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For the sixth consecutive year, a Lafayette team has earned first place in the Lehigh Valley Association of Independent Colleges Math Contest. For the second straight year, the College fielded squads that took three of the top four places in the competition, held Oct. 22 at Lafayette.
Finishing first in the 16th annual event was the team of Ekaterina Jager ’06 (Tashkent, Uzbekistan), a candidate for B.S. degrees in both mathematics and electrical & computer engineering; Jinjin Qian ’08 (Shanghai, China), an economics and business major; and Shiliang Cui ’09 (Shanghai, China).
In third place were Keming Liang ’08 (Zichuan District Zibo, China) and math majors Jordan Tirrell ’08(West Grove, Pa.) and Xue Ji’08 (Wuxi Jiangsu, China).
Taking fourth place were Marquis Scholars and math majors Brian Kronenthal ’07(Yardley, Pa.) and Maureen Jackson ’06 (Steubenville, Ohio), and Marquis Scholar and electrical and computer engineering major John Kolba ’06 (Chelmsford, Mass.).
The contest is open to students from Lafayette, Lehigh, Muhlenberg, Morvian, Cedar Crest, and DeSales. A free pizza lunch was provided during a “solution fest” afterward. Prizes will be given to the top three teams at a regional mathematics conference in February.
A sample problem from the test: Here is a game involving two players and a pile of integers from 1 to 10 in the middle of a table. Player A removes one of the integers from the pile, then player B removes one, and so on. The players alternate turns, each one removing an integer from the pile until the product of the integers that remain in the pile is no longer divisible by 6. When this happens, the game ends and the last player to move is the winner. Which player has a winning strategy, and what is that strategy?
Answer, courtesy of Thomas Yuster, associate professor of mathematics: Player A can always win. Select 3 or 9 on the first move. (All other first moves lose if B plays correctly – first move for B in those cases: take 6.)
As an example, let’s say player A takes 9. (The idea is the same for 3.)
Then there are two numbers left divisible by 3, namely 3 and 6. This is group one.
There are four numbers left that are divisible by 2 but not 3, namely 2, 4, 8, and 10. This is group two.
There are three other numbers, namely 1, 5, and 7.
Bad move x: Any player who removes a number from group one while there are still numbers in group two loses on the next turn when the opponent chooses the last number in group one.
Bad move y: Any player who removes the last number from group two while the number 6 is still in group one loses on the next turn when the opponent chooses 6.
No one can win until the opponent makes bad move x or bad move y. Player A wins by not making bad move x or bad move y. As soon as player B makes bad move x or y, player A wins on the next turn. But can player B be forced to make bad move x or bad move y?
If both players avoid bad moves x and y as long as possible, the position will eventually be as follows: a number from group 2 (2 for example), 6, and 9. Since seven numbers are gone, seven moves have been made to this point and it is player B’s turn. Player B is forced to make bad move x or bad move y.