MATHEMATICS CONTEST 2006 SOLUTIONS

Solution : a⁰ = 1 for any a ≠ 0, and 0^b = 0 for any b ≠ 0. Therefore, 2⁰ + 0⁶ = 1 + 0 = 1.

Solution: Perimeter equals 2L + 2W. If L = 800, 2W = 406, so W = 203 inches.

3. Express in simplest form :

Solution: The common denominator for these fractions is 120, so

Solution:

Solution: The “brute force” approach would be to divide each of these numbers by 9, and see
which have a remainder. There is a faster approach of you know that a number is divisible by 9
if (and only if) the sum of its digits is divisible by 9. For example, 108 is divisible by 9 because
1 + 0 + 8 = 9 is (obviously) divisible by 9. By this test, only 2006 is not divisible by 9, so the
answer is 1.

Solution: Follow standard order of operations: 12 + 3 × 4 − 6 + 8 = 12 + 12 − 6 + 8 = 26.

Solution: In the equation a ÷ b = c + d/b , often written as a = bc + d, the dividend is a, the
divisor b, the quotient c, and the remainder d. Therefore, the dividend is a = (11)(3)+4 = 37.

8. Which one of the following expressions is not equal to the other four?
(A) 3x(x + 2)(x − 1) (B) −(1 − x)(3x² + 6x) (C) 3(x³ + x² − 2)
(D) (x + 2)(3x² − 3x) (E) −6x + 3x² + 3x³

Solution: C is different from the rest. This observation can be made in a number of ways; for
example, by noting that C does not have a common factor of x (that is, all the other expressions
equal 0 when x = 0).

9. Express as a fraction in simplest form:

Solution 1: If , then so 100x − x = 99x = 12.2. Therefore,

Solution 2: Note that

Solution: Group like terms:
ax − by − bx + ay = ax + ay − by − bx = a(x + y) − b(x + y) = (a − b)(x + y).
Equivalently ,
ax − by − bx + ay = ax − bx − by + ay = x(a − b) + y(−b + a) = (a − b)(x + y).

Solution: The answer is B: The total degree measure of an n-gon is 180(n − 2). (For example,
the angles in a triangle sum to 180°; in a quadrilateral, they sum to 360°, etc.) Therefore, in a
regular n-gon, an individual angle measure is This gives 60° when n = 6 (a regular
hexagon), 90° for n = 4 (a square), 120◦ for n = 3 (an equilateral triangle), and 135° for n = 8
(a regular octagon).

12. Write in simplest form as an exponential expression with a single base:

Solution: Use the rule of exponents that says

Solution 1: Carry out polynomial long division.

Solution 2: Any time you perform the division a ÷ b and get quotient q and remainder r , those
four quantities are related by the equation a = b · q + r . (See the solution to problem #7
on the previous page .) For polynomial division, the remainder r should be a polynomial of
degree smaller than the degree of b. So we want r (a first-degree polynomial) in the equation
x³ + 3x − 2 = (x² − 2)q + r . In order to have an x³ term (but no x² term) on the right side, q
must be x, so x³ + 3x − 2 = (x² − 2)(x) + r, or r = (x³ + 3x − 2) − (x² − 2x) = 5x − 2.

14. Express as a single radical in simplest form:

Solution:

15. (See figure). is tangent to at A. If BC = 3, CD = 3, OD = 2, and AB = 6, find
the length of the radius of .

Solution: Extend segment to intersect the circle at E. For any tangent segment (like ) and
secant segment (like), the lengths are related by the equation AB² = (BC)(BE). Therefore,
BE = 12, and DE = 6.

Method 1: Extend into a diameter, intersecting the circle at points F and G. A formula similar to the one used above says that for any two secants (like and ) which intersect inside the circle, we have (CD)(DE) = (FD)(DG). Therefore, (3)(6) = (r − 2)(r + 2), so r2 = 22, and r =

Method 2: Let H be the midpoint of , and note that Let x = OH. Then by the
Pythagorean theorem (applied to ∆OHD), x² + 1.5² = 2², so x² = 1.75. By the Pythagorean
theorem (applied to ∆OHC), x² + 4.5² = r², so r² = 22, and r =