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Dividing the gems

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#1 unreality


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Posted 08 January 2008 - 11:29 PM

Priceless Gems

There are two different gemstones in the realm. Haxite, which is a deep red gem, and aquadium, a jewel of many beautiful blues and some green, which is worth 3x more than haxite.

This week the miners have been mining special ore carts for the three princes of the realm, and the princes are equal, thus they must all receive exactly the same worth in their ore cart. The Chief-Miner oversaw the operation, rubbing his haxite necklace for good luck. At the end of the last day, the miners got 14 aquadium crystals and 26 haxite chunks.

"How should we distribute them?" the Sub-Chief-Miner asked.

"That's easy!" laughed the Chief-Miner. "Split both gems' numbers in three and fill up the ore carts! And quickly, the princes come at dawn tomorrow!"

"But neither gem amount is divisible by three..." the Sub-Chief-Miner complained, but the Chief-Miner would hear none of it. He also ordered that ALL the gems that were mined MUST be used.

The Sub-Chief-Miner sighed and returned to the men- but then he got an idea. The next dawn the princes came to see each ore cart had the same wealth.

How did they split up equal worth into all three ore carts?

Try before you look at my spoiler, it's not that hard

Spoiler for Answer

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#2 bonanova



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Posted 08 January 2008 - 11:51 PM

Spoiler for How's this?

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The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
- Bertrand Russell

#3 spoxjox


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Posted 09 January 2008 - 01:02 AM

Spoiler for my analysis

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#4 unreality


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Posted 10 January 2008 - 12:06 AM

bonanova: good job :D

spoxjox: assume each gemstone is the same value and size, and the haxite necklace is one haxite stone
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#5 bhramarraj


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Posted 13 September 2012 - 01:41 PM

I know a story of a farmer, who before dying made a will in which 8 horses were to be divided among his three sons, such that each son gets 1/2 nos.
But the no of horses to be distributed were only 7.
Then a wise person solved their problem by adding his own horse to the 7 horses to be distributed.
Then he gave 1/2 nos. i.e. 4 horses to the eldest, then half of remaining i.e. 2 horses to the younger, and half of remaining i.e. 1 horse to the youngest son.
One horse remained after distribution which he took back. So without losing his own horse the problem was solved
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