Found problems: 248
2011 USAMO, 4
Consider the assertion that for each positive integer $n\geq2$, the remainder upon dividing $2^{2^n}$ by $2^n-1$ is a power of $4$. Either prove the assertion or find (with proof) a counterexample.
2011 USAMTS Problems, 2
Find all integers $a$, $b$, $c$, $d$, and $e$ such that
\begin{align*}a^2&=a+b-2c+2d+e-8,\\b^2&=-a-2b-c+2d+2e-6,\\c^2&=3a+2b+c+2d+2e-31,\\d^2&=2a+b+c+2d+2e-2,\\e^2&=a+2b+3c+2d+e-8.\end{align*}
2012 USAMTS Problems, 5
An ordered quadruple $(y_1,y_2,y_3,y_4)$ is $\textbf{quadratic}$ if there exist real numbers $a$, $b$, and $c$ such that \[y_n=an^2+bn+c\] for $n=1,2,3,4$.
Prove that if $16$ numbers are placed in a $4\times 4$ grid such that all four rows are quadratic and the first three columns are also quadratic then the fourth column must also be quadratic.
[i](We say that a row is quadratic if its entries, in order, are quadratic. We say the same for a column.)[/i]
[asy]
size(100);
defaultpen(linewidth(0.8));
for(int i=0;i<=4;i=i+1)
draw((i,0)--(i,4));
for(int i=0;i<=4;i=i+1)
draw((0,i)--(4,i));
[/asy]
2014 USAMTS Problems, 3a:
A group of people is lined up in [i]almost-order[/i] if, whenever person $A$ is to the left of person $B$ in the line, $A$ is not more than $8$ centimeters taller than $B$. For example, five people with heights $160, 165, 170, 175$, and $180$ centimeters could line up in almost-order with heights (from left-to-right) of $160, 170, 165, 180, 175$ centimeters.
(a) How many different ways are there to line up $10$ people in [i]almost-order[/i] if their heights are $140, 145, 150, 155,$ $160,$ $165,$ $170,$ $175,$ $180$, and $185$ centimeters?
2020 USAMTS Problems, 5:
Let $n \geq 3$ be an integer. Let $f$ be a function from the set of all integers to itself with the following property: If the integers $a_1,a_2,\ldots,a_n$ form an arithmetic progression, then the numbers
$$f(a_1),f(a_2),\ldots,f(a_n)$$
form an arithmetic progression (possibly constant) in some order. Find all values for $n$ such that the only functions $f$ with this property are the functions of the form $f(x)=cx+d$, where $c$ and $d$ are integers.
2023 USAMTS Problems, 1
In the diagram below, fill the $12$ circles with numbers from the following bank so that each number is used once. Two circles connected by a single line must contain relatively prime numbers. Two circles connected by a double line must contain numbers that are not relatively prime.
$$\text{Bank: } 20, 21, 22, 23, 24, 25, 27, 28, 30 ,32, 33 ,35$$
[asy]
real HRT3 = sqrt(3) / 2;
void drawCircle(real x, real y, real r) {
path p = circle((x,y), r);
draw(p);
fill(p, white);
}
void drawCell(int gx, int gy) {
real x = 0.5 * gx;
real y = HRT3 * gy;
drawCircle(x, y, 0.35);
}
void drawEdge(int gx1, int gy1, int gx2, int gy2, bool doubled) {
real x1 = 0.5 * gx1;
real y1 = HRT3 * gy1;
real x2 = 0.5 * gx2;
real y2 = HRT3 * gy2;
if (doubled) {
real dx = x2 - x1;
real dy = y2 - y1;
real ox = -0.035 * dy / sqrt(dx * dx + dy * dy);
real oy = 0.035 * dx / sqrt(dx * dx + dy * dy);
draw((x1+ox,y1+oy)--(x2+ox,y2+oy));
draw((x1-ox,y1-oy)--(x2-ox,y2-oy));
} else {
draw((x1,y1)--(x2,y2));
}
}
drawEdge(2, 0, 4, 0, true);
drawEdge(2, 0, 1, 1, true);
drawEdge(2, 0, 3, 1, true);
drawEdge(4, 0, 3, 1, false);
drawEdge(4, 0, 5, 1, false);
