Math Puzzle

March 11, 2004 at 10:26 pm | Posted in Fun Indoors | 2 Comments

I was going over some old class notes when I was taking an advanced controls class at UCLA when I stumbled on a flyer advertising the UCLA Math Students Society. On the back was a puzzle whose solution would qualify you for prizes like pocket protectors or slide rules đŸ™‚

Here’s the puzzle:

Q. If the remainder of polynomial P(x) is 2 when divided by x-1 and 1 when divided by x-2, what is the remainder of P(x) when divided by (x-1)(x-2)?

My answer is in the extended. There’s got to be an easier way to show this.

A. The solution involves expressing the first multiple divided by (x-1) in terms of (x-2)

   P(x) = f(x)(x-1) + 2
        = g(x)(x-2) + 1

Therefore,

   f(x)(x-1) + 1 = g(x)(x-2)     (*)

When x=2, f(2)(2-1) + 1 = 0. Therefore, the remainder of f(x) when divided by (x-2) is -1. In other words, f(2) = -1:

   f(x) = m(x)(x-2) - 1

Substitute back into equation (*):

   (m(x)(x-2) - 1)(x-1) + 1 = g(x)(x-2)

   m(x)(x-2)(x-1) - (x-1) + 1 = g(x)(x-2)

   m(x)(x-2)(x-1) - x + 2 = g(x)(x-2)

Therefore,

   P(x) = g(x)(x-2) + 1

        = m(x)(x-2)(x-1) - x + 2 + 1

        = m(x)(x-2)(x-1) + (3 - x)

The remainder of P(x) divided by (x-2)(x-1) is


   (3-x)

2 Comments

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  1. Excellent deductive solution compared to my inductive reasoning.

  2. Here is a simpler solution. The polynomial can be expressed as:

    P(x) = (x-1)(x-2)q(x) + r(x) (*)

    Since the order of r(x) is less than 2, r(x) can be expressed as: r(x) = ax + b, for some constants a and b.

    Now using (*) above,

    for x = 1, P(1) = a(1) + b = a + b = 2

    for x = 2, P(2) = a(2) + b = 2a + b = 1

    Solving for a and b gives r(x) = -x + 3


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