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• Another solution to North,
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  I believe anywhere along the latitude which is 1 mile away from the South pole are all also valid answers to this problem.
  
  Walking 1 mile South from this set of points will land you on the South pole. At the 2 poles, East and West are undefined, so moving 1 mile East will not move you at all. Think about it, moving East or West is defined as going clockwise or counterclockwise at a particular latitude, remaining at that latitude with only your longitude changing. Since 90 degrees South describes not a circle, but a point, moving East will have you remain on the same point.
  
  Then the final 1 mile North. Since moving North or South is defined as moving along the surface of the Earth towards the North or South pole, moving North from the South pole is equally valid in any direction. Why not go along the same mile you took in step 1, ending up at the original point 1 mile from the South pole?
  
  Clearly, this leaves an infinite set of solutions, since anywhere along that particular latitude will work. Any problems with this solution?

  Posted by EC_ on July 22, 2006 at 20:04:17 Pacific Time

• Alternate solution to Calendar Cubes
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  Suspicious of the wording in this puzzle (the examples of 1, 16 and 31 never specify that the given numbers are actually facing up, only that they exist on the cube), my roommate suggested using a non-decimal base to solve this problem. Base six is a good choice, since all values from 00 to 35 can be displayed.
  
  Using base six, both cubes would display 0-5. Generally, this does not contradict any of the clues. The 1st day of the month would display 01. The 31st would display 51, and could be oriented such that the sixes place has a 1 "appear on the front", and the ones place has a 3 "appear on the front." The clue for the 16th day poses a problem. On that day, the cubes would display 24. The ones place cube still "has a 1," but the sixes place cube does not display the numeral 6. However, the puzzle is worded such that this could be read as "having a digit with the value of 6," in which case the sixes place also "has a 1" which actually represents the value 6.

  Posted by LevRickards on June 24, 2006 at 08:46:02 Pacific Time

• Alternate solution to Coins on the table.
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  Assuming the table is thick enough on it's side...flip it so that it is balanced on it's side and prop the legs to hold it in place. balance a single quarter on the top edge of the table. Since the table is round any other quarter placed should roll off.
  
  Yah, I know there are lots of assumptions being made about other properties of the table (friction of the table surface, design of the table base, etc..) but I think I deserve a few points for thinking "non-horizontally"
  
  

  Posted by Spig68 on May 26, 2006 at 19:13:45 Pacific Time

• 1 and a bit north of south
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  Circumference is piD. Since We're wanting Radius, we have to solve for that too.
  Diameter, if we know the circumference is 1, has to be D/pi. and radius, thus, is 1/2pi.
  So, the distance north is 1+(1/2pi). or about 1.159 miles...roughly 1 mile and 840 feet.

  Posted by Sean_Roach on May 22, 2006 at 17:46:47 Pacific Time

• Coins on the Table solution is wrong.
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  The Coins on the Table answer is not 100% correct. This puzzle, and the solution, rely upon an assumption. The assumption here is that the kitchen table is at least big enough to place a quarter upon without violating the other rules. The size of the kitchen was never given so I am free to imagine any size. And when I imagine a kitchen table smaller than a quarter, the given solution fails.
  
  That's stupid you say? No kitchen table is that small. Wrong again. An assumption was made that we are talking about a kitchen table for us humans. I say there are hundreds of thousands of kitchen tables smaller than a quarter.
  
  Think doll house.
  
  It's a good idea to challenge the question before working on the answer.

  Posted by BJ on May 20, 2006 at 03:14:45 Pacific Time

• Coins on the Table solution is NOT wrong.
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  It's true that the problem didn't give the size of the table. And if the table is less than the size of a quarter, the first person to go would lose.
  
  However, the problem states that 'you sit down at the perfectly round kitchen table'. So, you are sitting at a *specific* table. Presumably, this kitchen table is for humans to use, and is larger than a quarter. Presumably.
  
  But the clincher is this: You want to go first because you have figured out a surefire way to win (by going first). Since you are seated at a specific table, this means that the size of that table must be bigger than a quarter.
  

  Posted by spiderbot on May 23, 2006 at 21:33:20 Pacific Time

• better method
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  Method for the laziest of the lazy ...
  
  
  Let 100! = (2^x)*(5^y)*N where N has no 2's or 5's as factors. For every trailing zero there is a pair made up of a 5 and a 2. So the number of trailing zeroes is equal to min(x,y) (the smaller of x or y).
  
  
  Half the numbers are even (that's 50), and one fifth are multiples of 5 (that's 20). So there are at least 20 trailing zeroes.
  
  
  Now count the 5's (since there are probably fewer of them):
  5*1 up to 5*20 -- that's 20
  25*1 up to 25*4 -- that's 4
  any 5 cubeds? Nope. We are done. It's 24.

  Posted by arouse on May 18, 2006 at 13:15:06 Pacific Time

• for the lazy programmer...
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' written in vbscript
' computes number of trailing zeros of n!

function fac2tz(n)
fac2tz = 0
if n < 5 then exit function
limit = 5
do
fac2tz = fac2tz + (n \ limit)
limit = limit * 5
loop until limit > n
end function

wscript.echo fac2tz(100)

'output for n = 100:

'21


  Posted by backtalker on May 17, 2006 at 14:53:06 Pacific Time

• for the lazy programmer...
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  typo:
  
  output is 24 for n=100, not 21

  Posted by backtalker on May 17, 2006 at 15:06:50 Pacific Time

• for the lazy programmer...
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  I think you mean to say for n<125.

  Posted by arouse on May 19, 2006 at 15:42:36 Pacific Time

• for the lazy programmer...
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  You are correct.
  
  I was thinking of this script:
  
  
' written in vbscript
' computes number of trailing zeros of n!

function fac2tz(n)
fac2tz = 0
if n < 5 then exit function
fac2tz = (n \ 5) + (n \ 25) + (n \ 125)
end function

wscript.echo fac2tz(100)

'output for n = 100:

'24

  
  Sorry about that, I guess I'm too lazy to check my own work...

  Posted by backtalker on May 21, 2006 at 01:56:48 Pacific Time

• for the lazy programmer...
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  perhaps I should have pointed out that the program will work for very many values of n, not just n=100.
  
  If the program only needed to be used for values of n=624 and less, it would be even simpler.
  
  
' written in vbscript
' computes number of trailing zeros of n!

function fac2tz(n)
fac2tz = 0
if n < 5 then exit function
fac2tz = (n \ 5) + (n \ 25)
end function

wscript.echo fac2tz(100)

'output for n = 100:

'24


  Posted by backtalker on May 18, 2006 at 20:11:19 Pacific Time

• not-in-my-head math
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  I used bc to calculate 100! and sure enough...
  
  9332621544394415268169923885626670049071
  5968264381621468592963895217599993229915
  6089414639761565182862536979208272237582
  51185210916864000000000000000000000000
  

  Posted by scayford on May 10, 2006 at 20:21:46 Pacific Time

• not-in-my-head math
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  I'm confused why you don't add another zero for every group of five twos. For example, I can see why you add two zeroes for 25, one for each 5 and a 2. But since there's so many twos, wont there be at least a few more zeroes added on?

  Posted by BarkingMouse on May 28, 2006 at 05:08:30 Pacific Time

• not-in-my-head math
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  No.
  
  When the 5 * (something) were counted it also counted numbers like 25 which has TWO factors of 5.
  
  So in the second pass, where 5^2's were counted, we count it again and gather up the reamining 5. And so on.

  Posted by arouse on June 14, 2006 at 13:02:24 Pacific Time