drawEdge(1, 1, 0, 2, false);
drawEdge(1, 1, 2, 2, false);
drawEdge(1, 1, 3, 1, false);
drawEdge(3, 1, 2, 2, true);
drawEdge(3, 1, 4, 2, true);
drawEdge(3, 1, 5, 1, false);
drawEdge(5, 1, 4, 2, true);
drawEdge(5, 1, 6, 2, false);
drawEdge(0, 2, 1, 3, false);
drawEdge(0, 2, 2, 2, false);
drawEdge(2, 2, 1, 3, false);
drawEdge(2, 2, 3, 3, true);
drawEdge(2, 2, 4, 2, false);
drawEdge(4, 2, 3, 3, false);
drawEdge(4, 2, 5, 3, false);
drawEdge(4, 2, 6, 2, false);
drawEdge(6, 2, 5, 3, true);
drawEdge(1, 3, 3, 3, true);
drawEdge(3, 3, 5, 3, false);
drawCell(2, 0);
drawCell(4, 0);
drawCell(1, 1);
drawCell(3, 1);
drawCell(5, 1);
drawCell(0, 2);
drawCell(2, 2);
drawCell(4, 2);
drawCell(6, 2);
drawCell(1, 3);
drawCell(3, 3);
drawCell(5, 3);
[/asy]
2009 USAMTS Problems, 4
Let $ABCDEF$ be a convex hexagon, such that $FA = AB$, $BC = CD$, $DE = EF$, and $\angle FAB = 2\angle EAC$. Suppose that the area of $ABC$ is $25$, the area of $CDE$ is $10$, the area of $EF A$ is $25$, and the area of $ACE$ is $x$. Find, with proof, all possible values of $x$.
2013 USAMTS Problems, 3
Let $A_1A_2A_3\dots A_{20}$ be a $20$-sided polygon $P$ in the plane, where all of the side lengths of $P$ are equal, the interior angle at $A_i$ measures $108$ degrees for all odd $i$, and the interior angle $A_i$ measures $216$ degrees for all even $i$. Prove that the lines $A_2A_8$, $A_4A_{10}$, $A_5A_{13}$, $A_6A_{16}$, and $A_7A_{19}$ all intersect at the same point.
[asy]
import graph;
size(10cm);
pair temp= (-1,0);
pair A01 = (0,0);
pair A02 = rotate(306,A01)*temp;
pair A03 = rotate(144,A02)*A01;
pair A04 = rotate(252,A03)*A02;
pair A05 = rotate(144,A04)*A03;
pair A06 = rotate(252,A05)*A04;
pair A07 = rotate(144,A06)*A05;
pair A08 = rotate(252,A07)*A06;
pair A09 = rotate(144,A08)*A07;
pair A10 = rotate(252,A09)*A08;
pair A11 = rotate(144,A10)*A09;
pair A12 = rotate(252,A11)*A10;
pair A13 = rotate(144,A12)*A11;
pair A14 = rotate(252,A13)*A12;
pair A15 = rotate(144,A14)*A13;
pair A16 = rotate(252,A15)*A14;
pair A17 = rotate(144,A16)*A15;
pair A18 = rotate(252,A17)*A16;
pair A19 = rotate(144,A18)*A17;
pair A20 = rotate(252,A19)*A18;
dot(A01);
dot(A02);
dot(A03);
dot(A04);
dot(A05);
dot(A06);
dot(A07);
dot(A08);
dot(A09);
dot(A10);
dot(A11);
dot(A12);
dot(A13);
dot(A14);
dot(A15);
dot(A16);
dot(A17);
dot(A18);
dot(A19);
dot(A20);
draw(A01--A02--A03--A04--A05--A06--A07--A08--A09--A10--A11--A12--A13--A14--A15--A16--A17--A18--A19--A20--cycle);
label("$A_{1}$",A01,E);
label("$A_{2}$",A02,W);
label("$A_{3}$",A03,NE);
label("$A_{4}$",A04,SW);
label("$A_{5}$",A05,N);
label("$A_{6}$",A06,S);
label("$A_{7}$",A07,N);
label("$A_{8}$",A08,SE);
label("$A_{9}$",A09,NW);
label("$A_{10}$",A10,E);
label("$A_{11}$",A11,W);
label("$A_{12}$",A12,E);
label("$A_{13}$",A13,SW);
label("$A_{14}$",A14,NE);
label("$A_{15}$",A15,S);
label("$A_{16}$",A16,N);
label("$A_{17}$",A17,S);
label("$A_{18}$",A18,NW);
label("$A_{19}$",A19,SE);
label("$A_{20}$",A20,W);[/asy]
1998 USAMTS Problems, 4
Show that it is possible to arrange seven distinct points in the plane so that among any three of these seven points, two of the points are a unit distance apart. (Your solution should include a carefully prepared sketch of the seven points, along with all segments that are of unit length.)
