Wednesday, January 17, 2007

2007: 17 and 25.

Day Seventeen

Difficulty: easy-schmeasy, if you subtract. More complicated if you try to add.

Problem 25 = 5².
(I wonder how often you can switch the digits and still write an equivalent expression.)

Difficulty: Medium. More damned logarithms, but there's a neat little prime factorization going on, too.

Standard instructions for this series: No calculator allowed. Express answers in reduced form. Rationalize denominators. Radicals must be reduced. All numbers are base ten unless otherwise specified. Do not approximate radicals or π. Leave such answers as 1025π or √39, for example. Source: UVM Math Contest

answer 17 and answer 25.


  1. 25 presents us with a gimmick - if they are asking for a value, they we don't necessarily need a general solution, just one specific set of a, b, c that makes the eqn true. We don't even have to worry about how many solutions there are - just find one, and done.

    There is a clean dissection (I was surprised) for 17. Choose the rectangle with coordinates (1,7) (1,3) (4,3) and (4,7), extend from each vertex vertically or horizontally to a vertex of the polygon, and drop a perpendicular from (5,10). No need to calculate any points, and we have 4 right triangles and two rectangles.

    Might be easier than the subtraction? (1 rectangle, 4 rt triangles, but subtraction is more prone to sloppy mistakes?) Still, I would subtract, because the cleanness of the cuts is not expected.


  2. I don't know about calling 25 a gimmick. I like it for its solid use of prime bases that requires the kids to know their exponent rules.

  3. By gimmick I meant that once I rewrote it (yes, good log practice), I just guessed at an answer, and didn't worry about the existence of other possibilities.