BMO Shortlist, 2022

Suppose that for some $k > 1$ and some placed cubelet there is a valid filling. In each unit cubelet (of the original cube) with coordinates $(x, y, z)$ where $0 \le x, y, z \le 2020$, we assign the complex number $\omega^{x+y+z}$ where $\omega = e^{\frac{2\pi i}{k}}$. We also assign the number $\omega^{a+b+c}$ in the additional cubelet in position $(a, b, c)$.

Since $1 + \omega + \omega^2 + \dots + \omega^{k-1} = \frac{\omega^k - 1}{\omega - 1} = 0$, then the sum of numbers in any $1 \times 1 \times k$ cuboid is equal to zero. So the sum of all assigned numbers is equal to
$$0 = (1 + \omega + \dots + \omega^{2020})^3 + \omega^{a+b+c}. \qquad (1)$$
This gives
$$1 = |-\omega^{a+b+c}| = |(1 + \omega + \dots + \omega^{2020})^3| = \left|\frac{1 - \omega^{2021}}{1 - \omega}\right|^3.$$
Thus $|1 - \omega| = |1 - \omega^{2021}|$ which means that $1$ is equidistant from the numbers $\omega$ and $\omega^{2021}$. Since $|\omega^{2021}| = 1$, this happens if and only if $\omega^{2021} = \omega$ or $\omega^{2021} = \omega^{-1}$. Then $\omega^{2020} = 1$ or $\omega^{2022} = 1$ which gives $k|2020$ or $k|2022$. However $k|2021^3 + 1$. Since $2021^3 + 1 \equiv 2 \mod 2020$, if $k|2020$ then $k|2$. So in any case we have $k|2022$. Now (1) gives
$$\omega^{a+b+c} = -(1 + \omega + \dots + \omega^{2020})^3 = -(-\omega^{2021})^3 = \omega^{6063} = \omega^{-3}.$$
So $a + b + c \equiv -3 \mod k$.
If we have a valid filling for $k$, then we have a valid filling for every prime factor of $k$. So we may assume that $k$ is prime and therefore $k \in \{2, 3, 337\}$.
Assume without loss of generality that the additional cubelet is at the bottom of the cube, i.e. $c = -1$. Then $a + b \equiv -2 \mod k$. By symmetry, if we have a valid filling for $(a, b, -1)$, then we have a valid filling for $(2022 - a, b, -1)$. So we must also have $2022 - a + b \equiv -2 \mod k$ and so $b - a \equiv -2 \mod k$. Since also $a + b \equiv -2 \mod k$ we get $a \equiv b \equiv -1 \mod k$ for $k \neq 2$ and $a \equiv b \mod 2$ for $k = 2$. We will now show that the above necessary conditions are also sufficient to have a valid filling.
We claim first that a square defined by coordinates $(x, y)$ where $0 \le x, y \le 2020$, with a removed cell $(a, b)$ satisfying the above restrictions can be covered by $1 \times k$ rectangles. This is because such a square can be covered by four rectangles of sizes $(a+1) \times b$, $(2020-a) \times (b+1)$, $(2021-a) \times (2020-b)$ and $a \times (2021-b)$, where each of these rectangles can be covered by $1 \times k$ rectangles. This follows since $k | a+1, b+1, 2021-a, 2021-b$ if $a \equiv b \equiv -1 \mod k$ and since $2|b, 2020-a, 2020-b, a$ if $a \equiv b \equiv 0 \mod 2$.
Now we fill the cube with the added cubelet as follows: The lowest $k-1$ "layers" of the original cube of side $2021$ are filled using the previous method together with a $1 \times 1 \times k$ cuboid covering the holes in these layers and the additional cubelet. The remainder is the $2021 \times 2021 \times (2022-k)$ cuboid which can be easily filled by $1 \times 1 \times k$ cuboids because $k | 2022-k$.
To complete the solution, we need to count the number of ordered pairs $(a, b)$ with $0 \le a, b \le 2020$, such that $a \equiv b \mod 2$, or $a \equiv b \equiv -1 \mod 3$ or $a \equiv b \equiv -1 \mod 337$. There are $1011^2$ choices with $a \equiv b \equiv 0 \mod 2$ and $1010^2$ choices with $a \equiv b \equiv 1 \mod 2$. If $a \equiv b \equiv -1 \mod 3$ but $a \neq b \mod 2$ then one of $a, b$ must be congruent to $2 \mod 6$ and the other to $5 \mod 6$. There are $2 \times 337 \times 336$ such choices. Finally, if $a \equiv b \equiv -1 \mod 337$ but is not yet accounted for, then one of them is equal to $\{336, 1010, 1684\}$ and the other to $\{673, 1347\}$. (Note that in this case the second one is definitely not congruent to $-1 \mod 3$.) There are $2 \times 3 \times 2$ such choices. In total we have $2268697$ choices for the pair $(a, b)$. Therefore, because of symmetry, the total number of ways is $6 \times 2268697 = 13612182$.

