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A man walks into a store with just enough money to buy exactly 30 balloons, he then he discovers the store has a buy 1 get, one 1/3 off, sale. (a rather ridiculous sale if I do say so myself) how many balloons can he get?

I don't know for sure if this is the correct way to say the AMC question but I at least have a good idea that it is. What is the answer? this question stumped me, surely one of you knows the proper way to execute the problem.

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    $\begingroup$ Is this a problem from the contest that was held today? Are there any restrictions on how long until you can discuss the problems? (Just curious...) $\endgroup$ – apnorton Feb 20 '14 at 3:45
  • $\begingroup$ You can't discuss it until the next day, but considering the time I'm sure it's fine. Regardless, it's not like this was one of the most difficult problems on the exam, so it doesn't give a huge advantage if anyone did see this prior to taking the exam. $\endgroup$ – MCT Feb 20 '14 at 3:46
  • $\begingroup$ @anorton, I didn't think about it, "haha" $\endgroup$ – Diamond Louis XIV Feb 20 '14 at 3:50
  • $\begingroup$ @user92774 Well now i just feel dumb "Hahah". $\endgroup$ – Diamond Louis XIV Feb 20 '14 at 3:51
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I think of it this way:

Let the price of a single balloon be $B$. The man, then, has $30B$. Every time he buys a full priced balloon, the next balloon he buys costs $\frac{2}{3}B$. Let's assume he buys the balloons in pairs, so every pair of balloons he buys costs $\frac{5}{3}B$.

To find the total amount of balloons he can buy, we solve $x\frac{5}{3}B = 30B$, where $x$ represents the number of pairs the man buys. To solve the equation, we cancel the $B$ on both sides, multiply by 3 and divide by 5 to get $x = 90/5 = 18$.

So he has enough money to buy 18 pairs of balloons—or 36 balloons.

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Hint: If he buys two and gets one of them 1/3 off, that's equivalent to getting each balloon at a price of $\frac{5}{6}$ of the original balloon. You may need to account for the fact that he needs to buy two to get the deal.

By the way, that test was today, so you probably shouldn't discuss it until tomorrow (that's the rules, though I doubt anyone is taking / hasn't taken it yet at this time)

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  • $\begingroup$ I never thought about it like that, thank you much. And you are right, i probably should not have put this up, i was not aware of the timing limitations. $\endgroup$ – Diamond Louis XIV Feb 20 '14 at 3:53
  • $\begingroup$ No problem. I was tempted to post a question about a problem I didn't get too. $\endgroup$ – MCT Feb 20 '14 at 3:56
  • $\begingroup$ Oh, well I definitely appreciate fact that you would have posted one perhaps. $\endgroup$ – Diamond Louis XIV Feb 20 '14 at 3:58

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