2005 USAMTS Problems, 2
George has six ropes. He chooses two of the twelve loose ends at random (possibly
from the same rope), and ties them together, leaving ten loose ends. He again chooses two loose ends at random and joins them, and so on, until there are no loose ends. Find, with proof, the expected value of the number of loops George ends up with.
2019 USAMTS Problems, 4
Princess Pear has $100$ jesters with heights $1, 2, \dots, 100$ inches. On day $n$ with $1 \leq n \leq 100$, Princess Pear holds a court with all her jesters with height at most $n$ inches, and she receives two candied cherries from every group of $6$ jesters with a median height of $n - 50$ inches. A jester can be part of multiple groups.
On day $101$, Princess Pear summons all $100$ jesters to court one final time. Every group of $6$ jesters with a median height of 50.5 inches presents one more candied cherry to the Princess. How many candied cherries does Princess Pear receive in total?
Please provide a numerical answer (with justification).
2016 USAMTS Problems, 1:
Shade in some of the regions in the grid to the right so that the shaded area is equal for each of the 11 rows and columns. Regions must be fully shaded or fully unshaded, at least one region must be shaded, and the area of shaded regions must be at most half of the whole grid.
[asy]
size(200);
defaultpen(linewidth(0.45));
real[][] arr = {
{0,0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,0,0,0},};
for (int i=0; i<11; ++i){
for (int j=0; j<11; ++j){
if(arr[10-j][i] == 1){
fill((i,j)--(i+1,j)--(i+1, j+1)--(i,j+1)--cycle, grey);
}
}
}
draw((0,0)--(4,0)--(4,3)--(3,3)--(3,1)--(1,1)--(1,2)--(2,2)--(2,1)--(2,3)--(3,3)--(1,3)--(1,2)--(1,3)--(0,3)--(0,0)--(0,5)--(1,5)--(1,4)--(4,4)--(4,3)--(6,3)--(6,2)--(5,2)--(5,1)--(4,1)--(4,0)--(6,0)--(6,3)--(9,3)--(9,0)--(8,0)--(8,2)--(7,2)--(7,1)--(8,1)--(7,1)--(7,0)--(6,0)--(11,0)--(11,2)--(10,2)--(10,1)--(9,1)--(9,3)--(11,3)--(11,2)--(11,4)--(8,4)--(8,3)--(8,4)--(6,4)--(6,3)--(6,4)--(3,4)--(3,5)--(2,5)--(2,6)--(1,6)--(1,5)--(1,7)--(0,7)--(0,5)--(0,8)--(2,8)--(2,6)--(3,6)--(3,7)--(4,7)--(4,6)--(3,6)--(4,6)--(4,5)--(3,5)--(5,5)--(5,4)--(5,5)--(7,5)--(7,4)--(7,5)--(9,5)--(9,4)--(9,5)--(11,5)--(11,4)--(11,6)--(10,6)--(10,7)--(9,7)--(9,5)--(9,7)--(8,7)--(8,5)--(8,6)--(7,6)--(7,7)--(8,7)--(6,7)--(6,6)--(7,6)--(6,6)--(6,5)--(5,5)--(5,6)--(4,6)--(4,8)--(2,8)--(2,9)--(0,9)--(0,8)--(0,10)--(1,10)--(1,9)--(1,10)--(2,10)--(2,9)--(3,9)--(3,8)--(3,11)--(0,11)--(0,10)--(0,11)--(5,11)--(5,10)--(3,10)--(4,10)--(4,8)--(5,8)--(5,6)--(5,7)--(6,7)--(6,8)--(5,8)--(5,9)--(4,9)--(6,9)--(6,10)--(5,10)--(5,11)--(8,11)--(8,10)--(6,10)--(7,10)--(7,9)--(6,9)--(7,9)--(7,8)--(6,8)--(7,8)--(7,7)--(8,7)--(8,8)--(7,8)--(8,8)--(8,10)--(9,10)--(9,9)--(8,9)--(10,9)--(10,8)--(9,8)--(9,7)--(10,7)--(10,6)--(11,6)--(11,8)--(10,8)--(11,8)--(11,10)--(9,10)--(11,10)--(11,11)--(8,11));
[/asy]
You do not need to prove that your answer is the only one possible; you merely need to find an answer that satisfies the constraints above. (Note: In any other USAMTS problem, you need to provide a full proof. Only in this problem is an answer without justification acceptable.)
