Sharing Crayons
Three brothers shared 24 crayons. Each had a quantity equal to his age 3 years ago. A trade was suggested by the youngest brother ... that he keep half his crayons and divide the other half between his two brothers. Then he proposed, the middle aged brother, after receiving the crayons, do the same thing. To complete his proposal, the older brother, after receiving crayons from the middle aged brother, would perform the same sharing procedure.
After some discussion, the brothers agreed to the swap. Upon completion, each brother had 8 crayons.
What are the ages of the brothers?
After some discussion, the brothers agreed to the swap. Upon completion, each brother had 8 crayons.
What are the ages of the brothers?
posted by Zaux at
6:27 AM
6 Comments:
7, 10 and 16 years old. Easy one... :)
Hi Miguel ... you are right.
What a relief there weren't 4 or 5 of them ;)
This post has been removed by the author.
Hi Cam, previous post was a bit too brief. I'd pop the equations into wolframalpha, and present the solution.i.e. Using,
Solve[{41/64 x +5/16 y+1/4 z==8,13/64 x+9/16 y+1/4 z==8,5/32 x +1/8 y+1/2 z ==8},{x,y,z}]
http://www.wolframalpha.com/input/?i=Solve%5B%7B41%2F64+x+%2B5%2F16+y%2B1%2F4+z%3D%3D8%2C13%2F64+x%2B9%2F16+y%2B1%2F4+z%3D%3D8%2C5%2F32+x+%2B1%2F8+y%2B1%2F2+z+%3D%3D8%7D%2C%7Bx%2Cy%2Cz%7D%5D
Result => x,y,z = 4,7,13
Try it. I just hope that wolframaplpha is with us forever.
PS If "Solve" doesn't behave well, you could try "Reduce".
You might like to have a play on that site. Although the results don't always present in a consistent way, you often get extra stuff that's quite nice e.g. plots of curves.
Go there and try e.g.
Plot[x^2], Plot[x^2,{x,0,10}],
Plot[y=x^2] and Plot[y=x^2,{x,0,10}]
or Sum[1/x^2,{x,1,infinity}]
You'll love the last one.
Also try RiemannZeta[2] and ZetaZero[3] ;)
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