Computer Science Canada another riddle |
| Author: | mo [ Fri Feb 01, 2008 6:49 pm ] |
| Post subject: | another riddle |
you have 10 pennies each penny identical, but one weighs more than the others so 9 are the same and one is different you also have a scale(no numbers or anything) where it has two sides and you put the coins on each side to see which pile weighs more you can only use the scale twice figure out which penny is different and if you have seen this riddle somewhere else please dont answer it |
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| Author: | Zampano [ Fri Feb 01, 2008 6:54 pm ] |
| Post subject: | Re: another riddle |
You made a mistake I think. That's 9 pennies, not 10. When there's more 3 to the power of the number of weighings, it fails. |
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| Author: | mo [ Fri Feb 01, 2008 6:58 pm ] |
| Post subject: | RE:another riddle |
sorry folks question is 9 pennies instead of 10 my bad |
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| Author: | fishtastic [ Fri Feb 01, 2008 6:59 pm ] |
| Post subject: | RE:another riddle |
10 pennies?? how do you solve for 10? |
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| Author: | fishtastic [ Fri Feb 01, 2008 7:00 pm ] | ||
| Post subject: | Re: another riddle | ||
OIC..
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| Author: | mo [ Fri Feb 01, 2008 7:01 pm ] |
| Post subject: | RE:another riddle |
explain more thoroughly |
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| Author: | pistolpete [ Sun Feb 17, 2008 11:43 pm ] |
| Post subject: | RE:another riddle |
You divide the 9 coins into 3 piles, place any 2 on either end of the scale. If the scale is unbalanced you know the side which contains the heavier coin, and if it stays balanced you know the pile that you left alone contains the coin. Take whichever pile you know contains the heavy coin and repeat the same process to find the heavy coin in that pile of 3. zomg its just like recursion  |
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| Author: | Mackie [ Mon Feb 18, 2008 12:12 am ] |
| Post subject: | RE:another riddle |
Did you just say 'Zomg'? |
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| Author: | ericfourfour [ Mon Feb 18, 2008 1:02 am ] |
| Post subject: | RE:another riddle |
0 = regular coin 1 = heavier coin Randomize coins: 000010000 Divide into groups: Group 1 = 000 Group 2 = 010 Group 3 = 000 Scale groups 1 and 2: 1 < 2. Group 2 contains heavier coin. Divide into groups: Group 1 = 0 Group 2 = 1 Group 3 = 0 Scale groups 1 and 2: 1 < 2. Group 2 contains heavier coin. |
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