A cube of cheese is divided into 27 subcubes.  A mouse starts at one corner and eats through every subcube.  C
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It cannot end in the middle.\n\nLet's look at the cube.  \nIt's divided 3x3 to give you 27 subcubes.\nThe middle cube is therefore only one deep from any side.\nThere are eight subcubes on each corner of the cube.\nThere are six subcubes in the middle of each face of the cube.\nWe'll put all these subcubes into a group.\nCall it group A. 8+6=14 subcubes.\n\nAnd there are all the other cubes.\nThere are 27 subcubes.  14 are in group A.\nSo there are 27-14=13 of these.  \nWe'll group them together and call it group B.  13 subcubes.\n\nSome things become clear at this point.\nThe mouse could progress always ahead. So...\nEach time the mouse eats one of the group A subcubes, it must go to one of the group B subcubes.\n\nEach time it eats one of the group B subcubes, it must go to one of the group A subcubes.\n\nEven if it goes sideways, to a subcube in the same group (that's only possible for group B subcubes because they are the only subcubes where members of the group can be adjacent), what happens?\n\nThe number of subcubes in group B will reduce by two instead of one.\nThe number of subcubes in group A will remain the same as before.\n\nHowever the mouse moves and eats, there will always be more group A subcubes than there are group B subcubes.\n\nThe middle subcube belongs to group B.\nIt's one of the group of 13 subcubes.\n\nWhichever group the mouse eats from on the first day, he will always have one of group A subcubes left.\nBecause there are always more group A subcubes than group B subcubes.\n\nI'll repeat all that.\n27 subcubes, one per day = 27 days. \n13 subcubes in group B.\n14 subcubes in group A.\n\nIt is impossible for the mouse to have eaten the last group B subcube and there be no group A subcubes left.\nBecause there are always more group A subcubes than there are group B subcubes.\n\nSo the last subcube the mouse eats will always be one of group A.\nA corner subcube or a subcube that's in the middle of one of the faces.\nNot the subcube in the middle.\nQED, I think.
797253
12,000,198 microscopical creatures\n\nit's 7 x 7 x 7 x 7 x 17 x 7 x 6 x 7 = 12,000,198\n\nBy the way, why is this in Words and Wordplay not Math?\nAm I missing something?
46428
1, 2 3, crunch\nTHREE!
547564
Estimate it=use about 70 sq meters, as supplied by Mars, and figure 1.5 candies per centimeter long, and 2 per centimeter high-so 6 candies per cubic cm. I would estimate about 5, though, because I dont think they are as small as I estimated.  So at 1 million cub cm/cub M, that would get you about 5 X 1 million, X 70, or about 350 million.  Because they are oblate spheroids, they would pack slightly tighter than the flat rectangular boxes, so I'd add a fudge factor of about 20%.  Say anything from a low end of about 200 million to a high end of about 500 million.
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http://computer.howstuffworks.com/mouse.htm
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Here is a way to cut a perfectly circular pizza into sector-shaped slices for N people some of whom demand small slices, some large, and the rest will accept any size: First choose any irrational number s . Cut the pizza from its center east to its edge; then turn the pizza through an angle s*pi clockwise and cut it again from its center east to its edge; repeat this turn-and-cut until the pizza has been cut into N sectors. Prove that they have just two or three different sizes. \n\nMuch harder problem: how do you choose s?
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The Cube has 43,252,003,274,489,856,000 different possible configurations.\n\nI cheated and looked up the answer, but to figure it out you'll want to use combinatorics.  I'll have to work on that part of it...\n\nIf you're taking the cube apart and counting how many different ways you can put it back together, this is a relatively simple problem (although it doesn't match the above answer).\n\nThe center cubes don't move at all, so there is only one possible configuration of them.\n\nThere are 12 edge cubes, which each have 2 outside edges and therefore two orientations for each location.  The number of possible combinations of these edges is (2*12)*(2*11)*...*(2*2)*(2*1) = (2^12)*12!\n\nThere are 8 corner cubes, which each have 3 outside edges and therefore three orientations for each location.  The number of possible combinations of these corners is (3*8)*(3*7)*...*(3*2)*(3*1) = (3^8)*8!\n\nMultiply the number of possible combinations of edge cubes by the number of possible combinations of corner cubes and you end up with 519,024,039,293,878,300,000 combinations.  \n\nThis is different than the answer I looked up because there are possible combinations when you take the cube apart that cause it to no longer be solvable.  I can't think of an easy way to find the number of combinations that keep solvability.
