[/quote]
,
and 
.
be a positive integer. Consider the following equation: ![\[ \{x\}+\{2x\}+ \dots + \{nx\} = \lfloor x \rfloor + \lfloor 2x \rfloor + \dots + \lfloor 2nx \rfloor\]](http://latex.artofproblemsolving.com/8/2/9/8299b3ceb1650418c0a373536cb1d2931900efd9.png)
a) For 
,
.
real solutions.
and 
intersect at two points 
and 
.
be the line tangent to these circles at 
and 
,
be the line parallel to 
on 
on 
and 
meet at 
;
and 
;
and 
.
.
is such that 
for any positive integers 
.
such that 
for all positive integers 
.
.
satisfy that 
.
with odd integer side lengths is divided into small rectangles with integer side lengths. Prove that there is at least one among the small rectangles whose distances from the four sides of 
be real numbers . Prove that : There exist positive integer 
such that 
Where 
be an isosceles triangle 
)
.
passes through 
and 
.
and circumcircle of 
and 
,
be the midpoint of 
and 
be the midpoint of 
.
,
and 
are concurrent.
,
in the array can be changed into either 
or 
so that the row-sums and column-sums remain unchanged. (Note that 
Y by mkhayech, rashah76, itslumi, mathematicsy, Adventure10
A domino is a 
or 
tile.
Let
be an integer. Dominoes are placed on an 
board in such a way that each domino covers exactly two cells of the board, and dominoes do not overlap. The value of a row or column is the number of dominoes that cover at least one cell of this row or column. The configuration is called balanced if there exists some 
such that each row and each column has a value of 
. Prove that a balanced configuration exists for every , and find the minimum number of dominoes needed in such a configuration.
or 
tile.Let
be an integer. Dominoes are placed on an 
board in such a way that each domino covers exactly two cells of the board, and dominoes do not overlap. The value of a row or column is the number of dominoes that cover at least one cell of this row or column. The configuration is called balanced if there exists some 
such that each row and each column has a value of 
. Prove that a balanced configuration exists for every , and find the minimum number of dominoes needed in such a configuration.This post has been edited 3 times. Last edited by microsoft_office_word, Feb 18, 2020, 9:47 PM
,
and you need 
dominoes.
,
and you need 
dominoes.
.
;
dominoes then 
.
.
.
.
,
,
,
,
,
and not a multiple of 
and you need 
/
should be 
,
which implies the right lower bounds.
.
:
grid in four 
subgrids and fill each of them with the 
construction for 
.
(put k 
squares and 1 
on the diagonal);
(similarly put k of 
and 1 of 
)
(k of 
)
(k of 
,
using 
,![\[3d = 2nk.\]](http://latex.artofproblemsolving.com/0/e/2/0e252d054cbd4bba69c5873408dd6ed17343eedf.png)
Hence if 
so 
is forced.
,![[asy]
import graph;
size(4cm,4cm);
void makeSquare(pair ll, real s) {
pair r = (1,0);
pair u = (0,1);
draw(ll -- (ll + s*r) -- (ll + s*(r+u)) -- (ll + s*u) -- cycle);
}
void makeGrid(int n) {
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
makeSquare((2*i,2*j),2);
}
}
}
void drawDomino(pair beg, pair end) {
pair diff = end - beg;
pair alt = rotate(90)*diff;
pair pt1 = beg - (0.2)*diff;
pair pt2 = end + (0.2)*diff;
real sc = 0.25;
filldraw((pt1 - sc*alt) -- (pt2 - sc*alt) -- (pt2 + sc*alt) -- (pt1 + sc*alt) -- cycle, grey, black);
