[jdevious22]

You are at Mcdonald's and are ordering chicken nuggets. There are 6, 9, and 20 piece options available. What is the largest order you can make that cannot be fulfilled by combining these three options.

[bwtyree]

I believe the answer is 6.

[bwtyree]

Or, it could also be 35.

[galactus]

35 IS obtainable. 9+6+20=35.

Go a little higher.

[Unknown008]

But it says:

What is the largest order you can make that cannot be fulfilled by combining these three options.

:confused:

[ebaines]

I'm going to assume that the question as written: "what's the largest order you can make that can not be fulfilled by combining these three options" is better asked as: "what's the largest integer that can not be constructed by adding up some combination of these three numbers." With that understanding - I believe the answer is 43. Here's why:

Notice that by using combinations of 6 and 9 you can get to any number that is a mutiple of 3, with the exception of 3 itself. For example: 15 = 9 +6, 21 = 9 + 2*6, 24 = 2*9+6, etc. So for numbers less than 20, any that are not divisible by 3 are candidates: 19 is the largest of these.

Now consider numbers between 21 and 39. We need to see which can be constructed using one 20 plus a combination of 6's and 9's, or by 6's and 9's alone. Let's define the number as having digits A and B: for example, for the number 31, A=3 and B = 1. This number is a multiple of 3 if A+B is divisible by 3. We already know that if A+B is divisible by 3 then the number can be constructed from 6's and 9's. If not, then we try subtracting 20 and see if the result is a multiple of 3. This turns the original number into (A-2)B. For example, we started with 31, we subtract 20 and get 13. (A-2)B is divisible by 3 if A-2+B is divisible by 3. In our example we have 3 - 2 + 1, which is not divisible by 3, so 31 can not be constructed from the numbers 6, 9 and 20. In this manner we find that 37 is the largest number between 20 and 39 that meets both criteria; that is, 3+7 is not divisible by 3, and 3-2+7 is also not divisible by 3.

For numbers greater than 40 we need to check three conditions: A+B is not a multiple of 3, A-2+B is not a multiple of 3, and A-4+3 is not a multiple of 3. But for A = 4 or greater, one of these must be a multiple of 3, because for any number greater than or equal to 4 either that number is a multiple of 3, or it's 2 greater than a multiple of 3, or it's 4 greater than a multiple of 3. So at first it seems that there can be no numbers for which A = 4 or greater that can not be constructed from 6, 9 and 20.

But wait - recall that 3 is an exception. This means that for (A-2)B =3 or (A-4)B = 3, the number can't be constructed from 6, 9 and 20. So the only number greater than 40 that can not be constructed from the 3 options are those where A+B is not a multiple of 3, A-2+B is not a multiple of 3, and (A-4)B = 3. Hence A=4, B = 3.

For all other numbers greater than this it must be the case that A+B. A-2+B, or A-4+B is a multiple of 3. Hence no other larger numbers need be considered.

Conclusion: the answer is 43.

[Unknown008]

At last a good explanation! :) I'm still confused on how numbers with tenth digit greater than or equal to 4 are obtainable... Never mind, I'll try understand it later. I'll post if I still have problems. Thanks ebaines! :)

[ebaines]

I'm still confused on how numbers with tenth digit greater than or equal to 4 are obtainable

Maybe this will help:

Every number greater than 40 is either:

a) Divisible by 3, OR
b) is 20 more than a number that is divisible by 3, OR
c) is 40 more than a number that is divisible by 3.

This is based on the fact that any number that is divisible by 3 has the property that the sum of its digits is also divisible by 3. For any number if you add the digits together you get a number, let's call it N, which when divided by 3 gives a remainder R of either 0, 1 or 2.

a) If R is 0 then clearly N is a multiple of 3, and so the original number is also a multiple of 3.
b) If R = 2 then you can subtract 20 from the original number and you have a new number whose digits add together to a multiple of 3, so the new number is a multiple of 3. Hence the original number minus 20 is a multiple of 3.
c) And if R is 1 then you can subtract 10 from the original number to get to a multiple of 3, and then subtract an additional 30 to get to yet another new number that is again a multiple of 3. Hence the original number minus 40 is a multiple of 3.

Once you have gotten to a number that is a multiple of 3, you can combine always 6's and 9's to build to that number and then add either zero, one, or two 20's (depending on whether case a, b, or c respectively applies) to get the original number. The only exception is if the number you get in step a, b, or c above is 3 itself. That's why the number 43 works.

[Unknown008]

Ohh! Ok, The remainder explanation was as crystal clear to me! I understood that! Thanks! :)

[aerovan]

20,9 and 6