Riddler's Gauntlet-reloaded

[dlingeverything]

Originally posted by tite-wad@29 June 2003 - 14:53
This kinds works!

• 

Look closely...

It sorta looks like the word NINE

LOL

[tite-wad]

Originally posted by 2nd gen noob@29 June 2003 - 15:31
... i'm just gonna paste in the answer in 5 minutes ...

It's been an hour already...

What's the "answer" noob?

[WeeMouse]

2..4..3

4..3..2

3..2..4

[tite-wad]

Originally posted by WeeMouse@29 June 2003 - 16:48
2..4..3

4..3..2

3..2..4

Where's the 28th penny?

[WeeMouse]

I lost it down the back of the sofa........

[Illuminati]

Did ya? I lost mine (and the other 27 ) in a game of Keno in Vegas

[tite-wad]

Originally posted by WeeMouse@29 June 2003 - 16:56
I lost it down the back of the sofa........

• 

• 

[2nd gen noob]

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."

[dlingeverything]

WTF?

i didnt bother to read that

[tite-wad]

@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