}
int GRIDSIZE = 3;
makeGrid(GRIDSIZE);
int DOMS = 2;
pair[] fst = {(1,1), (5,3)};
pair[] snd = {(3,1), (5,5)};
for (int k = 0; k < DOMS; ++k) {
drawDomino(fst[k], snd[k]);
}
[/asy]](http://latex.artofproblemsolving.com/6/1/4/61416d14c8feba6cb50d23f4880e176872540362.png)
we have![[asy]
import graph;
size(4cm,4cm);
void makeSquare(pair ll, real s) {
pair r = (1,0);
pair u = (0,1);
draw(ll -- (ll + s*r) -- (ll + s*(r+u)) -- (ll + s*u) -- cycle);
}
void makeGrid(int n) {
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
makeSquare((2*i,2*j),2);
}
}
}
void drawDomino(pair beg, pair end) {
pair diff = end - beg;
pair alt = rotate(90)*diff;
pair pt1 = beg - (0.2)*diff;
pair pt2 = end + (0.2)*diff;
real sc = 0.25;
filldraw((pt1 - sc*alt) -- (pt2 - sc*alt) -- (pt2 + sc*alt) -- (pt1 + sc*alt) -- cycle, grey, black);
}
int GRIDSIZE = 4;
makeGrid(GRIDSIZE);
int DOMS = 8;
pair[] fst = {(1,1), (3,1), (5, 5), (7, 5), (1,5), (1,7), (5, 1), (5, 3)};
pair[] snd = {(1,3), (3,3), (5, 7), (7,7), (3,5), (3, 7), (7, 1), (7,3)};
for (int k = 0; k < DOMS; ++k) {
drawDomino(fst[k], snd[k]);
}
[/asy]](http://latex.artofproblemsolving.com/d/8/b/d8bfffb97309293134edb56998023622502527f5.png)
we have![[asy]
import graph;
size(4cm,4cm);
void makeSquare(pair ll, real s) {
pair r = (1,0);
pair u = (0,1);
draw(ll -- (ll + s*r) -- (ll + s*(r+u)) -- (ll + s*u) -- cycle);
}
void makeGrid(int n) {
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
makeSquare((2*i,2*j),2);
}
}
}
void drawDomino(pair beg, pair end) {
pair diff = end - beg;
pair alt = rotate(90)*diff;
pair pt1 = beg - (0.2)*diff;
pair pt2 = end + (0.2)*diff;
real sc = 0.25;
filldraw((pt1 - sc*alt) -- (pt2 - sc*alt) -- (pt2 + sc*alt) -- (pt1 + sc*alt) -- cycle, grey, black);
}
int GRIDSIZE = 5;
makeGrid(GRIDSIZE);
int DOMS = 10;
pair[] fst = {(1,1), (3,1), (7,1), (7,3), (7,5), (1,5), (5,3), (1,7), (3,7), (5,7)};
pair[] snd = {(1,3), (5,1), (9,1), (9,3), (9,5), (3,5), (5,5), (1,9), (3,9), (5,9)};
for (int k = 0; k < DOMS; ++k) {
drawDomino(fst[k], snd[k]);
}
[/asy]](http://latex.artofproblemsolving.com/0/b/e/0be6545d67369ff9180ae64b3330c6e8b854dc42.png)
![[asy]
import graph;
size(4cm,4cm);
void makeSquare(pair ll, real s) {
pair r = (1,0);
pair u = (0,1);
draw(ll -- (ll + s*r) -- (ll + s*(r+u)) -- (ll + s*u) -- cycle);
}
void makeGrid(int n) {
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
makeSquare((2*i,2*j),2);
}
}
}
void drawDomino(pair beg, pair end) {
pair diff = end - beg;
pair alt = rotate(90)*diff;
pair pt1 = beg - (0.2)*diff;
pair pt2 = end + (0.2)*diff;
real sc = 0.25;
filldraw((pt1 - sc*alt) -- (pt2 - sc*alt) -- (pt2 + sc*alt) -- (pt1 + sc*alt) -- cycle, grey, black);
}
int GRIDSIZE = 7;
makeGrid(GRIDSIZE);
int DOMS = 14;
pair[] fst = {(1,1), (5,1), (3,5), (1,9), (1,13), (5,5), (5,11), (7,3), (9,7), (9, 11), (11, 7), (11,1), (11,5), (13,11)};
pair[] snd = {(1,3), (5,3), (3,7), (3,9), (3,13), (7,5), (7,11), (9,3), (9,9), (9,13), (11, 9), (13,1), (13,5), (13,13)};
for (int k = 0; k < DOMS; ++k) {
drawDomino(fst[k], snd[k]);
}
[/asy]](http://latex.artofproblemsolving.com/e/e/f/eeffe76852e3716124510f5f32b2c70bbb306779.png)
for some integer 
,
along the main diagonal of the 
grid.
is not a multiple of 
,
subgrid, and then induct on the remaining 
grid to finish.
