i have a problem of the week for my college prep analysis class. It says: knowing that chicken mcnuggets can only be packaged in containers of 6, 9, and 20, what is the most number of nuggets that someone can order that can't be packaged evenly? example, 15 isnt the answer cause you can do one 6 and one 9. 101 isnt the answer because you can pack nine 9's and one 20.
all u need to no is the greatest common denominator of all the numbers..figure that out and u can get the answer
43. I know this because ive done this problem before. Heres sort of an explanation, but I cant word it well enough to make a lot of sense. Look it up on google or something. The number is not divisible by 3. Otherwise you could make it with 6s and 9s. If your number isn't divisible by 3, then use a box of 20. If you need a certain number more that is divisible by 3, then you can make it. If you can't, use another box of 20. There is no possible way you will need a third box of 20. The number therefor has 2 boxes of 20. You can make any combination of 6s and 9s to get 6 9 12 15 18 21 24 27 ... and so on. 3 is missing, however. You have to add 3 to the maximum number of 20s you can use, so the answer is 43. If I was posed with this problem I would just use a computer program. An easy loop with a couple mod commands could get you your max number in no time.