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Tartaglia's Riddle, Redux

Date: 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

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 
Associated Topics:
Middle School Word Problems

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