# 2010 Puzzlers

January 8, 2010

I asked my middle schoolers to design problems involving the number 2010 to share with a partner. I ended up being the logical person to be my 13yo son’s partner. This is what he gave me. I haven’t solved them yet. ;-) Feel free to take a shot at them.

Using the following integers and an unlimited number of +, -, ÷, ×, and parentheses, create an expression that is equal to 2010. (You must use all the numbers in the set exactly the number of times they are listed.)

- {2, 2, 2, 2, 3, 3, 5, 5, 17}
- {1, 3, 3, 3, 7, 7, 7, 7}
- {1, 1, 1, 1, 3, 3, 3, 3, 3, 5, 5, 5}

If you like puzzles like this, also check out the 2010 game over at Let’s Play Math!

5 Comments
leave one →

Oooh! I solved the first one. There may be more than one way because my son doesn’t have his solutions here but he thinks his may have been different than mine.

Oooh! I solved the second one too! These aren’t quite as impossible as I imagined they’d be. ;-)

Great article, thanks!

Hey, you ever think of doing any cross-posting on other blogs? I have a pretty popular blog and I’d love to have you do a guest post on my blog.

You can move on to solving Fermat’s Last Theorem. The binomial expansion of (p +q)^n -(p -q)^n contains 2 as a common factor of all its terms. Where n is an integer greater than 1, the nth root of 2 is always irrational. This irrationality is corrected in the binomial expansion if p equals q, but persists as uncorrected if p and q are unequal. This proves Fermat’s Last Theorem.