Japan Mathematical Olympiad

$\boxed{3}$

For $1\le i\le 2021$ and $1\le j\le 2021$, let $(i,j)$ denote the cell in the $i$-th row and the $j$-th column, and let $f(i,j)$ denote the number filled in $(i,j)$. We also define $g(i,j)$ as $$g(i,j)=\{\begin{array}{ll}f(i,j)& (i+j\text{ is even}),\\ 4-f(i,j)& (i+j\text{ is odd}).\end{array}$$

It is easy to check that $g(i,j)\in \{1,2,3\}$ and $g$ satisfies, for $1\le i\le 2020$ and $1\le j\le 2020$, $$g(i+1,j)-g(i,j)=g(i+1,j+1)-g(i,j+1).(\ast )$$ Indeed, since $f(i,j)+f(i,j+1)+f(i+1,j)+f(i+1,j+1)=8$, if $i+j$ is even, $$\begin{array}{rl}g(i+1,j)-g(i,j)& =(4-f(i+1,j))-f(i,j)\\ & =f(i+1,j+1)-(4-f(i,j+1))\\ & =g(i+1,j+1)-g(i,j+1)\end{array}$$ and if $i+j$ is odd, $$\begin{array}{rl}g(i+1,j)-g(i,j)& =f(i+1,j)-(4-f(i,j))\\ & =(4-f(i+1,j+1))-f(i,j+1)\\ & =g(i+1,j+1)-g(i,j+1).\end{array}$$ On the other hand, if we assign $g(i,j)$ for $1\le i\le 2021$ and $1\le j\le 2021$ in such a way that $g(i,j)\in \{1,2,3\}$ and $g$ satisfies (), $$\tilde{f}(i,j)=\{\begin{array}{ll}g(i,j)& (i+j\text{ is even}),\\ 4-g(i,j)& (i+j\text{ is odd}).\end{array}$$ satisfies the restriction of the original problem. Hence, $A$ equals to the number of cases of defining $g(i,j)$ under the above condition.

Let $M$ and $m$ denote the maximum and the minimum of $\{g(1,1),\dots ,g(1,2021)\}$, respectively. Having () for every $1\le i\le 2020$ and $1\le j\le 2020$ is equivalent to the condition that $g(k+1,\ell )-g(1,\ell )$ is constant for $1\le \ell \le 2021$ for each $1\le k\le 2020$. We denote this constant value by $d_k$. Since $1\le g(i,j)\le 3$ for every $(i,j)$ is equivalent to $1\le m+d_k$ and $M+d_k\le 3$, or $1-m\le d_k\le 3-M$, for every $1\le k\le 2020$, there are $(3+m-M{)}^{2020}$ possibilities for assigning $g(i,j)$ when $g(1,1),\dots ,g(1,2021)$ are given.

$M-m=0$.

In this case we have $g(1,1)=\cdots =g(1,2021)$ and there are only $3$ possibilities for $(g(1,1),\dots ,g(1,2021))$. For each possibility, we have $3^{2020}$ ways of assigning the rest of $g(i,j)$, thus the number of cases is $3\cdot 3^{2020}=3^{2021}$.

$M-m=1$.

In this case we have $m=1$ or $2$. For each $m$, we have $2^{2021}-2$ possibilities for $(g(1,1),\dots ,g(1,2021))$. Thus, the number of cases is $2\cdot ((2^{2021}-2)\cdot 2^{2020})=2^{4042}-2^{2022}$.

* $M-m=2$.

In this case we have $3^{2021}-2\cdot (2^{2021}-2)-3$ possibilities for $(g(1,1),\dots ,g(1,2021))$. Thus, the number of cases is $(3^{2021}-2\cdot (2^{2021}-2)-3)\cdot 1^{2020}=3^{2021}-2\cdot (2^{2021}-2)-3$.

$$A=3^{2021}+(2^{4042}-2^{2022})+(3^{2021}-2\cdot (2^{2021}-2)-3)=2\cdot 3^{2021}+2^{4042}-2^{2023}+1.$$

To obtain the remainder after dividing $A$ by $100$, we compute $A(\mathrm{mod}4)$ and $A(\mathrm{mod}25)$. Since $3^{2021}\equiv 3(\mathrm{mod}4)$, we have $A\equiv 2\cdot 3+0-0+1\equiv 3(\mathrm{mod}4)$. And since $2^{20}\equiv 3^{20}\equiv 1(\mathrm{mod}25)$ from Euler's totient theorem, $$A\equiv 2\cdot 3\cdot (3^{20}{)}^{101}+2^2\cdot (2^{20}{)}^{202}-2^3\cdot (2^{20}{)}^{101}+1\equiv 6+4-8+1\equiv 3(\mathrm{mod}25).$$ This concludes that $A\equiv 3(\mathrm{mod}100)$.