The probability of getting a matched pair (either both white or both blue) is given by the sum of the probabilities found in parts (a) and (b).  You should find that the probability is 17/33 = 0.51515151....\n\nFor the third problem, you need to know that the expected "payoff" of a game is given by the sum of the products of the prize value times the probability of winning that prize.  In this case, the probabilities of winning the various prizes are:\n\n$100 : 1/500\n$50 : 3/500\n$25 : 5/500\n\nSo the expected payoff (i.e., the "mean" or "average" winnings for a single ticket) is given by:\n\nP = $100*(1/500) + $50*(3/500) + $25*(5/500)\nP = $100/500 + $150/500 + $125/500\nP = $375/500\nP = $0.75
595633
When you are dividing by something you are trying to find out how many of a certain amount (say, 3) is in another amount (say, 12). The result for my example, is of course 4, since 3 goes into the number 12 four times.\n\nNow the number 0 is nothing, literally. So if you're dividing into, say 12, you're wondering how much nothing is in 12. That's just not possible, because 12 is all "something".\n\nI hope that's clear enough...I tried.\n\nEdit: The guy above me is wrong.
1095382
If I'm understanding your question correctly....\n\nIf you have to divide the elephants this way, you know that each person is gonna be getting at least a portion of their elephants from the butcher, right?  :(  poor elephants...  you couldn't have done this same question with, say, apples???  Just lovely....\n\nWell, from the mathematical perspective however, the person to get half the elephants would get 8 1/2 elephants.  The person to get one third would get 5 2/3 elephants.  The person to get a ninth of the elephants would get 1 4/5 elephants.\n\nSo who gets the 1 1/10 elephants left over?  The butcher?  :(\n\nSince you mention "no add no less," and that seems impossible, it would seem that this is supposed to be one of those math riddles?  But even if you made elephant soup, I have no clue how this could be done without leftovers!  The way they're to be divided between the three people, there's 1/18 of the elephants left...  if this is some kinda riddle, I have no clue how that fact is to be resolved.
871673
So call the number\nB = {b_0,b_1,b_2,b_3,b_4,b_5,b_6,b_7,b_8,b_9}\n(this reverse indexing makes things easier).\n\nYou must have b_0 != 0, since you were told it was a ten-digit number. So b_0 is between 1 and 9.\nSo b_0 = 0 - (number of the b_i which are non-zero)\n\nAlso note that Sum_i=0..9 (b_i) = 10 since B has ten digits, and each b_i has to have a single value between 0..9\nHence Sum_i=1..9 (b_i) = (10 - b_0) for any particular choice of b_0\n\nTo prove the number is unique, see below:\n\nIt's easy to show 6210001000 is one possible answer.\nb_0 = 6 since there are six 0's in B\nb_1 = 2 since b_2, b_6 =1\nb_6 = 1 since b_0 = 6\nAll other b_i are 0 since there are no other non-zero b_i.\n\nTo prove the number is unique, the easiest way seems to be to show that the other b_0 between 1..5 and 7..9 are impossible by (proof by contradiction).\n\nb_0 = 9 is impossible since then b_9 must be 1, which gives a contradiction giving '9000000001', which only has 8 zeros, also b_1 must be 1, et.c\n\nb_0 = 8 gives '8000000010', but then b_1 must be 1, and it has one less zero, so that becomes '7100000100'...\nbut that has two ones, so b_1++, and b_2 must be 1, and you have again one less zero so b_0=6, giving '6210001000', which finally works.\n\nHarder to show that b_0 = 1..5 are impossible.\n\nLet's start from the opposite extreme, b_0=1.\nThen Sum_i=1..9 (b_i) = (10 - b_0) = 9\nThis constraint means each b_i can only be 0, 1 or 2.\nSince we set b_0=1, this means we have one zero and eight ones e.g. B='11111111110', doesn't matter where you put the 0. But that doesn't work since clearly b_1 must >=8, thus Sum_i=1..9 (b_i) = 8+7>9 which is a contradiction.\n\nIt should be trivial to crank out the cases b_0=2..4, maybe there is a more compact proof.
