LOLQuote:
Originally posted by tite-wad@29 June 2003 - 14:53
:P This kinds works! :P Look closely... :geek:
It sorta looks like the word NINE
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LOLQuote:
Originally posted by tite-wad@29 June 2003 - 14:53
:P This kinds works! :P Look closely... :geek:
It sorta looks like the word NINE
http://members.roadfly.com/tite-wad/laugh.gif
http://members.roadfly.com/tite-wad/wtf.gif It's been an hour already...Quote:
Originally posted by 2nd gen noob@29 June 2003 - 15:31
... i'm just gonna paste in the answer in 5 minutes ...
What's the "answer" noob?
2..4..3
4..3..2
3..2..4
Where's the 28th penny?Quote:
Originally posted by WeeMouse@29 June 2003 - 16:48
2..4..3
4..3..2
3..2..4
I lost it down the back of the sofa........ :P
Did ya? I lost mine (and the other 27 ;)) in a game of Keno in Vegas :P
Quote:
Originally posted by WeeMouse@29 June 2003 - 16:56
I lost it down the back of the sofa........ :P
:lol:
i forgot to give the answer :(
i'm not going senile honest, please don't put me in that nursing home
please :(
anyway, here's the answer...
(i copied this from another page, i don't claim the credit for solving it by the way)
"I think I have solved this riddle if I have interpreted it right, it is a bit hard to work out what is meant from the wording... as it sounds as if the authors are not particularly good at english, i had to reword it. if this is wrong then my solution is wrong. as i understand it,there are 8 piles, three pennies in each pile arranged like
000
0 0
000 a square with nothing in the middle
if you add across or down any row except the middle one, you get 9 pennies as it says now we want a square exactly like this, but we must use 28 pennies, not 24 so, after thinking for a while I set this problem up as a series of simple math equations first letter the board like this
abc
d e
fgh
now let each letter be a variable representing the numbe of coins in that pile
set up the equations
a + b + c + d + e + f + g + h = 28 // must have 28 coins
a + b + c = 9 // top row = 9
a + d + f = 9 // left vertical = 9
f + g + h = 9 // bottom row = 9
c + e + h = 9 // right vertical = 9
solving these equations simultaneously, a moderately long process that will not be repeated here, gives the simple equation:
b + g = 10
this cannot be reduced further and suggests that there are multiple solutions to the problem, probably at least 4 or 5. nevertheless, choosing the most obvious values, b = 5, g = 5 we now have the following square
?5?
? ?
?5?
none of the equations relate b and g to a third variable explicitly, so we have another choice, and more solutions. Still, the top row = 9, ie a + b + c = 9, ie a + 5 + c = 9, ie a + c = 4
choose again a common value, both equal to 2 and we have the following square
252
? ?
?5?
next one of the equations that can be gotten by writing out and solving the equations simultaneously is 9 - b - c + d + f = 9, ie d + f = b + c, ie d + f = 7 again, arbitrarily choose two values, I chose d = 4, f = 3 now we have the square
252
4 ?
35?
obviously the bottom right hand corner must be 1
252
4 ?
351
and obviously the right middle square must be 6
add all the squares up and the total is ... 28
add all the rows and or columns up, and the total is .. 9
therefore a solution to the problem is
252
4 6
351
This is not a unique solution. Unless I have misinterpreted the question, this is one of the many solutions to this riddle."
WTF?
i didnt bother to read that
:huh: @noob, the solution you posted doesn't match the original question.
Why didn't you elaborate when we were asking about the "piles" and "rows"?
Ornery bugger! <_<
Edit: typo