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MATHEMATICS CONTEST 2003 SOLUTIONS
PART I: 30 Minutes; NO CALCULATORS
Section A. Each correct answer is worth 1 point.
1. Find the sum of the first six prime numbers.
Solution : 2 + 3 + 5 + 7 + 11 + 13 = 41 (the number 1 is not a prime).
2. Give the official ( and more common ) name for a regular
quadrilateral.
Solution: “Regular” means equal sides and equal interior angles, so it’s
a square.
3. Express the repeating decimal
as a ratio of two positive integers in simplest form.
Solution 1: Let x =
;
then 10x =
,
and 10x − x = 1.777777 . . . − 0.177777 . . . = 1.6.
If 9x = 1.6, then x = 1.6/9 = 16/90 = 8/45 .
Solution 2: Recall that 7/9 = , so we want 7/9 − 3/5. (Or 1/10 + 7/90, or 1/6 + 1/90 .)
4. If 2x + 1 = 2003, find the value of 3x − 1000.
Solution: Solve for x: If 2x + 1 = 2003, then 2x = 2002, so x = 1001.
Then 3x − 1000 =
3(1001) − 1000 = 2003.
5. A nonisosceles triangle has integral sides of 4, 5,
and x. Find all possible values of x.
Solution: The triangle inequality theorem says that the sum of any two
sides must be greater than
the third side, so 5−4 < x < 4+5, or 1 < x < 9. But the sides are integral (they
are integers),
so x must belong to the set {2, 3, 4, 5, 6, 7, 8}. Additionally, the triangle is
nonisosceles (no
two sides can be equal). Therefore, x is in {2, 3, 6, 7, 8}.
6. Using some or all of the digits 0–9 (no digit more than
once ), construct the largest possible
sixdigit odd number with a 9 in the tens place.
Solution:
9 must be in the tens place.
Biggest possible number.
Biggest possible number.
Biggest possible number, number must be odd.
7. 7 + (−7) = 0. This is an example of what basic property
of addition ?
Solution: Additive inverse property or Opposites property.
Section B. Each correct answer is worth 2 points.
8. If (x^{5} + x^{4} + x − 5) is divided by (x + 1), find the
remainder.
Solution 1 (Long division):
Solution 2 (Synthetic division): Works when
dividing by x −c. Rewrite x +1 as x −(−1), then
write out the coefficients , multiply and add ; the remainder is the last number
on the bottom.
Solution 3 ( Math knowledge ): The remainder theorem
says you can just plug in −1 to get the
remainder: (−1)^{5} + (−1)^{4} + (−1) − 5 = −6
9. Find the 2003rd digit after the decimal point in the
decimal representation of 4/7.
Solution: Rational numbers (such as fractions) either stop or repeat. If
you do long division far
enough, the pattern is clear:
4 ÷ 7 = 0.57142857142857... or
So it repeats every six numbers. Now divide 6 into 2003, and observe that the
remainder is 5,
so we choose the 5th repeated digit: The answer is 2.
10. In the figure on the right, the length of tangent
is 12,
PD = 8, and chord
bisects chord
.
If EB = 3, find
the length of
Solution: (AP)(AP) = (PD)(PC), so 12^{2} = (8)(PC),
or PC = 18. Then CD = 10, and because
bisects
,
CE = ED = 5. Now (AE)(EB) = (CE)(ED), so
(AE)(3) = (5)(5), and we conclude that AE = 25/3 .
11. Express in simplest radical form (no radicals in the
denominator ):
Solution: This is sometimes called “rationalizing the denominator”:
12. Write the exact numerical value of x if log_{8} 128 = x.
Express in simplest form.
Solution 1: Transform to exponential form: 8^{x} = 128. Now rewrite both 8 and 128
as powers of
2: 8 = 2^{3} and 128 = 2^{7}, so 8^{x} = (2^{3})^{x} = 2^{3x} = 2^{7}. Therefore, 3x = 7, or x = 7/3.
Solution 2: Use the changeofbase formula and properties of logarithms:
Section C. Each correct answer is worth 3 points.
13. The ellipse x^{2} + 2y^{2} + 12y − 10x − 57 = 0 has a major axis with two
endpoints. Find the
coordinates of the endpoint that lies in quadrant IV. Express in ordered pair
form, (x, y).
Solution: Put in standard form by completing the squares:
The ellipse is centered at (5,−3), and the major axis is therefore horizontal
with length 2a = 20;
the endpoint in quadrant IV is (5 + 10,−3) or (15,−3).
14. Softball player Berni Williams has 120 hits in 300 atbats for a current
batting average of .400.
In today’s game, she will have 5 atbats. What is the probability that she will
get exactly 2 hits?
Solution: The number of hits among her next five atbats has a binomial
distribution (more on this
below), so P(exactly 2 hits) =
(0.4)^{2}(0.6)^{3}, where
is “5 choose 2” (sometimes written
as
),
which is equal to 10. This evaluates to
If you have never heard of the binomial distribution , here are two ways to think
about it:
1. Write out every possible sequence of two hits and three outs:
HHooo, HoHoo, HooHo, HoooH, oHHoo, oHoHo, oHooH, ooHHo, ooHoH, oooHH
There are 10 such sequences, and each has probability (0.4)^{2}(0.6)^{3}.
2. The binomial distribution gets its name from the terms in the expansion of
the binomial
(q + p)^{n}, where p is the probability of success (a hit), q = 1 − p is the
probability of failure
(an out), and n is the number of attempts. Thus:
(0.6 + 0.4)^{5} = 0.6^{5} + 5(0.6^{4}0.4^{1}) + 10(0.6^{3}0.4^{2}) + 10(0.6^{2}0.4^{3}) + 5(0.6^{1}0.4^{4}) + 0.4^{5}.
Each term in this sum corresponds to the probability of a certain number of
failures and successes;
we want the third term.
15. Solve the inequality
and graph its solution on the given number line.
Solution: Either
, so that x > 1 or x < −5. This can be drawn either
of these ways:
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