Pizza and Problems/Pizza and Problems 3 (10/17/08)

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Use this page to post solutions to our third Pizza and Problems session, held 10/17/08. Examine the original problem set.

1 Student Solutions to the Problems in Pizza and Problems Set #3

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Student Solutions to the Problems in Pizza and Problems Set #3

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Problem 1: Solution by:

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Problem 2: Solution by: Bruce Wagner

I did not see a more clever way to do this problem, but a direct calculation works:

$229 = (2 10)(2 10)(2 9) = (1024)(1024)(512) = 536870912$ , so the answer is 4.

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Problem 2: Notes by: David Arnold

On the way home tonight, Dan Gent shared some remarkable insight. For example, the sum of the digits 0 through 9 is 45. The number $229$ is a nine digit number, all unique. Suppose we leave out a 3. Then the sum of the digits is 42, a sum that is divisible by 3. Hence, $229$ is divisible by 3, a contradiction. A similar argument works for leaving out a 0, 3, 6, or 9. Hence, each of these must be included in the nine-digit number.

The argument is not yet complete, but this is a good start.

Very nice idea, Dan.

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Problem 3: Solution by: David Arnold

Let

$x=\cfrac{1}{2+\cfrac{1}{2+\cfrac{1}{2+\cdots}}}.$

Note that the continued fraction contains a copy of itself. That is,

$x=\cfrac{1}{2+x}.$

Cross multiply.

$x=\frac{1}{2+x}$

Make one side zero.

$x 2 + 2 x - 1 = 0$

The quadratic formula provides two answers.

$x=-1\pm\sqrt{2}$

But our original continued fraction is definitely a positive number. Hence,

$\cfrac{1}{2+\cfrac{1}{2+\cfrac{1}{2+\cdots}}}=-1+\sqrt{2}.$

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Problem 4: Solution by: Bruce Wagner

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Problem 4: Demo by: David Arnold

Here is an image, built in Geometer's Sketchpad:

If you have the Geometer's Sketchpad installed on your system, you can try the demo out. Right click on the link ThreeCircles.gsp, then do a File Save As, but don't let your system append .txt to the end of the filename. Save the file as ThreeCircles.gsp, then open in Geometer's Sketchpad. Drag the point Drag Me #2 with your mouse until the circle centered at O aligns with the circle centered at D.

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Problem 8: Solution by: Bruce Wagner

The probability is 1. Consider the first two points, A and B. Let P be the plane containing 0 (the center of the sphere), A, and B. The plane P divides the sphere into two equal hemispheres whose border is a great circle which contains A and B. The point C must then lie in one of the two hemispheres.

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