2010 March Problems
Thursday, March 4, 2010 23:34The March problems are intended to help students prepare for the 2010 camp selection problems, used to choose students to attend our week-long residential training camp in Christchurch in January.
In recent years the camp selection problems have been known as the “September Problems”, as they were made available in September. This year we’re going to trial moving the selection problems earlier in the year, releasing them in July and moving the due date to August. This will allow more time for pre-camp training, building up to Round One of the British Mathematical Olympiad in December.
The solutions will be posted in about two month’s time, but can be obtained before then by email if you write to me with evidence that you’ve tried the problems seriously.
Good luck!
(Looking for something more challenging? Try the squad number theory assignment.)
Joseph Huang says:
May 4th, 2010 at 8:50 pm
Hi Chris,
I was doing the 2010 March Problem 1. It seemed that B has a winning strategy, but not A. Can you mail me the solution of it? Thanks a lot. Joseph
Chris says:
May 5th, 2010 at 5:52 pm
Dear Joseph,
B can always make sure the number isn’t divisible by 2 or 5, but if she does so A can force it to be divisible by 3. Do you know how to test for divisibility by 3? Have another think, and I’ll send you the solutions if you still can’t see how A should play.
Best wishes,
Chris
Joseph Huang says:
May 12th, 2010 at 3:03 pm
Dear Chris, B makes the final move, and like you said, B can always make sure the number isn’t divisible by 2 or 5, so how can A has a winning strategy? Regards, Joseph
Chris says:
May 12th, 2010 at 6:15 pm
Remember, to win B has to make sure the number isn’t divisible by 3 as well, which means that she has to choose the last number so that the sum of the six digits isn’t divisible by 3 (since a number is divisible by 3 if and only if the sum of its digits is). A can exploit this to ensure that no matter what B picks on her last move, the number will be divisible by at least one of 2, 3 and 5.
Best wishes,
Chris