daijobu Posted December 21, 2013 Share Posted December 21, 2013 I know I should put this to the AOPS site, but those kids are so smart, this would be way too easy for them. (It isn't in their archives, and when I've asked easier questions in the past I've been ignored.) Besides, I don't need the answer, I need to figure out a teachable approach to the problem other than guess and check. It's from MathCounts Nationals Team Round 2010: "The mean, median, unique mode and range of a set of eight positive integers are each 10. What is the largest positive integer that could be part of the set?" The answer is 17, and can be achieved with this set of numbers: 7,8,9,9,10,10,10,17 I want to teach how one would solve this sort of problem. I can see that in order to maximize the largest number, the other numbers need to be as small as possible, but I falter after that. Quote Link to comment Share on other sites More sharing options...

Tattarrattat Posted December 21, 2013 Share Posted December 21, 2013 First of all, I'm not sure if you noticed, but the median of your set is 9.5, not 10. 7, 8, 8, 10, 10, 10, 10, 17 works though. First, the set must have 2 10s, because it is the unique mode, and median. So the set can be: a, b, c, 10, 10, d, e, f We know that, because the mean is 10, that a+b+c+d+e+f=60. Also, f-a=10 because the range is 10. We can substitute f=a+10 too. Notice that if we want f to be maximum, we want a to be close to 10. Also, since we want the closest integers, to maximize a+b+c we need to minimize d+e, so d=e=10. The set now is: a, b, c, 10, 10, 10, 10, a+10 and we know 2a+b+c=30. Finally, we want to maximize a, so we minimize b+c. In that case, we would need a=b=c or at least a+1=b=c. The former fails, so we try the latter.Finally, substituting we find 4a=28, so a=7, and our set is 7, 8, 8, 10, 10, 10, 10, 17. Disclaimer: Kid wrote the above, and didn't want me to look at it. Quote Link to comment Share on other sites More sharing options...

daijobu Posted December 24, 2013 Author Share Posted December 24, 2013 Sorry it took me so long to sit down and read through this very elegant solution. So many of my approaches to these sorts of problems are rather haphazard. Thanks so much, and happy holidays! Quote Link to comment Share on other sites More sharing options...

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