SMT 2013
General Tiebreaker Solutions
February 2, 2013
1. Compute all real nonzero values of b such that bx2 bx b = 0 has b as a solution.
1 5
Answer:
2
Solution: Plugging in x = b, we have b3 b2 b = b(b2 b 1) = 0. From the quadratic
1 5
2
as solutions.
f

SMT 2013
General Test Solutions
February 2, 2013
1. Robin goes birdwatching one day. He sees three types of birds: penguins, pigeons, and robins.
2
1
3 of the birds he sees are robins. 8 of the birds he sees are penguins. He sees exactly 5 pigeons.
How ma

SMT 2013
Advanced Topics Tiebreaker Solutions
February 2, 2013
1. Andrew flips a fair coin 5 times, and counts the number of heads that appear. Beth flips a fair
coin 6 times and also counts the number of heads that appear. Compute the probability Andrew

SMT 2013
Advanced Topics Test Solutions
February 2, 2013
1. How many positive three-digit integers a b c can represent a valid date in 2013, where either a
corresponds to a month and b c corresponds to the day in that month, or a b corresponds to a
month