Round 1 is Nearly Over (You Have Till 5am)
… And rather than posting the solutions like last year, it’d probably be more productive to use the time to discuss how in the world I / you / us / we found the answers. I’ll link to the official answers I suppose for those who are just following along for the ride, but otherwise it would be best to focus on what we tried that worked and / or didn’t work, and how we ultimately got the answer we did.
Hopefully expect some small notes on how I attacked the problems by tomorrow afternoon (spoiler: a couple of small brute force scripts on small numbers for the two annoying problems [you know which two]).
Have a nice night, happy submitting.
2010 Math Challenge Has Begun
The text version of the questions has once again been reposted here for anyone interested.
Answers are due by March 1, 2010 (as always double check the official site for changes!) and we hope to once again post solutions after that date. Please remember to refrain from discussing the problems until then.
Good luck to all.
Round 1: Problem 5
Begin with a set of distinct positive integers. A new positive integer may be constructed and added to the set so long as it has the form (a+b)/(a-b) where a and b are already in the set. (For example, if 9 and 6 are already in the set, then the number 5 may be added.) The original set of integers is called “prolific” if every positive integer can eventually be constructed and added to the set. What is the smallest size that a prolific set can have? Prove your answer.
Round 1: Problem 4
A basketball coach has 10 boys and 10 girls on his roster. In how many ways can the coach partition the group into 4 teams of 5 students such that there are at least 2 boys and 2 girls on each team?
Round 1: Problem 3
Alice and Bob are bored and want to play a game. “I have an idea,” Alice chimes in. “How about you flip this coin 2009 times and I’ll flip it 2010 times and whoever gets more Heads wins?”
Bob replies, “No, that’s not fair! You’re probably going to win since you get more flips!”
“Fine!” answers Alice. “How about this? You flip the coin 2009 times and I’ll flip it 2010 times and if I get more Heads, I win. If you get more Heads, you win. And, if there’s a tie, we’ll say that you win too.” Bob shrugs his shoulders and agrees to play.
What’s the probability that Alice wins the game? Prove your answer.
Round 1: Problem 2
The figure to the right is made of 16 congruent arcs. Each arc is a quarter of a circle of radius 1. What is the area of this figure?
Round 1: Problem 1
The numbers 1,2,3,…,2010 are listed in this order. What is the least non-negative result that you can form by placing a “+” or “-” sign in front of each number and summing the values? Prove your answer.
Round 5: Problem 6
Let’s play a game.
A fair die is rolled repeatedly until two consecutive rolls are equal, at which point the game ends.
If we let S be the sum of the results of all of the rolls, is S more likely to be even or odd, or is it equally likely to be both?
Round 5: Problem 5
Each of nine lines cut a unit square into two quadrilaterals whose areas are in a ratio of 2:1. Prove that at least three lines must be concurrent (i.e. they intersect at a common point).
Round 5: Problem 4
Let be an infinite arithmetic progression of positive integers with the special property that the sum of the first n terms of is a perfect square for all n. If 2009 is the k-th term of the sequence, how small can k be?