Tartaglia's Riddle, ReduxDate: 09/22/2012 at 10:23:59 From: Siobhan Subject: Logic Question Logically, if half of 4 were 3, then what would one-third of 50 equate to? One-third of 50 is not a whole number. Do you have to find a common denominator? Do you start by saying that half of 8 would be 6? Date: 09/24/2012 at 17:59:13 From: Doctor Peterson Subject: Re: Logic Question Hi, Siobhan. This is not really a logic question; rather, it is asking you to be creatively illogical. The creators of this brain teaser are using words such as "half" and "equate" in ways that do not adhere to their conventional definitions. Therefore, you have to invent meanings that would work -- and stay open to the possibility that this stumper will not have one, single correct answer. This problem reminds me of one I have seen in the book _Mathematical Puzzles for Beginners and Enthusiasts_, by Geoffrey Mott-Smith. He calls it "Tartaglia's Riddle" (though I know of no historical basis for the name). It goes like this: In ancient times, the neophyte in logic was posed such questions as: If half of 5 were 3, what would a third of 10 be? The answer given for this is 4, on the assumption that answers are being scaled by some factor. Since 2.5 is scaled to 3 (scale factor 3/2.5 = 1.2), then 10/3 would be scaled to 4 (= 1.2 * 10/3). Mott-Smith's answer, in his own words, is Four. Set up this proportion: 5/2 : 3 = 10/3 : x Solve for x. The argument is: Whatever mysterious factor causes 1/2 * 5 to give the result 3 must also be introduced into the product 1/3 * 10. This factor is expressed by the ratio 5/2 : 3. If you solve your riddle the same way, you'll get a whole number (try it), so I suspect that is what was intended. But since nothing in the problem says that such a proportional scaling is in view, this is only the probable intended answer, not the "correct" answer. Logically speaking, if you change the rules of arithmetic, anything is possible! - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/