phj

11 ปีที่แล้วdiscuss

I work in one of these labs. I suppose if you really need to do work, a pair of headphones suffice. But the open space makes it really easy to have visitors (there are a lot of them), as well as the aforementioned interacting w/ neighbors. It's certainly more interesting than working alone in an office or at home.

The meeting offices are mentioned briefly, but those see quite a bit of use as well.

phj

12 ปีที่แล้วdiscuss

The version I learned was that Franklin did all the experimental work, while Crick did the Fourier transforms in his head to interpret Franklin's results. Not sure about Watson's contribution, though.

phj

12 ปีที่แล้วdiscuss

I think this also makes a good tie-in for teaching algebra.

Once you figure out that the most significant digits should be the largest ones, and the 3rd and 4th largest digits should be in the 2nd most significant place, you end up with a situation like this:

      A C E
    *   B D

To make it more clear what to do next, you can just append a zero onto the number "BD" and still solve the same problem, because instead of multiplying "ACE" * "BD", you are multiplying "ACE" * "BD0" = ("ACE" * "BD") * 10:

      A C E
    * B D 0

Now, to figure out which of the two greatest digits are A and B, and which of the next two greatest are C and D, you can apply the identity (x + y) * (x - y) = x^2 - y^2 to this. Since "ABC" * "BD0" = (x + y) * (x - y), then equating "ABC" = x + y and "BD0" = x - y, you can solve to get x = ("ACE" + "BD0") / 2, which is the same number no matter the order between A and B, or between C and D. Then maximizing (x + y) * (x - y) means minimizing y^2; or, making the two numbers "ACE" and "BD0" as close together as possible, which leads to the given solution.

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