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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.)

  1. {2, 2, 2, 2, 3, 3, 5, 5, 17}
  2. {1, 3, 3, 3, 7, 7, 7, 7}
  3. {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 →
  1. January 9, 2010 9:58 pm

    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.

  2. January 9, 2010 11:17 pm

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

  3. February 23, 2010 9:55 pm

    Great article, thanks!

  4. March 22, 2010 10:30 am

    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.

  5. Peter L. Griffiths permalink
    August 8, 2013 10:58 am

    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.

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