[Guest]

If a diamond miner had 24 jewels, and three daughters, help him to dvide the jewels in a way that all three daughters get the same worth and jewel type. THe catch is that, the worth of all the jewels together is 720$. Also, the jewels are of 3 prices. Now help him......

[EDM]

Isuru.....wrong section again.....

Edited April 11, 2010 by EDM

[Guest]

A 80 $ ; 3 jewels

B 40 $ ; 6 jewels

C 16 $ ; 15 jewels

[rookie1ja]

Off to New Logic/Math Puzzles. Again

[Guest]

There are way too many solutions to this.

Each daughter gets 8 Jewls worth 240$, let's say 1 from type A which is worth 1$, one from type b which is worth 5$, amd 6 of type C which by calculation should worth 39$...

There are probably more than 20 more possible combos...