A cube of cheese is divided into 27 subcubes.  A mouse starts at one corner and eats through every subcube.  Can it finish in the middle?
210889
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.
453545
here you go...\n(it'll work with up to 255 different disks, if you need more, you'll have to replace chars with ints)\n\nm1(char *a,b) {memove(b+1,b,strlen(b)+1); *b=*a; memmove(a+1,a, strlen(a);}\nm(char *a,b,c) {if (*a)  {m(a+1,b,c);m1(a,c);m(b,c,a);}
294218
There will be 325 marbles in the 25th row.\nSolution- As we move a layer(downward)the number of marbles in that row will get added to its consecutive number(keeping in vie ho much has been added before).For eg-\nif the 1st row has 1 marble,the 2nd row will have 1+2=3 marbles.Similarly, the third row will have 3+3=6.The third row will have 6+4=10, 5th row-10+5=15 and so on.....
369457
The nth term is n cubed.\nThe third tem is 27.
450766
Since the piece of cheese is at rest. and no net force is acting. by Newton's First Law of Motion, it would continue in its initial 'state of motion'  which is rest here (since it was initially at rest) ....all other choices would violate the law!. got it dear!
71386
move the mouse to the center and leave it there and they go away eventually
286133
I can do it... but let me think a little on it.  Can I take the full 4 months?  :-)\n\nOkay, I think I have a solution... took me about 5 minutes I think:\n\nIf I move the dot from B1 -> B2 and C1 -> C2, then B and C will still have 4 dots, and row four will still have 4 dots, but now row two will have 4 dots.\n\nYou didn't say they couldn't share the same square!!!!\n------------\nI'm back\n\nYou didn't exactly state what a "row" was.  I would have called that a "line" but maybe that's what you meant.\n\nOkay, I have a couple revised solutions.  A "row" is now defined as 4 dots that are colinear (on the same line) but each point must be distinct.  This invalidates my previous answer.\n\nOne solution matches with what has already been stated, which is:\nA4 --> B5  and  D4 --> C5.  This makes two pairs of overlapping lines each with 4 dots.\n\nHowever, my new solution is to take C1 and place it halfway between columns B and C and halfway between rows 2 and 3.\n\nThen you have a line on row 4, a line on column B, and two diagonal lines!!!!\n\nOne dot moved, 4 lines created!!!!\n\nAnd if you *must* move two dots, then just move B1 --> B5 and you still have a valid answer.
403089
27 cubes, then 81, then 243, then 729...
60373
Here's the algorithm\n\n    solve ( current, final )\n    {\n 1      let max be the number of disks\n 2      let dest be the final place of max\n 3      let disk = max\n        repeat\n 4          while disk > 0 do\n 5              if disk is already on dest,\n 6              or, moving it succeeds then\n 7                  if disk = max then\n 8                      decrement max by 1\n 9                      if max = 0 then\n10                          return    // done\n                        end if\n11                      let dest be the final place of max\n                    end if\n                else\n12                  let dest be the alternative place between dest and\n                    the current place of disk\n                end if\n13              decrement disk by 1\n            end while\n14          let p and q be the places different of dest\n15          let disk be the smaller of the disks on top of p and q\n16          let dest be the place between p and q with greater disk on top\n        end repeat\n    } \n\n\nif u need more help..\n Just let me know...
182273
Let 1 2 3 4 5 6 7 8 9 A B C D be the dots, with D in the middle. (You say 1 in the middle)\n\n1 2 3 4\n5 -- -- 6\n-- -D- --\n7 -- -- 8\n9 A B C\n\nIT CAN BE DONE WITH 5 LINE SEGMENTS AND DOESN'T NEED 6.\n\nConnect 7-5-1-2-3-4-D-9-A-B-C-8-6 in a Z-shape\n\nSo I think if it needs six lines, your question was meant to be regarding 16 dots, 4 of them in the middle. It would then be similar to your other question with 9 dots.\n\n1 2 3 4\n5 6 7 8\n9 A B C\nD E F G\n? -- -- !\n\nConnect 1-2-3-4-8-C-G-! and then proceed as for 9=dot problem, F-A-5-6-7-8-B-E-?-D-9. Six lines needed.\n\nGeneralising, an n x n square of n^2 dots will require 2n-2 lines. (for n>2)\n\nImagine a further 9 dots wrapped around the 16-dots as follows\n\nH I- J K L\n1 2 3 4 M\n5 6 7 8 N\n9 A B C 0\nD E F G P\n\nConnect G-F-E-D-0-5-1-H-I-J-K-L-  M-N-O-P and then proceed as for the 9-dot problem:\n\nC-7-2-3-4-M-8-B-E-A-6\n\nwhich is 8 lines.
93094
Actually I had something to help you but x=0 which makes it pretty difficult.\n\nI guess you could try to talk about tomatoes or something.  Say Tom (X) has 1/2 the amount of tomatoes that Jill has, plus four.  Or in other words, if you take Jill's amount of tomatoes, multiply it by two, and take Tom's amount and multiply it by three, there total amount of tomatoes will equal 12.  How many tomatoes does Jill have?  How many does Tom have?
481089
You start from the end and work backward.  If Dad ate half the cookies and left two, he must have eaten two so when he got there there were 4.  Ashley had just eaten 1/3 of the cookies, so she had left 2/3 in the jar.  Well, four is two/thirds of six so when she got there there were six cookies.  Charles had eaten 1/4 meaning he had left 3/4 of the cookies.  Well, 6 is 3/4 of 8 so originally there were 8 cookies in the jar.  Good luck, hope that helped.
116784
3454.3615403812755165920718499567 centimetres cubed
40793
With this given data, it is not possible\n\nalthough the area is totalling to be 70 for both cases (7*10 and two pieces of 8*8 + 1*6) but with one division only, this is not possible\n\nPLEASE DO LET ME KNOW IF THERE IS SOME SOLUTION (i found it very interesting)
208898
This is very tricky. Start off with the basics. Use Red and Green, 2 cubes and count sides of color.\n\ncube1: gggggg\ncube2: rrrrrr\n\nrrrrrr - gggggg\nrrrrrr - gggggr\nrrrrrr - ggggrr\n.\n.\nrrrrrg - gggggg\nrrrrrg - gggggr\n\nTake out all the similiar ones to find the count.\n\nInteresting question. Probably easiest to run a recursive computer simulation to find out.\n\nTotal for 1 cube = 10\ngggggg - 1 way\nrggggg - 1 way\nrrgggg - 2 ways\nrrrggg - 2 ways\nrrrrgg - 2 ways\nrrrrrg - 1 way\nrrrrrr - 1 way
52724
Hey asker, this is a great puzzle.  Could you provide the correct answer after a couple of days?  I am really curious.  Thanks.
210599
keep it together, cut it in half, the stack the halves.  then cut it twice.  you will have 3 stacks of 4