,
subgrid, and then induct on the remaining 
grid to finish.
Where 
we show 
dominoes:
we see 
by divisibility. We show 
we describe the construction explicitely:
we go by induction. If 
then we place a 
grid in the top-left corner and a 
grid in the bottom-right corner. We can do this as 
and 
.
then we place a 
grid in the top-left corner and a 
grid in the bottom-right corner. We can do this as 
and 
(as 
)
.
.
so 
.
for divisibility, so 
.![[asy]
size(6cm);
real d = 0.1;
int n = 6;
path rect(int a, int b, int x, int y)
{
return (a-1+d,b-1+d)--(x-d,b-1+d)--(x-d,y-d)--(a-1+d,y-d)--cycle;
}
for (int i = 0; i <= n; ++i)
{
draw((i,0)--(i,n));
draw((0,i)--(n,i));
}
fill(rect(1,4,1,5),red);
fill(rect(2,6,3,6),red);
fill(rect(4,1,4,2),red);
fill(rect(5,3,6,3),red);
[/asy]](http://latex.artofproblemsolving.com/3/5/c/35cf5890a7d473e15a8f53ff017977dabcd16b03.png)
times).
not divisible by 
except 
has a balanced configuration with ![[asy]
size(4cm);
real d = 0.1;
int n = 4;
path rect(int a, int b, int x, int y)
{
return (a-1+d,b-1+d)--(x-d,b-1+d)--(x-d,y-d)--(a-1+d,y-d)--cycle;
}
for (int i = 0; i <= n; ++i)
{
draw((i,0)--(i,n));
draw((0,i)--(n,i));
}
fill(rect(1,1,1,2),red);
fill(rect(2,1,2,2),red);
fill(rect(1,3,2,3),red);
fill(rect(1,4,2,4),red);
fill(rect(3,1,4,1),red);
fill(rect(3,2,4,2),red);
fill(rect(3,3,3,4),red);
fill(rect(4,3,4,4),red);
[/asy]](http://latex.artofproblemsolving.com/9/d/a/9da41b18b42d3a95e3a4d06a64beee62a2bf0fa7.png)
![[asy]
size(5cm);
real d = 0.1;
int n = 5;
path rect(int a, int b, int x, int y)
{
return (a-1+d,b-1+d)--(x-d,b-1+d)--(x-d,y-d)--(a-1+d,y-d)--cycle;
}
for (int i = 0; i <= n; ++i)
{
draw((i,0)--(i,n));
draw((0,i)--(n,i));
}
fill(rect(1,1,1,2),red);
fill(rect(1,3,1,4),red);
fill(rect(1,5,2,5),red);
fill(rect(2,2,3,2),red);
fill(rect(2,3,3,3),red);
fill(rect(3,1,4,1),red);
fill(rect(4,3,5,3),red);
fill(rect(4,4,4,5),red);
fill(rect(5,4,5,5),red);
fill(rect(5,1,5,2),red);
[/asy]](http://latex.artofproblemsolving.com/8/1/c/81cec2b3aa3a5fb8414886e34eca50208bff7c6e.png)
![[asy]
size(7cm);
real d = 0.1;
int n = 7;
path rect(int a, int b, int x, int y)
{
return (a-1+d,b-1+d)--(x-d,b-1+d)--(x-d,y-d)--(a-1+d,y-d)--cycle;
}
for (int i = 0; i <= n; ++i)
{
draw((i,0)--(i,n));
draw((0,i)--(n,i));
}
fill(rect(1,2,1,3),red);
fill(rect(1,5,1,6),red);
fill(rect(1,7,2,7),red);
fill(rect(2,1,2,2),red);
fill(rect(2,4,2,5),red);
fill(rect(3,1,4,1),red);
fill(rect(3,4,4,4),red);
fill(rect(3,6,3,7),red);
fill(rect(4,5,5,5),red);