1070909
Five and seven\n\nBy giving one, it's four and eight (twice)\n\nBy givng one, it's six and six
924084
Start with the fixed part... 2.1 is 21/10.\n\nThe rest of the number is 0.036363636...\nLet's call that x.\n\nx = 0.036363636...\nNotice what happens if you multiply by 100.  I picked 100 because there are two repeating decimals.  If there were 1 you would pick 10, if 3, you would pick 1000, if 4 pick 10000, etc.\n100x = 3.6363636...\n.......x = 0.0363636...\n\nSubtract and the repeating decimals are cancelled:\n99x = 3.6\n\nMultiply by 10 to get rid of the last remaining decimal:\n990x = 36\n\nNow solve for x:\nx = 36/990\n\nNow you just need to add your two fractions:\n21/10 + 36/990\n\nGet a common denominator:\n21*99/990 + 36/990\n\n= (2079 + 36)/990\n= 2115/990\n\nNow reduce:\n= 235 / 110\n= 47 / 22\n= 2 3/22\n\nSo your answer is 2 and 3/22.\n\nYou can also do this in one step, without having to do the two parts:\n100x = 213.63636...\n.......x = .....2.13636...\n\nSubtracting:\n99x = 211.5\n\nMultiply by 10 to remove the decimal:\n990x = 2115\n\nSolve for x:\nx = 2115/990\nx = 47/22\nx = 2 3/22\n\nA final way, if you now understand the underlying math is to just do the following:\n\nPull out the part in front of the decimal (2)... just keep that for later.\n\nThe remaining part is 0.136363636... \nSince we have two repeating decimals, use two nines (99).  However, since it is offset by one fixed digit, add a trailing zero (990).\nTake the fixed digit (1) and subtract it from your number (136, ignoring anything but the first repeat).  This leaves 135.\n\nSo the fraction should be 135/990.  This reduces to 3/22.\n\nSo it becomes 2 and 3/22.\n\nAlways check that the final result makes sense.  Obviously the first part of the mixed fraction should match the decimal (2) and it does.  If you put 3/22 in your calculator, you get 0.1363636...\n\nThere's 3 different methods and they all give you 2 and 3/22.
216872
on a domino tile there can be 0,1,2,3,4,5 or 6 points on each half. so there are 7 tiles with 0 on one half, 7 with 1 and so on..\n\nI found two ways to count them: \none is 7+6+5+4+3+2+1=28\nanother 7*7-6-5-4-3-2-1=28\n\nyou can take seven tiles with 0 on one half and don't put them back. then you can find only six with 1 on one half, because (0,1) is already taken(and don't put them back). then only five with 2 on one half, because (0,2) and (1,2) are already taken. and so on..\n\nmaybe there is a formula, but you don't need it in this case. or you can use sum of arythmetic progression:}}
1011386
I tried and I found that 100! is perfectly divisible by 350 as 350=2*5*5*7=50*7.\n\n100!= 100*99*98*97!\nnow 100 is divisible by 50 and 98 is divisible by 7.\n\nSo the least number is 1.
1118957
The first number has to be (x+6) and the second number is (x).  There are four possible combinations: 6,0 (60); 7,1 (71); 8,2 (82); 9,3 (93).  60 plus 8 (68) does not equal 06, or 6.  71 plus 8 does not equal 17.  82 plus 8 does not equal 28.  93 plus 8 does not equal 39.\nUnfortunately, this problem is impossible.\nSorry I couldn't be of more help.
801584
If the number begins with 5, there are still three of the odd digits available to fill the last place.  If a 3 is used for the last place, the others are filled using 1, 2, 3, 4, and so there are 4! arrangements.\nIf 1 is used for the last place, that leaves 2, 3, 3, 4 for the others, and so the number of arrangements is  4!/2! because of the repeated 3.  So that gives \n4! + 4!/2! = 36\n\nIf it begins with a 3, the last place can be filled with 1, 3, or 5, and  the numbers left to fill the remaining four places are all different and so the number of possibilities is\n3*4! =72\n\nIf it begins with a 2 or 4 ( two choices), and the last with 1 or 5 (two choices), there are 4!/2! ways of filling the other places because 3 is repeated.\n2*2*4!/2! = 48\n\nIf it begins with 2 or 4 and ends with 3, then the other four digits are all different, and the number of possibilities is\n2*4! = 48\n\nAdding all these together gives 204.  Yes.