Since there are a total of $2021^3 + 1$ cubelets and each cuboid covers $k$ of them, we must have $k|2021^3 + 1$. We have
$$2021^3 = 2022 \cdot (2021^2 - 2021 + 1) = 2 \cdot 3 \cdot 337 \cdot 3 \cdot 7 \cdot 31 \cdot 6271$$
as a product of prime factors.
If we have a valid filling for $k$, then we have a valid filling for every prime factor of $k$. So we may assume that $k$ is prime and therefore $k \in \{2, 3, 7, 31, 337\}$. (The case $k = 6271$ is easily seen to be impossible.)
We colour the cubelet at position $(x, y, z)$ for $0 \le x, y, z \le 2020$ by the colour $x+y+z \bmod k$. So every $1 \times 1 \times k$ cuboid covers 1 cubelet of each colour.
If $k = 2$ then it is easy to see that we have one more cubelet of the form $0 \bmod 2$ than of the form $1 \bmod 2$. So assuming that the additional cubelet is in position $(a, b, -1)$, we must have $a+b \equiv 0 \bmod 2$ as in Solution 1.
If $k = 3$ then, since it is each to cover all cubelets with $1 \times 1 \times k$ cuboids except those of the form $(x, y, z)$ with $0 \le x, y, z \le 2$, we see that there is one less cubelet of the form $0 \bmod 3$ than of the form $1, 2 \bmod 3$. So assuming that the additional cubelet is in position $(a, b, -1)$, we must have $a+b \equiv -2 \bmod 3$. Exploiting the symmetry as in Solution 1 we get $a \equiv b \equiv -1 \bmod 3$.
If $k = 7$ then note that (since $7|2016$) it is easy to cover all cubelets with $1 \times 1 \times k$ cuboids except those of the form $(x, y, z)$ with $0 \le x, y, z \le 4$. Note that exactly 19 of these cubicles have the colour $0 \bmod 7$. (3 when $z = 0$, 3 when $z = 1$, 4 when $z = 2$, 5 when $z = 3$ and 4 when $z = 4$.) These are more than $\frac{5^3+1}{7} = 18$ so one of these will remain uncovered. So $k = 7$ is impossible.
If $k = 31$ then note that (since $31|2015$) it is easy to cover all cubelets with $1 \times 1 \times k$ cuboids except those of the form $(x, y, z)$ with $0 \le x, y, z \le 5$. Note that only one of those cubelets has the colour $0 \bmod 31$. So even if the additional cubelet has the same colour it is still less than the $\frac{6^3+1}{31} = 7$ which are expected in a proper covering. So $k = 31$ is impossible.
If $k = 337$ then note that the square containing all cells of the form $(x, y)$ with $0 \le x \le 2020$ and $0 \le y \le 2021$ can be covered with $1 \times k$ rectangles and so contains equal number of cells of each colour (thinking of $z = 0$). In the column with $y = 2021$, the cells have in order the colours $-1, 0, 1, \dots, -3 \bmod k$. So the colour $-2 \bmod k$ appears one time less in that column and therefore one time more in the square of the form $(x, y)$ with $0 \le x, y \le 2020$. In the 'layer' above the extra colour is $-1 \bmod k$, then $0 \bmod k$ and so on until $2020 - 2 \equiv -4 \bmod k$. So in the original cube all colours appear an equal number of times except $-3 \bmod k$ which appear one time less and must be the colour of the additional cubelet $(a, b, -1)$. Thus $a+b \equiv -2 \bmod k$. Exploiting the symmetry as in Solution 1 we get $a \equiv b \equiv -1 \bmod 337$.
So we get the same necessary conditions as in Solution 1. The sufficiency of these conditions is proved in a similar way as in Solution 1.