In McDonald's restaurants, we can order Chicken McNuggets in packages of 6, 9, or 20. (For example, we can order 21 pieces because $21=6+6+9$, but we cannot get 19 pieces in any way.) What is the largest number of pieces that we cannot order?
The number of chicken pieces ordered will henceforth only be integers that can be expressed as the sum of multiples of 6, 9, and 20.

The sums of multiples of 6 and 9 are numbers of the form $6a + 9b = 3(2a + 3b)$, and among these, every number of the form $3k (k \geq 2)$ appears, since numbers of the form $3r, 3r+1, 3r+2$ can all be written in the form $2a + 3b$.

Therefore, numbers of the form $20 + 3k$ for $k \geq 2$ can be expressed in the desired way, and the same is true for numbers of the form $40 + 3k$. Starting from 46, every number can appear, because they are either multiples of 3, or if they are of the form $20 + 3k$, they give a remainder of 2 when divided by 3, while if they are of the form $40 + 3k$, they give a remainder of 1 when divided by 3.

The question is whether every number between 26 and 46 can be expressed? We claim that no, 43 is missing, and this is the largest number that cannot be written as the sum of multiples of 6, 9, and 20. The number 45 is a multiple of 9, and the numbers of the form $20 + 3k$ $(k \geq 2)$ are:

$$
\begin{array}{llllllllll}
26, & 29, & 32, & 35, & 38, & 41, & 44, & 47, & \ldots
\end{array}
$$

while the numbers of the form $40 + 3k$ $(k \geq 2)$ start with 46. The other (missing) numbers are either multiples of 3 or are less than 43.

Indeed, 43 cannot be broken down in the desired way, because if one 20 is included, then $43 = 20 + 23$, but 23 does not appear among the numbers of the form $20 + 3k$ for $k \geq 2$. If, however, 43 includes two 20s, then $43 = 20 + 20 + 3$, but there is no 3-piece package.

Therefore, the largest number of pieces that cannot be ordered is 43.