fill(rect(5,2,6,2),red);
fill(rect(5,3,5,4),red);
fill(rect(6,1,7,1),red);
fill(rect(6,3,7,3),red);
fill(rect(7,6,7,7),red);
[/asy]](http://latex.artofproblemsolving.com/2/a/6/2a63b4cb571c4404d5393eac2831682b2a5bce22.png)
and 
have this property, then just construct 
by appending the construction for 
:![[asy]
size(11cm);
real d = 0.1;
int n = 11;
path rect(int a, int b, int x, int y)
{
return (a-1+d,b-1+d)--(x-d,b-1+d)--(x-d,y-d)--(a-1+d,y-d)--cycle;
}
for (int i = 0; i <= n; ++i)
{
draw((i,0)--(i,n));
draw((0,i)--(n,i));
}
fill(rect(1,8,1,9),red);
fill(rect(2,8,2,9),red);
fill(rect(1,10,2,10),red);
fill(rect(1,11,2,11),red);
fill(rect(3,8,4,8),red);
fill(rect(3,9,4,9),red);
fill(rect(3,10,3,11),red);
fill(rect(4,10,4,11),red);
fill(rect(5,2,5,3),red);
fill(rect(5,5,5,6),red);
fill(rect(5,7,6,7),red);
fill(rect(6,1,6,2),red);
fill(rect(6,4,6,5),red);
fill(rect(7,1,8,1),red);
fill(rect(7,4,8,4),red);
fill(rect(7,6,7,7),red);
fill(rect(8,5,9,5),red);
fill(rect(9,2,10,2),red);
fill(rect(9,3,9,4),red);
fill(rect(10,1,11,1),red);
fill(rect(10,3,11,3),red);
fill(rect(11,6,11,7),red);
[/asy]](http://latex.artofproblemsolving.com/c/9/8/c98806d1a5f3cd4394ba51e263c6d9d4dd98657d.png)
and 
have this property, by the Chicken McNugget Theorem, all integers larger than 
have this property. So it suffices to check 
have this property. We have already constructed 
and have 
,
and 
boards such that both the configurations have the same value of 
board with the same value of 
and 
and according to the 
and 
,
with 
are already given in the above solutions.
.
.
.
.
.
.
![[asy]
size(2cm);
draw((0,0)--(3,0));
draw((0,1)--(3,1));
draw((0,2)--(3,2));
draw((0,3)--(3,3));
draw((0,0)--(0,3));
draw((1,0)--(1,3));
draw((2,0)--(2,3));
draw((3,0)--(3,3));
filldraw((0,2)--(0,3)--(2,3)--(2,2)--cycle, grey);
filldraw((2,0)--(3,0)--(3,2)--(2,2)--cycle, grey);
[/asy]](http://latex.artofproblemsolving.com/4/7/6/476e0b325d250a2976c0a06a85993f372d9dadc3.png)
![[asy]
size(2cm);
add(grid(4,4));
filldraw((0,0)--(1,0)--(1,2)--(0,2)--cycle,grey);
filldraw((1,0)--(2,0)--(2,2)--(1,2)--cycle,grey);
filldraw((2,2)--(3,2)--(3,4)--(2,4)--cycle,grey);
filldraw((3,2)--(4,2)--(4,4)--(3,4)--cycle,grey);
filldraw((0,2)--(2,2)--(2,3)--(0,3)--cycle,grey);
filldraw((0,3)--(2,3)--(2,4)--(0,4)--cycle,grey);
filldraw((2,0)--(4,0)--(4,1)--(2,1)--cycle,grey);
filldraw((2,1)--(4,1)--(4,2)--(2,2)--cycle,grey);
[/asy]](http://latex.artofproblemsolving.com/c/d/f/cdfe90d6b2b22c82bdeba3df49f4ccc78901f41c.png)
for 
.
board of 
.
