2023 SSMO Accuracy Round Problems/Problem 9

Revision as of 22:22, 15 December 2023 by Pinkpig (talk | contribs) (Created page with "==Problem== Consider a <math>2023 \times 2023</math> grid called <math>A = P_{2023}</math>. We take one of the four smaller <math>2022 \times 2022</math> grids located in <mat...")
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

Problem

Consider a $2023 \times 2023$ grid called $A = P_{2023}$. We take one of the four smaller $2022 \times 2022$ grids located in $P_{2023}$ as $P_{2022}$. We repeat the process of taking smaller grids until we eventually converge at the unit square $P_1.$

[asy] size(7cm); filldraw((0,0)--(0,10)--(10,10)--(10,0)--cycle, opacity(0.2)+lightblue, blue); fill((0,0)--(0,9)--(9,9)--(9,0)--cycle,  opacity(0.1)+lightblue); draw((0,9)--(9,9)--(9,0),  blue); fill((1,0)--(1,8)--(9,8)--(9,0)--cycle, opacity(0.1)+lightblue); draw((1,0)--(1,8)--(9,8), blue);  label("$A = P_{2023}$", (8.3, 9.52)); label("$P_{2022}$", (6.8, 8.52)); label("$\dots$", (4.78, 7.52)); [/asy]

Of the $4^{2022}$ distinct tuples of shrinking grids $(P_{2023}, P_{2022}, \dots P_1)$, let $T$ be the number of these tuples such that their last element is the center square of the original grid $A$. Find the largest integer $a$ such $2^a \mid T.$

Solution