Dividing the gems

[unreality]

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

See if you can figure it out yourself first!

14 aquadium, 26 haxite

14a, 26h

14*3=42, 42+26=68. Add the Chief-Miner's haxite necklace and that's 69. That's divisible by 3 so that will work. Split up the 27 haxite (26+1, the 1 from the necklace) among the ore carts, 9-9-9. Then split the aquadium the best you can, 4-4-4. Two aquadium are left over that must be used. Add them both to the first ore cart, which is now 2a, or 6h, ahead, so take 4 haxite from it and give 2 to each of the other ore carts. So that's it like this:

a: 6-4-4

h: 5-11-11

multiplying the a by 3, just to check:

18-12-12

5-11-11

all add to 23 which is 69/3. It works.

Unfortunately the poor Chief-Miner has to let go of his trusty haxite necklace...

[bonanova]

Use abbreviations to simplify.

H = Haxite; A = Aquadium.

A = 3H.

They mined 14A + 26H = 14[3H] + 26H = 68H.

68 is not a nice number.

So add theChief-Miner's Haxite necklace = 69H.

Each prince should get 23H.

23H = 4A + 11H = 5A + 8H.

Prince 1 gets 4A + 11H

Prince 2 gets 5A + 8H

Prince 3 gets 5A + 8H.

Chief Miner gets shafted, but lives to mine another day.

[Guest]

(1) ...aquadium...is worth 3x more than haxite...

(2) ...The Chief-Miner oversaw the operation, rubbing his haxite necklace for good luck.

(3) ...the miners got 14 aquadium crystals and 26 haxite chunks.

(Directive) "Split both gems' numbers in three and fill up the ore carts!"

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

The three relevant facts, the directive and the problem appear in the quote above. Note that the problem, which is to divide value, is actually different from the directive, which is to divide numbers. Nevertheless, let's assume that's what the sub-chief meant.

My first thought was to convert everything to haxite value and find out how much each prince gets. Thus, 14 x 3 + 26 = 42 + 26 = 68. Since 68 is not evenly divisible by 3, the problem is unsolvable without more information.

Based on the apparently passing reference found in #2, I assume the "real" answer has to do with "borrowing" the Chief Miner's necklace to increase the total value to 69 haxite and divide up the bootie, with the supposition that the Chief Miner gets his necklace back from the miners' work tomorrow. But this resolution suffers from some problems:

1. No gem that I have ever heard of has a single monolithic value for each stone or crystal. This assumption would require that haxite and aquadium gemstones are all perfectly uniform in their natural state, a condition that is neither stated in the problem nor reasonable to assume. It is more reasonable to assume that statement #1 means that aquadium is three times as valuable as haxite per unit weight, or some equivalent treatment.

2. A "haxite necklace" may well consist of more than just a single stone pendant, just as a "pearl necklace" might be a string of pearls and not just a single pearl. Based on the wording of the problem, we cannot safely assume that the "haxite necklace" is a single stone.

3. We do not know that the Chief Miner was even present when the division was made, or that he would be willing to give up a valuable (we assume) necklace just so that his underling could divide things up.

4. Somehow, introducing outside value into the pot to be divided seems to violate the spirit of what's trying to be done. Sure, if you can't divide things out evenly, you can always throw some of your own money in to make things work out. You might even do this if you're a parent and your kids are fighting. But this doesn't seem like a reasonable course of action for a manual laborer.

A better solution, I think, is to choose the largest aquadium gemstone and the largest haxite gemstone and split each of them in two. A fraction of a gemstone is still a gemstone, and doing this would give you a number of each gemstone that is evenly divisible by three, thus allowing the Chief Miner's original instructions to be followed to the letter.

[unreality]

bonanova: good job

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

[bhramarraj]

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