,
)![[asy]
size(4cm);
add(grid(11,11));
filldraw((4,6)--(6,6)--(6,7)--(4,7)--cycle, grey);
filldraw((6,4)--(6,6)--(7,6)--(7,4)--cycle, grey);
draw((4,4)--(4,7)--(7,7)--(7,4)--cycle, linewidth(1.5));
draw((6,7)--(6,11),linewidth(1.5));
draw((7,5)--(11,5),linewidth(1.5));
draw((5,4)--(5,0),linewidth(1.5));
draw((4,6)--(0,6),linewidth(1.5));
[/asy]](http://latex.artofproblemsolving.com/a/9/9/a99a1a654afd9c321d6c24e87dbef38bd08d602f.png)
dominoes like so: starting in the bottom left corner with a vertical domino, continuing to place dominoes in a staircase fashion until the top of the grid is reached, and finishing with a horizontal domino.![[asy]
size(4cm);
add(grid(11,11));
filldraw((4,6)--(6,6)--(6,7)--(4,7)--cycle, grey);
filldraw((6,4)--(6,6)--(7,6)--(7,4)--cycle, grey);
draw((4,4)--(4,7)--(7,7)--(7,4)--cycle, linewidth(1.5));
draw((6,7)--(6,11),linewidth(1.5));
draw((7,5)--(11,5),linewidth(1.5));
draw((5,4)--(5,0),linewidth(1.5));
draw((4,6)--(0,6),linewidth(1.5));
filldraw((0,6)--(1,6)--(1,8)--(0,8)--cycle, grey);
filldraw((1,7)--(2,7)--(2,9)--(1,9)--cycle, grey);
filldraw((2,8)--(3,8)--(3,10)--(2,10)--cycle, grey);
filldraw((3,9)--(4,9)--(4,11)--(3,11)--cycle, grey);
filldraw((4,10)--(6,10)--(6,11)--(4,11)--cycle,grey);
[/asy]](http://latex.artofproblemsolving.com/5/6/b/56beb2aefed7e3667683ee7c54de4fea020f18d8.png)
![[asy]
size(4cm);
add(grid(11,11));
filldraw((4,6)--(6,6)--(6,7)--(4,7)--cycle, grey);
filldraw((6,4)--(6,6)--(7,6)--(7,4)--cycle, grey);
filldraw((0,6)--(1,6)--(1,8)--(0,8)--cycle, grey);
filldraw((1,7)--(2,7)--(2,9)--(1,9)--cycle, grey);
filldraw((2,8)--(3,8)--(3,10)--(2,10)--cycle, grey);
filldraw((3,9)--(4,9)--(4,11)--(3,11)--cycle, grey);
filldraw((4,10)--(6,10)--(6,11)--(4,11)--cycle,grey);
filldraw((7,0)--(8,0)--(8,2)--(7,2)--cycle, grey);
filldraw((8,1)--(9,1)--(9,3)--(8,3)--cycle, grey);
filldraw((9,2)--(10,2)--(10,4)--(9,4)--cycle, grey);
filldraw((10,3)--(11,3)--(11,5)--(10,5)--cycle, grey);
filldraw((5,0)--(7,0)--(7,1)--(5,1)--cycle, grey);
filldraw((0,4)--(1,4)--(1,6)--(0,6)--cycle, grey);
filldraw((3,0)--(5,0)--(5,1)--(3,1)--cycle, grey);
filldraw((2,1)--(4,1)--(4,2)--(2,2)--cycle, grey);
filldraw((1,2)--(3,2)--(3,3)--(1,3)--cycle, grey);
filldraw((0,3)--(2,3)--(2,4)--(0,4)--cycle, grey);
filldraw((9,7)--(11,7)--(11,8)--(9,8)--cycle, grey);
filldraw((8,8)--(10,8)--(10,9)--(8,9)--cycle, grey);
filldraw((7,9)--(9,9)--(9,10)--(7,10)--cycle, grey);
filldraw((6,10)--(8,10)--(8,11)--(6,11)--cycle, grey);
filldraw((10,5)--(11,5)--(11,7)--(10,7)--cycle, grey);
draw((6,7)--(6,11),linewidth(1.5));
draw((7,5)--(11,5),linewidth(1.5));
draw((5,4)--(5,0),linewidth(1.5));
draw((4,6)--(0,6),linewidth(1.5));
draw((4,4)--(4,7)--(7,7)--(7,4)--cycle, linewidth(1.5));
[/asy]](http://latex.artofproblemsolving.com/7/d/3/7d32ba4a9a997d10b6b3d4edb3f2ea71bb082209.png)
be the greatest power of two that divides 
.
is odd; if 
,
grid of 
,
grid of 
sub-grids. As we have shown, it is possible to tile a 
tiles, we have 
.
when 
otherwise.
