Thread: 62(1+2) = ?
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Old April 29th, 2011, 12:33 AM   #67 (permalink)
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I answered '1.'

It was a Malcolm Gladwell "Blink" answer. I didn't ponder my decision at all. Doesn't make my answer correct, but it does prevent me from getting sucked into a self debate. Here's my logic:

We make the equation a bit more algebraic without changing the operators and parens:

c/2(a+b) where a=1, b=2, and c=6

If you assume c/2 comes after evaluating (a+b), you are essentially putting (a+b) in the numerator of your division:

c/2 * (a+b)/1 --> c(a+b)/2

So if the equation was written as c(a+b)/2 ((a+b) is in the numerator), your final answer is 9. But the equation was definitely not written in this way. So 9 is wrong.

As the original equation was written, (a+b) is clearly in the denominator because I proved above it can't be in the numerator without a drastic re-write of the equation. In other words, c/2(a+b) can only imply that (a+b) is in the denominator without re-writing the equation. Therefore, as the original equation was written, the answer is the following:

eval parens first: (a+b) = 3
3 is part of the denominator, so 6/2(3) = 6/6 = 1

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