88 = 8 × 11 253 = 23 × 11 184 = 8 × 23 Answer = 23 × 88 = 2022

miyukina wrote: 88 = 8 × 11 253 = 23 × 11 184 = 8 × 23 Answer = 23 × 88 = 2022 uhh $23\times 88=2024$? (which is indeed the correct answer)

we do the funniest thing possible As mentioned by miyukina, miyukina wrote: 88 = 8 × 11 253 = 23 × 11 184 = 8 × 23 This basically means that we can partition the board into $11 \times 11$ squares. In each $11 \times 11$ square, every row and column only has $1$ purple square. This works because there will be $8$ purple squares in every row and $23$ in every column which satisfies our conditions. Because of this, there are $11$ purple squares in our $11 \times 11$ square. Adding them altogether, we get $8\cdot 23\cdot 11= 2024$ purple squares.

The answer is $\boxed{2024}$. Construction: Sub-divide the $88\times 253$ grid into a $11\times11$ grid of $8\times 23$ grids. Then color all sub-grids along the diagonal of the resulting grid purple. There is a total of $11\cdot8\cdot23=2024$ purple cells and $x(c)y(c)=23\cdot 8=184$. [asy][asy]int i, j; for(i=0; i<12; i=i+1) { draw((23i,0)--(23i,88)); } for(j=0; j<12; j=j+1) { draw((0,8j)--(253,8j)); } for(i=0; i<11; i=i+1) { fill((23i,8i)--(23+23i,0+8i)--(23+23i,8+8i)--(0+23i,8+8i)--cycle,purple); } [/asy][/asy] Bound: Let $P$ be the number of purple cells in the grid. A simple counting argument gives $$P=\frac{1}{253}\sum_{c}{x(c)}=\frac{1}{88}\sum_{c}{y(c)}$$Now be the Cauchy-Shwartz Inequality $$\left(\sum_{c}{\sqrt{184}}\right)^2\leq\left(\sum_{c}{\sqrt{x(c)y(c)}}\right)^2\leq \sum_{c}{x(c)}\cdot\sum_{c}{y(c)}=(253P)(88P)$$So we have $11^2\cdot8^2\cdot23^2\leq P^2\Rightarrow 2024\leq P$.

This is the only problem I enjoy this day and it’s suitable for it position.

This problem is fairly similar to ISL 2012 C3 The answer is $\boxed{2024}$. Construction. The key observation is $88 = 8\times 11$, $253 = 23\times 11$, and $184 = 8\times 23$. With that, we subdivide the $88\times 253$ grid to $11^2$ blocks of size $8\times 23$. Then, color each block along the diagonal purple. For any cells, $x(c)=23$, and $y(c)=8$. There are $8\times 23\times 11 = 2024$ cells. Bound. This is the part that is similar to the shortlist problem. We first note that for any yellow cell $c$, $$\frac{x(c)}{23} \cdot \frac{y(c)}8 \geq 1 \implies \frac{x(c)}{23} + \frac{y(c)}8 \geq 2.$$Let $x_1,\dots,x_{88}$ be the number of purple cells in each row, and let $y_1,\dots,y_{253}$ be the number of purple cells in each column. Let $T$ be the total number of purple cells. We now sum the above displayed equation across all yellow cells. Observe that the term $\tfrac{x_i}{23}$ is counted $253-x_i$ times, while the term $\tfrac{y_i}{8}$ is counted $88-y_i$ times. Thus, we get that \begin{align*} 2\cdot \#(\text{yellow cells}) &\leq \sum_{i=1}^{88} \frac{x_i(253-x_i)}{23} +\sum_{j=1}^{253} \frac{y_i(88-y_i)}{8} \\ &= 22T - \frac 1{23}\sum_{i=1}^{88} x_i^2 - \frac 1{8}\sum_{j=1}^{253} y_i^2 \\ &\leq 22T - \frac{T^2}{2024} - \frac{T^2}{2024}, \end{align*}so we have $88\cdot 253 - T \geq 11T - \tfrac{T^2}{2024}$, which gives $T\geq 2024$.

Here is my solution: Construction Here, we will show that $2024$ purple cells are enough to make a construction. This construction part is simple, let's say that we divide the $88\times 253$ table into $184$ subtables all having the size $11\times 11$. Color the diagonal of each $11\times 11$ subtable to be purple as below: [asy][asy] size(120); int n = 11; // Number of rows and columns for (int i = 0; i <= n; ++i) { draw((0,i)--(n,i)); draw((i,0)--(i,n)); } for (int i = 0; i < n; ++i) { fill((i,n-1-i)--(i+1,n-1-i)--(i+1,n-i)--(i,n-i)--cycle, purple); } [/asy][/asy] If we pick any row $r$, there would be exactly $23$ cells that colored purple in $r$, and if we pick any row $c$, there would be exactly $8$ cells that colored purple in $c$. This means $x(t)y(t)=23\cdot 8\geqslant 184$ for every yellow cell $t$ as desired. Bounding Let $j$ be the number of purple cells. Suppose on the contrary, that $j<2024$ and we can still make the desired construction. By an average principle, there exists a row with at most $\left\lfloor\frac{j}{88}\right\rfloor\leqslant 22$ purple cells. It is easy to see that there are at least $253-22=231$ yellow cells in the same row as above. Consider each yellow cells (call it $g$), so that $$x(g)y(g)\geqslant 184\Rightarrow x(g)\geqslant \frac{184}{y(g)}\geqslant \frac{184}{22}\Rightarrow x(g)\geqslant 9$$So we should have at least $9\cdot 231=2079$ purple cells which is a contradiction. Hence, it should be at least $2024$ purple cells.