,
,
![\[
\begin{cases}
\frac{2n}{3} &\text{for } n\equiv 0\pmod{3} \\
2n &\text{for } n\not\equiv 0\pmod{3}
\end{cases}
\]](http://latex.artofproblemsolving.com/a/b/9/ab99303b9dcbd234effb94f4b5bf6c2e2a01129a.png)
![[asy]
size(5cm);
int n = 3;
real d = 0.1;
path rect(int a, int b, int x, int y)
{
return (a+d,b+d)--(x-d,b+d)--(x-d,y-d)--(a+d,y-d)--cycle;
}
for (int i = 0; i <= n; ++i)
{
draw((i,0)--(i,n));
draw((0,i)--(n,i));
}
fill(rect(0, 0, 1, 2), black);
fill(rect(1, 2, 3, 3), black);
[/asy]](http://latex.artofproblemsolving.com/3/8/6/386f7b1f2573e85bbfe3669f952302273990f8b1.png)
![[asy]
size(5cm);
int n = 6;
real d = 0.1;
path rect(int a, int b, int x, int y)
{
return (a+d,b+d)--(x-d,b+d)--(x-d,y-d)--(a+d,y-d)--cycle;
}
for (int i = 0; i <= n; ++i)
{
draw((i,0)--(i,n));
draw((0,i)--(n,i));
}
fill(rect(0, 3, 1, 5), black);
fill(rect(1, 5, 3, 6), black);
fill(rect(3, 0, 4, 2), black);
fill(rect(4, 2, 6, 3), black);
[/asy]](http://latex.artofproblemsolving.com/1/a/0/1a03a2665b535f9d79eeef8cc139351a69a15c6d.png)
![[asy]
size(5cm);
int n = 4;
real d = 0.1;
path rect(int a, int b, int x, int y)
{
return (a+d,b+d)--(x-d,b+d)--(x-d,y-d)--(a+d,y-d)--cycle;
}
for (int i = 0; i <= n; ++i)
{
draw((i,0)--(i,n));
draw((0,i)--(n,i));
}
fill(rect(0, 2, 1, 4), black);
fill(rect(1, 2, 2, 4), black);
fill(rect(2, 3, 4, 4), black);
fill(rect(2, 2, 4, 3), black);
fill(rect(0, 0, 2, 1), black);
fill(rect(0, 1, 2, 2), black);
fill(rect(2, 0, 3, 2), black);
fill(rect(3, 0, 4, 2), black);
[/asy]](http://latex.artofproblemsolving.com/c/3/d/c3d25cd465a78bccd3d920e37db8959c502db4ab.png)
.
.
.![[asy]
size(5cm);
int n = 5;
real d = 0.1;
path rect(int a, int b, int x, int y)
{
return (a+d,b+d)--(x-d,b+d)--(x-d,y-d)--(a+d,y-d)--cycle;
}
for (int i = 0; i <= n; ++i)
{
draw((i,0)--(i,n));
draw((0,i)--(n,i));
}
fill(rect(0, 0, 1, 2), black);
fill(rect(0, 2, 2, 3), black);
fill(rect(0, 3, 1, 5), black);
fill(rect(1, 4, 3, 5), black);
fill(rect(1, 0, 3, 1), black);
fill(rect(2, 2, 3, 4), black);
fill(rect(3, 1, 4, 3), black);
fill(rect(3, 3, 5, 4), black);
fill(rect(3, 4, 5, 5), black);
fill(rect(4, 0, 5, 2), black);
[/asy]](http://latex.artofproblemsolving.com/a/2/d/a2dbef890014dc4f39edc4343599e211a63ae783.png)
![[asy]
size(5cm);
int n = 7;
real d = 0.1;
path rect(int a, int b, int x, int y)
{
return (a+d,b+d)--(x-d,b+d)--(x-d,y-d)--(a+d,y-d)--cycle;
}
for (int i = 0; i <= n; ++i)
{
draw((i,0)--(i,n));
draw((0,i)--(n,i));
}
fill(rect(0, 0, 2, 1), black);
fill(rect(0, 1, 2, 2), black);
fill(rect(0, 6, 2, 7), black);
fill(rect(2, 0, 4, 1), black);
fill(rect(2, 2, 3, 4), black);
fill(rect(2, 4, 3, 6), black);
fill(rect(4, 0, 6, 1), black);
fill(rect(5, 1, 6, 3), black);
fill(rect(5, 5, 6, 7), black);
fill(rect(6, 1, 7, 3), black);
fill(rect(6, 3, 7, 5), black);
fill(rect(6, 5, 7, 7), black);
fill(rect(3, 3, 5, 4), black);
fill(rect(3, 4, 5, 5), black);
[/asy]](http://latex.artofproblemsolving.com/9/0/8/908158fc434cb754daf1f6461b7d1d6c507449d1.png)
be the prime that we are trying to reach a construction for and let two numbers 
and 
such that 
and 
.
