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Mmm... nuggets. (Posted on 2002-10-10) Difficulty: 3 of 5
At a certain nation-wide fast food chain, you can get chicken nuggets in boxes of 6, 9, and 20. What is the maximum number of nuggets you could ask for such that no combination of the 3 sizes will give you exactly what you wanted?

  Submitted by Happy    
Rating: 4.0000 (5 votes)
Solution: (Hide)

If the number is divisible by 3, then just using the 6 and 9 boxes can fulfill the order. Fill as much as you can with the 6 boxes, and if there's 3 left over, take away one box and replace with 9. The only exception to this is 3 nuggets, of course.

If the number is not divisible by 3, then first use a box of 20. If the remaining number of nuggets is divisible by 3, then use the first strategy to fill the rest.

If the number is STILL not divisible by 3 after using the box of 20, then use a second box of 20. Now the remaining number MUST be divisible by 3 (otherwise it should've met one of the above two criterias). Use 6 and 9's to fill the rest of the order.

However, if the remaining number was 3, then 6 and 9 can't be used. So the maximum number that can't be fulfilled is 2*20+3 = 43 nuggets.

(source: Math Puzzles Index)

Comments: ( You must be logged in to post comments.)
  Subject Author Date
AnswerK Sengupta2008-10-26 01:18:59
re(3): transformationsHappy2002-10-10 09:41:04
re(2): transformationsCheradenine2002-10-10 07:17:48
SolutionNo SubjectTomM2002-10-10 06:19:49
re: transformationsHappy2002-10-10 05:59:07
Some ThoughtstransformationsCheradenine2002-10-10 05:53:41
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