1979 AMC 12 Problem 27

Attempt Problem 27 of the 1979 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1979 AMC 12 solutions, or check the answer key.

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27.

An ordered pair (b,c)(b,c) of integers, each of which has absolute value less than or equal to five, is chosen at random, with each such ordered pair having an equal likelihood of being chosen. What is the probability that the equation x2+bx+c=0x^2+bx+c=0 will not have distinct positive real roots?

106121\frac{106}{121}

108121\frac{108}{121}

110121\frac{110}{121}

112121\frac{112}{121}

none of these

Answer: E
Concepts:quadraticcomplementary probabilitysystematic listing
Difficulty rating: 2200
Small Hint:

Distinct positive roots require b<0,b\lt0, c>0,c\gt0, and b2>4cb^2\gt4c

Big Hint:

Count the qualifying cc values for each b=1,b=-1, b=2,b=-2, ,\ldots, b=5,b=-5, then take the complement among 121121 pairs

Solution:

There are 112=12111^2=121 ordered pairs. Distinct positive roots require b<0,b\lt0, c>0,c\gt0, and b2>4c.b^2\gt4c. For b=1b=-1 and b=2b=-2 there are no choices; for b=3,b=-3, b=4,b=-4, and b=5b=-5 there are respectively 2,2, 3,3, and 55 choices of c.c. Thus only 1010 pairs produce distinct positive roots, so the requested probability is 110121=111121, 1-\frac{10}{121}=\frac{111}{121}, which is not listed.

Therefore, the correct answer is E.

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