I'm lovin' it

Observe that $\gcd(6,9,20) = 1$ and consider the integer $43$ . If $43$ can be written as a nonnegative integer combination of $6,9,$ and $20$ , then it must have $0,1,$ or $2$ multiples of $20$ . So $43, 23,$ or $3$ would have to be representable as a nonnegative integer combination of $6$ and $9$ . This cannot be done since $gcd(6,9) = 3,$ which does not divide $43$ or $23$ , and since $3 < \min(6,9)$ . Furthermore,

$\begin{aligned} 44 & = 1 \cdot 20 + 0 \cdot 9 + 4 \cdot 6\\ 45 &= 0 \cdot 20 + 3 \cdot 9 + 3 \cdot 6\\ 46 & = 2 \cdot 20 + 0 \cdot 9 + 1 \cdot 6\\ 47 & = 1 \cdot 20 + 3 \cdot 9 + 0 \cdot 6\\ 48 & = 0 \cdot 20 + 0 \cdot 9 + 8 \cdot 6\\ 49 & = 2 \cdot 20 + 1 \cdot 9 + 0 \cdot 6, \end{aligned}$

and every number larger than 49 can be written as one of these numbers plus a multiple of $6$ . So $g(6,9,20) = 43$ .

After 6 all numbers divisible by 3 can be ordered (because they can all be expressed as a sum of 6's and 9's). After 26, all numbers divisible by three when subtracted by 20 can be obtained. After 46, all numbers divisible by three when subtracted by 40 can be obtained. After 46, all numbers fit into one of these 3 categories, so all numbers can be obtained. 43 is the last number that doesn't fall into one of these categories (44 = 20 + 6 * 4, 45 = 6 * 6 + 9).

According to Dr. Warm's theorem , since $LCM(9, 20) = 180$ and $6|180$ , $g(6, 9, 20) = LCM(6, 9) + LCM(6, 20) -6 -9 -20 = 18 + 60 -35 = \boxed{43}$ .

First let the total number you take be $x$

Now we have three cases:

(Case 1) $x\equiv0\pmod3$

It is obvious that you can take all multiples of 3 except 3 (in other words, 6,9,12,15...) since you only have packets of 6 and 9 to choose from. So the maximum number you cannot take in this case is just 3.

(Case 2) $x\equiv1\pmod3$

For this case to work we must have at least 2 packets of 20, so choose the minimum. From the previous case we proved that 3 is the largest you cannot take from the packets of 6 and 9. Therefore, the maximum number you cannot take in this case is $2\times{20}+3=43$ .

(Case 3) $x\equiv2\pmod3$

Similarly, we must have at least one packet of 20. Thus the maximum number you can't take in this case is $1\times{20}+3=23$ .

$\therefore \boxed{43}$

With just 6 and 9 you can make any multiple of 3 except 3. So all orders which are valid are of the form 3a + 20b.

Let's take away all the 20s and we will be left with some multiple of 3 between 1 and 20, these are (excluding 3): {6, 9, 12, 15, 18} So if a number mod 20 is equal to any of the above, and that number is > 3 we know we can order that many pieces of chicken.

Suppose b ≥ 1 (which means the number is ≥ 20), then there is at least 1 twenty we can not remove. This means we can take away all but one twenty, and be left with a number from 20-40. Multiples of 3 between 20-40 are: 21, 24, 27, 30, 33, 36, 39 If we take mod 20 of all of these numbers, then we get: {1, 4, 7, 10, 13, 16, 19} So if a number mod 20 is equal to any of the above, and that number is ≥ 20, then we know that we can get that many pieces of chicken.

Suppose b ≥ 2 (which means the number is ≥ 40), then we can not remove two twenties. So now we're left with a number from 40-60. Multiples of 3 between 40-60 are: 42, 45, 48, 51, 54, 57 If we take mod 20 of all of these numbers, then we get: {2, 5, 8, 11, 14, 17} So if a number mod 20 is equal to any of the above and that number is ≥ 40, we know that we can get that many pieces of chicken.

Suppose b ≥ 3, then we get multiples between 60-80, which basically gives us all the values from the first set, but this time including 3: {3}

So basically, if a number is ≥ 60 (and thus ≥ 20 and ≥ 40) then the possible values (mod 20) we can create an order for are (union of all sets above): {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19}. Which covers any value. So we know past 60 there's no number of chicken pieces we can't order. So consider the range 40-60: A number in this range is ≥ 20 and ≥ 40, so by taking the first three sets, we get cover everything except for {3}, giving us the answer, 43.

You don't actually have to list out all the possibilities, it's possible to arrive at 43 by just realising that since 20 is one less than a multiple of 3 (21), for every multiple of 20 you cross you get an extra set of values mod 20 that can be created and that after 60 anything can be made.

Run the code: Frobenius number {6, 9, 20}

I don't know any other solutions though