it is easy to see that such 
works. Then we can string together the 
board and the 
board along the diagonal and we would have a valid configuration. 
is composite, then we can pick some 
such that 
and string together 
boards along the diagonal. 
.
it is easy to see that 
then notice that the minimum 
is wrong, clearly the value of some rows is 4 and some others 3
dominoes if 
dominoes on the board; then, double counting pairs 
where 
is a row or column and 
For this to be a positive integer, we clearly have the above constraints on 
grid and bottom-right 
grid of the 
.
dominoes in the top-left and 
dominoes in the top right, we can guarantee a consistent 
always exist due to our base cases.
and 
and 
.
for 
there are 
dominoes.
the smallest 
giving a lower bound of 
the smallest 
giving a lower bound of 
in this case.
Observe the below obvious configuration for 
![[asy]
import olympiad;
unitsize(20);
real d = 0.1;
path r(int a, int b, int x, int y)
{
return (a-1+d,b-1+d)--(x-d,b-1+d)--(x-d,y-d)--(a-1+d,y-d)--cycle;
}
for (int i = 0; i <= 3; ++i)
{
draw((0,i)--(3,i));
draw((i,0)--(i,3));
}
fill(r(1,2,1,3));
fill(r(2,1,3,1));
[/asy]](http://latex.artofproblemsolving.com/3/b/9/3b9eff5ad3def4115893d4dcef365a52f0912d23.png)
dominoes. Extend the 
board into a 
board by constructing 
This gives a working construction using 
dominoes for 
completing the induction.
Observe the below configurations for 
and 
![[asy]
import olympiad;
unitsize(18);
real d = 0.1;
path r(int a, int b, int x, int y)
{
return (a-1+d,b-1+d)--(x-d,b-1+d)--(x-d,y-d)--(a-1+d,y-d)--cycle;
}
for (int i = 0; i <= 4; ++i)
{
draw((0,i)--(4,i));
draw((i,0)--(i,4));
}
fill(r(1,4,2,4));
fill(r(1,3,2,3));
fill(r(1,1,1,2));
fill(r(2,1,2,2));
fill(r(3,3,3,4));
fill(r(4,3,4,4));
fill(r(3,2,4,2));
fill(r(3,1,4,1));
[/asy]](http://latex.artofproblemsolving.com/9/9/5/9957919daf26fa40915e1e2a2ee8b8bdda8a25c5.png)
![[asy]
import olympiad;
unitsize(18);
real d = 0.1;
path r(int a, int b, int x, int y)
{
return (a-1+d,b-1+d)--(x-d,b-1+d)--(x-d,y-d)--(a-1+d,y-d)--cycle;
}
for (int i = 0; i <= 5; ++i)
{
draw((0,i)--(5,i));
draw((i,0)--(i,5));
}
fill(r(1,1,1,2));
fill(r(1,3,2,3));
fill(r(3,2,3,3));
fill(r(2,1,3,1));
fill(r(1,4,1,5));
fill(r(2,4,2,5));
fill(r(3,4,3,5));
fill(r(4,1,5,1));
fill(r(4,2,5,2));
fill(r(4,3,5,3));
[/asy]](http://latex.artofproblemsolving.com/d/a/a/daadfae6c2d5923856fdeff2832a5f16a1aa4f89.png)
![[asy]
import olympiad;
unitsize(18);
real d = 0.1;
path r(int a, int b, int x, int y)
{
return (a-1+d,b-1+d)--(x-d,b-1+d)--(x-d,y-d)--(a-1+d,y-d)--cycle;
}
for (int i = 0; i <= 7; ++i)
{
draw((0,i)--(7,i));
draw((i,0)--(i,7));
}
fill(r(1,1,2,1));
fill(r(1,2,1,3));
fill(r(1,4,1,5));
fill(r(3,1,4,1));
fill(r(5,1,5,2));
fill(r(5,3,5,4));
fill(r(4,5,5,5));
fill(r(2,5,3,5));
fill(r(2,6,2,7));
fill(r(3,6,3,7));
fill(r(4,6,4,7));
fill(r(6,2,7,2));
fill(r(6,3,7,3));
fill(r(6,4,7,4));
[/asy]](http://latex.artofproblemsolving.com/0/4/c/04c4eaecc59278e1b14a22579a9991921b461d81.png)
Observe that given two constructions of 
and 
with 
and 
respectively, appending them in a diagonal block fashion creates a valid construction for 
with 
By the Chicken McNugget theorem, the largest number that cannot be expressed as a linear combination of 
It remains to check valid configurations with 
But observe 
completing the proof.
