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By Loop123 (Loop123) on Friday, July 16, 2004 - 03:00 pm: Edit |

In a distant land, there were 10 princes who paid homage to a single king. Every year, each prince would give the king 10 bags of gold, each bag containing one pound of gold. All of the bags were identical. This particular year, one prince decides to cheat the king and give him only 9 pounds of gold. Each of his bags would contain only 9/10 of a pound of gold. The king's attendant informed the king of this, and the king responded with the following:

"Do not worry, we shall catch the scoundrel in a

single weighing."

All of the princes arrive in the kingdom with their ten bags of gold in hand. How, using a standard weight scale (not a balance scale), can the king discover which of the ten princes was trying to cheat him? Keep in mind that the weighing must be done simultaneuously (the bags cannot be placed on the scale in increments, and there can only be one weighing). Also, there is no limit to the amount of bags that the scale can hold.

By Thermodude (Thermodude) on Friday, July 16, 2004 - 03:12 pm: Edit |

Nice riddle...I have a solution...the king can just tell each prince to place their bags on the scale one by one...when the "bad" prince places his first bag on the scale...the scale will only increase by 9/10'ths instead of 1 pound...so then the king will know who he is. Notice that since no bags will be removed...this is all done in one weighing.

By Loop123 (Loop123) on Friday, July 16, 2004 - 03:14 pm: Edit |

Note that I wrote in parentheses "the bags cannot be placed on the scale in increments."

Good try, though.

By Thermodude (Thermodude) on Friday, July 16, 2004 - 03:14 pm: Edit |

ahhghgh...ignore my previous post...i misread the riddle..didn't see how the weighing must be done simultaneously

By Loop123 (Loop123) on Friday, July 16, 2004 - 03:34 pm: Edit |

The clock is ticking...MUAHAHAHAHA

By Phantom (Phantom) on Friday, July 16, 2004 - 03:36 pm: Edit |

oooh, excellent riddle, but i've heard one similar to it before and won't give it away.

By Lildv (Lildv) on Friday, July 16, 2004 - 03:37 pm: Edit |

Now I'm really curious to know what the answer is because I am totally perplexed. I want someone to figure it out so I can know.

By Loop123 (Loop123) on Friday, July 16, 2004 - 03:39 pm: Edit |

Say I if you DON'T want the answer revealed.

By Calkidd (Calkidd) on Friday, July 16, 2004 - 03:39 pm: Edit |

Have each of the ten princes place a different number of bags on the scale (prince # 1 = 1 bag, prince # 2 = 2 bags, etc). Then take the total weight. Every prince that doesn't cheat the king will be putting down a multiple of 10 pounds, which won't contribute to the ones digit of the total weight. On the other hand, the prince that did cheat the king will be placing down some multiple of nine. Since the first ten multiples of 9 (i.e. 1x9, 2x9, 3x9, 4x9, etc) all produce different ones digits (9,8,7,6...), the king will know which prince cheated him based on the ones digit of the scale.

By Loop123 (Loop123) on Friday, July 16, 2004 - 03:40 pm: Edit |

Excellent job!

By Seleucus26 (Seleucus26) on Friday, July 16, 2004 - 03:40 pm: Edit |

u have one prince put one bag, the next out 2, the next put 3, etc. and the last prince has put all ten bags on. (the placing was simultaneous)

the weight should be 55. 55 minus the amount weighed gives u x. 10x is the culprit's number.

if the culprit were prince #1 x would be 1/10 because he contributed 1 bag with 9/10 instead of 1.

By Seleucus26 (Seleucus26) on Friday, July 16, 2004 - 03:41 pm: Edit |

dammit

calkid answered while i was posting

By Mehere (Mehere) on Saturday, July 17, 2004 - 01:37 am: Edit |

Here is a theoretical and impractical solution using physics.

Number the princes from 1 to 10 and put their numbers on their bags. Then, place one bag of each prince at different heights on the scale. Place the bags so far apart that g (Earth's gravitational constant) is different for each bag, but is calculatable for each bag. Then calculate the weight as if all bags are the same weight. Then find the actual weight. By calculating the difference, it is possibel to find which bag is the 9/10 pound one.

Of course that is theoretical and impractical......

By Jimster0489 (Jimster0489) on Saturday, July 17, 2004 - 12:09 pm: Edit |

Hey! We had this problem in algebra.. isn't this from IMP (Interactive Math Program)?

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