What can I say? I didn't know the "hidden trick".

Bob Hansen

On Sep 29, 2012, at 1:23 AM, Wayne Bishop <wbishop@calstatela.edu> wrote:

My guess is that the approach Dave had in mind was more obvious:
a^2 - (a-1)^2 = (a - (a-1)) (a + (a-1)) = (1)(2a - 1)


At 08:36 PM 9/28/2012, Robert Hansen wrote:
a^2 - (a-1)^2 = a^2 - (a^2 - 2a + 1) = 2a - 1 = 13.04822...

This is a very good point, work the algebra FIRST. I wonder how many algebra teachers are even capable of concocting such a problem? Working backwards and making sure that 2*a doesn't require any more than single digit math (no digit greater than 4). I have to think about this. This works well with large integers as well, with the same condition on the digits.

Lou, what do you say to problems like this with regards to our prior art discussion?

Ha, calculators are allowed. Calculators with 70 digits of precision I suppose.:)

Bob Hansen

On Sep 28, 2012, at 5:51 PM, "Dave L. Renfro" <renfr1dl@cmich.edu> wrote:

> Robert Hansen wrote:
> http://mathforum.org/kb/message.jspa?messageID=7897638
>> I want to try something different. I want everyone to contribute
>> problems for a hypothetical algebra 2 exam. You can contribute
>> just topics if you wish though I would like see examples as well.
>> I am going with algebra 2 rather than algebra 1 because I think
>> the line is more well defined.
> I only have a few moments before I need to leave to tutor
> someone, but here's a somewhat silly one off the top of
> my head:
> Determine the exact value of a^2 - b^2 if
> a = 7.0241132301442003123012230341430201
> b = 6.0241132301442003123012230341430201
> Calculators are allowed.
> Dave L. Renfro