that satisfies that. First, we look at the 
for some positive integer 
grid such that no two cells of distinct 
,
such that 
,
and 
.
rows, respectively, and each row is covered 
,
.
or 
.
is always an integer, so the minimal 
)
is attainable by placing 
:
.
Since each domino is in two rows and one column or two columns and one row, if there are 
is an integer. For 
we can have one domino in each row/column, with the following construction:
For 
we have 3 dominoes in each row/column. After finding the construction 
we can arrange those arrangements once along the diagonal with a repeated arrangement of
![\[(2n) \cdot k = 3 \cdot d \implies k = \frac{3d}{2n}.\]](http://latex.artofproblemsolving.com/e/8/9/e8947be4aa8678b3f369c219358a396269ca98f7.png)
![\[\boxed{\begin{cases} k=1, \quad d=\frac{2n}{3} & \text{ if } 3 \mid n \\ k=3, \quad d=2n & \text{ if } 3 \nmid n \end{cases}}\]](http://latex.artofproblemsolving.com/7/2/b/72b3181d86fc6d35fd411479aff8fedc13e138ec.png)
.
and 
while leaving the rest of the board blank.
:
,
:
,
:
,
![[asy]
size(100);
draw((0,0)--(1,0)--(1,1)--(0,1)--cycle);
draw((.3,0)--(.3,1)^^(0,.7)--(1,.7));
label("$A$", (.15,.85), 0*dir(0));
label("$B$", (.65,.35), 0*dir(0));
[/asy]](http://latex.artofproblemsolving.com/3/f/f/3ffb105763f083f9f860d24e2121c407357121a2.png)
.
.
,
= 
.
,
and 
,
.
.
.
,
.
,
.![[asy]
size(3cm); // Default size you prefer
void drawGrid(int n) {
for (int i = 0; i <= n; ++i) {
draw((0, i) -- (n, i), gray); // Horizontal lines
draw((i, 0) -- (i, n), gray); // Vertical lines
}
}
void highlightDomino(pair p, bool vertical=false) {
path domino;
if (vertical)
domino = p -- (p + (0,2)) -- (p + (1,2)) -- (p + (1,0)) -- cycle;
else
domino = p -- (p + (2,0)) -- (p + (2,1)) -- (p + (0,1)) -- cycle;
fill(domino, lightblue);
draw(domino, blue);
}
// Draw the grid
drawGrid(6);
// Highlight dominos forming a stair-step pattern
highlightDomino((0,5));
highlightDomino((3,2));
highlightDomino((2,3), true);
highlightDomino((5,0), true);
[/asy]](http://latex.artofproblemsolving.com/2/0/3/203def42eb9a2fc3135f7b0b1b48a2ea08c01b28.png)
is an integer iff 
.
.
— from here, we can recursively generate boards with 
board with two 
board with two 
board with a 
board. The Chicken McNugget Theorem tells us that every integer greater than 
can be represented as 
for some nonnegative integers 
,
dominoes if we count by row and column across all rows and columns. If 
,
(which exists by just combining two 
.
