Problem 3
Is number 2009·2011 - 48 composite? Explain.
Solution
The first problem - the less standard of the two - was solved by 50% of the participants, whereas the second problem was solved by only 19%. It is hard to explain why more students found the second problem more difficult. I tweeted both problems on twitter.com; several solutions were submitted for the second problem, none for the first. For this reason, I place the solutions below.
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Problem 1
Numbers a and b satisfy a² + b² - 2a + 6b + 10 = 0. Find a2009 - 2009b.
Solution
Rewrite the expression by completing the squares:
a² + b² - 2a + 6b + 10 = (a - 1)² - 1 + (b + 3)² - 9 + 10,
implying
(a - 1)² + (b + 3)² = 0.
Since both terms on the left are never negative, for the identity to hold, both must be 0:
(a - 1)² = (b + 3)² = 0.
Thus we may conclude that a = 1 and b = -3. We now need to carry out a substitution. For a = 1 and b = -3,
a2009 - 2009b = 1 + 3·2009 = 6028.
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Copyright © 1996-2012 Alexander Bogomolny
