At McDonald’s you can order Chicken McNuggets in boxes of 6, 9, and 20.
What is the largest number of nuggets that it is not possible to obtain by purchasing some combination of boxes?
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11 guesses to McNuggets |
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October 14th, 2009 | Category: 
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You can buy any number of nuggets that is evenly divisible by three except for three just by using combinations of boxes of 6 and boxes of 9. (Use at most one box of 9, then multiples of boxes of 6.) If the number is not divisible by three, use a box of 20. If, after a box of 20 is purchased, the remaining number is divisible by three, you’re all set. Otherwise, use a second box of 20. The remaining number will necessarily be divisible by 3, and you’re all set.
So the largest number that cannot be purchased would be one that requires two boxes of 20 before the remainder is reduced to a number divisible by three. Since three is the only number evenly divisible by three that cannot be purchased, the largest impossible number is 3 + 20 + 20 = 43.
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The number is 43 and thats all you need to know figure out why and tell me why. I’ll tell you if you’re right or night, The Riddle Dude signing off bye.
largest number you can not get is 43
1) 6 =6
2) 9 =9
3) 20 =20
4) 6+6 =12
5) 6=9 =15
6) 9+9 =18
7) 6+20 =26
8) 9+20 =29
9) 20+20 =40
10) 20+6+6 =32
11) 20+9+6 =35
12) 20+9+9 =38
Any number that is possible to get would be a combination or multiple of these of these 12.
Largest number you can not get is 43.
Thats what I came up with until I saw Dude’s explantion which is also correct, just a different and maybe easier way of looking at it
43
chicken nuggets are relly god so tell me answer plez
this makes nosense
you can have any number that you cant create with 20 ,9 and 5
mcdonalds sells mcnuggets in boxes of 8,15, and 20. pieces. what is the largest number of pieces that cannot be bought?
43