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9/10/2015 12:16:05 AM
3
Algebra riddle (help)
I got this off Raymond Smullyan's "The Riddle of Scherazade". It is question #39 - A Hindu Puzzle
Beautiful maiden, with beaming eyes, tell me what is the number that, multiplied by 3, then increased by three fourths of the product, then divided by 7, then diminished by one third of the quotient, then multiplied by itself, then diminished by 52, then the square root taken, then increased by 8, then divided by 10, gives the number 2?
I get about halfway and then I get stuck. By doing the problem in reverse, I get √((((2*10) - 8)²) + 52) = 14
I'm not sure where to go from there. Am I doing 14 + (1/3 * 14) for the next step?
The end result to the whole problem should equal 28. It's bugging me that I can't get that number.
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1 ReplyEdited by Neno is drunk: 9/10/2015 11:36:23 PMWow, what a pain in the ass. Try plugging the answer into the x of this function ((sqrt ((((((3x)+((3/4)(3x))/7)-(((3x)+((3/4)(3x)+((3/4))(3x))/7)(1/3)(((3x+((3/4)(3x))/7)-(((3x)+((3/4)(3x))/7)(1/3)))-52)+8)÷10=2 I would have done the algebra part and plugging it in part, but I can't be assed to write everything down for 10 min. Besides, finding the function seems like the most important part anyway. [spoiler]It should be right but idk. There's a chance I might have missed a parenthesis somewhere.[/spoiler]
