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Monday, March 2, 2015

Problem of the Week

Happy Monday to all of our readers, regular and new. We've been working on a lot of great new content for the blog that should be starting this month, including the continuation of the interview series. But for now, take a look at our newest Problem of the Week. Like usual, I've hosted the solution below. Here's what was posted to our Facebook, Twitter, and Google +:
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A proof! This one shouldn't be too difficult. It will act as a good brain teaser to kick off the week. Here's how I solved the problem.

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This wasn't too bad so far. I made my original supposition because one expression must be larger than (or equal to) the other, and we lost no generality by assuming the first term was larger rather than the second.
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Just this past Fall I took a class on mathematical reasoning and proofs. One of my favorite homework proofs that we had was resolved by realizing the expression was the product of three consecutive numbers. That proof will stick with me in future classes, and the Problem of the Week was a welcome refresher.

If you have any questions about my solution or if you solved it a different way, please share in the comments section.

1 comment:

  1. I was wondering if there are any proofs for your last couple of statements?
    "A set with a multiple of 3 is divisible by 6"
    "The set can be rewritten as..."