## Two Problems from a Serbian 7 Grade Math Competition

I received a surprising statistics from Vladimir Nikolin concerning a 7 grade mathematics competition that was run in Belgrade, Serbia, on April 9, 2011. The unexpected result of the competition concerns two problems:

### Problem 1

Numbers a and b satisfy a² + b² - 2a + 6b + 10 = 0. Find a^{2009} - 2009b.

### Problem 3

Is number 2009·2011 - 48 composite? Explain.

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 a^{2009} - 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^{2009} - 2009b = 1 + 3·2009 = 6028.

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Is number 2009·2011 - 48 composite? Explain.

### Solution

We twice use the well know formula a² - b² = (a - b)(a + b):

2009*2011 - 48 = (2010² - 1²) - 48 = 2010² - 7² = 2003·2017.

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Copyright © 1996-2018 Alexander Bogomolny