1145054
dim c(5) as integer\ndim i as integer\n\nfor i = 0 to 4\n    i = i+1\n    c(i)=3\nnext i
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american cheese, goat chesse, cheddar cheese, string cheese, swiss cheese
992764
64.\n\n1 cubed = 1\n2 cubed =8\n3 cubed =27\n4 cubed =64\n5 cubed = 125 etc.
964273
If you number the tpopping from 1 to 8, you can have the following combinations of different toppings (numbered for simplicity):\n1) 1, 2, 3\n2) 1, 2, 4\n3) 1, 2, 5\n4) 1, 2, 6\n5) 1, 2, 7\n6) 1, 2, 8\n7) 1, 3, 4\n8) 1, 3, 5\n9) 1, 3, 6\n10) 1, 3, 7\n11) 1, 3, 8\n12) 1, 4, 5\n13) 1, 4, 6\n14) 1, 4, 7\n15) 1, 4, 8\n16) 1, 5, 6\n17) 1, 5, 7\n18) 1, 5, 8\n19) 1, 6, 7\n20) 1, 6, 8\n21) 1, 7, 8\n22) 2, 3, 4\n23) 2, 3, 5\n24) 2, 3, 6\n25) 2, 3, 7\n26) 2, 3, 8\n27) 2, 4, 5\n28) 2, 4, 6\n29) 2, 4, 7\n30) 2, 4, 8\n31) 2, 5, 6\n32) 2, 5, 7\n33) 2, 5, 8\n34) 2, 6, 7\n35) 2, 6, 8\n36) 2, 7, 8\n37) 3, 4, 5\n38) 3, 4, 6\n39) 3, 4, 7\n40) 3, 4, 8\n41) 3, 5, 6\n42) 3, 5, 7\n43) 3, 5, 8\n44) 3, 6, 7\n45) 3, 6, 8\n46) 3, 7, 8\n47) 4, 5, 6\n48) 4, 5, 7\n49) 4, 5, 8\n50) 4, 6, 7\n51) 4, 6, 8\n52) 4, 7, 8\n53) 5, 6, 7\n54) 5, 6, 8\n55) 5, 7, 8\n56) 6, 7, 8\n\nThis means that there are 56 different topping combinations. Since there are 3 types of pizza, 168 (which is 56 * 3) is the number of different pizzas you can choose from here.\n\nAnswer: 168 different pizza type and topping posibilities.
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the answer is 56 cu.in i.e. choice (D)\nVol of the cube is side*side*side and so is 6*6*6 cu.in i.e.216 cu.in..........................(1)\nVol of the rectangular solid is length*breadth*height and so is 10*4*4 cu.in i.e.160 cu.in................................(2)\nComparing (1) and (2) it can be seen that the cube-shaped box is bigger than the rectangular box by\n216 less 160 cu.in i.e.50 cu.in\nHence the choice (D)
55702
Here's a table:\n___Joan________Peg_____________\n     3.........................1\n     4.........................2\n     5.........................3\n    n+2.......................n
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You can divide the cube into 9 1x3x3 slices.\n\nEach one contains the following prisms:\n1x1x2: 12\n1x1x3: 6\n1x2x2: 4\n1x2x3: 4\n1x3x3: 1\nSubtotal 27 x 9 = 81\n\nYou can divide the cube into 6 2x3x3 slices.\nEach one contains the following prisms that weren't counted above:\n\n2x2x2: 4\n2x2x3: 4\n2x3x3: 1\n\nSubtotal 9 x 6 = 54\n\nThat's 135.\n\nIf you count the 3x3x3, you have 136.
837284
Maybe a cheese contest.
773136
the strip of cheease next to the mouse
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Wrong--if you count the 16 cubes on all four sides, you forget that you are counting edge pieces twice and corner pieces four times that way.  Look at it as two opposite faces, each with 16 cubes (although I though Rubik's cubes always were 3X3, but I guess there are 4X4s too).  Then you have 32 plus 8 on each of two other pieces that aren't counted yet.  The last two sides only have 4 cubes that aren't counted yet.  That gives 16*2 plus 8*2 plus 4*2 or 32+16+8=56 visible cubes.  If you want to count all cubes internal and otherwise and assumed it was a block of cubes 4X4X4 (Rubik's cubes obviously don't have cubes on the inside), then you would just have 64.  Basically there would theoretically be a block (2X2X2) in the inside that you can't see (8 cubes is 64-56).  I hope this is the answer you're looking for and didn't simply confuse you further.