2019 USAMTS Problems, 3
Circle $\omega$ is inscribed in unit square $PLUM$, and points $I$ and $E$ lie on $\omega$ such that $U,I,$ and $E$ are collinear. Find, with proof, the greatest possible area for $\triangle PIE$.
2015 USAMTS Problems, 4
Let $\triangle ABC$ be a triangle with $AB<AC$. Let the angle bisector of $\angle BAC$ meet $BC$ at $D$, and let $M$ be the midpoint of $\overline{BC}$. Let $P$ be the foot of the perpendicular from $B$ to $\overline{AD}$. Extend $\overline{BP}$ to meet $\overline{AM}$ at $Q$. Show that $\overline{DQ}$ is parallel to $\overline{AB}$.
2019 USAMTS Problems, 1
Fill in each square with a number from $1$ to $5$; some numbers have been given. If two squares $A$ and $B$ have equal numbers, then $A$ and $B$ cannot share a side, and there also cannot exist a third square $C$ sharing a side with both $A$ and $B$.
There is a unique solution, but you do not need to prove that your answer is the only one possible. You merely need to find an answer that satisfies the constraints above. (Note: In any other USAMTS problem, you need to provide a full proof. Only in this problem is an answer without justification acceptable.)
[asy]unitsize(30);
int a[][] = {
{9, 9, 0, 0, 0, 9, 0, 0, 0},
{4, 0, 0, 2, 0, 0, 0, 2, 0},
{0, 1, 0, 0, 0, 2, 0, 0, 5},
{0, 0, 0, 9, 0, 0, 0, 9, 9}
};
for (int i = 0; i < a.length; ++i) {
for (int j = 0; j < a[0].length; ++j) {
if (a[i][j] < 9)
draw(shift(j, -i-1) * unitsquare);
label((a[i][j] >= 1 && a[i][j] <= 5) ? string(a[i][j]) : "", (j+.5, -i-.5), fontsize(24pt));
}
}[/asy]
2011 USAMTS Problems, 5
Miss Levans has 169 students in her history class and wants to seat them all in a $13\times13$ grid of desks. Each desk is placed at a different vertex of a 12 meter by 12 meter square grid of points she has marked on the floor. The distance between neighboring vertices is exactly 1 meter. Each student has at most three best friends in the class. Best-friendship is mutual: if Lisa is one of Shannon's best friends, then Shannon is also one of Lisa's best friends. Miss Levans knows that if any two best friends sit at points that are 3 meters or less from each other then they will be disruptive and nobody will learn any history. And that is bad. Prove that Miss Levans can indeed place all $169$ students in her class without any such disruptive pairs.
2023 USAMTS Problems, 1
Fill in the grid with the numbers 1 to 6 so that each number appears exactly once in
each row and column. A horizontal gray line marks any cell when it is the middle cell of
the three consecutive cells with the largest sum in that row. Similarly, a vertical gray line
marks any cell when it is the middle of the three consecutive cells with the largest sum in
that column. If there is a tie, multiple lines are drawn in the row or column. A cell can have
both lines drawn, with the appearance of a plus sign.
[asy]
// Change this to see the solution
bool DRAW_SOLUTION = true;
int n = 6;
real LINE_WIDTH = 0.3;
void drawHLine(int x, int y) {
fill((x,y+0.5-LINE_WIDTH/2)--(x,y+0.5+LINE_WIDTH/2)--(x+1,y+0.5+LINE_WIDTH/2)--(x+1,y+0.5-LINE_WIDTH/2)--cycle, gray(0.8));
}
void drawVLine(int x, int y) {
fill((x+0.5-LINE_WIDTH/2,y)--(x+0.5+LINE_WIDTH/2,y)--(x+0.5+LINE_WIDTH/2,y+1)--(x+0.5-LINE_WIDTH/2,y+1)--cycle, gray(0.8));
}
void drawNum(int x, int y, int num) {
label(scale(1.5)*string(num), (x+0.5,y+0.5));
}
void drawSolNum(int x, int y, int num) {
if (DRAW_SOLUTION) {
drawNum(x, y, num);
}
}
drawHLine(2,0);
drawHLine(4,1);
drawHLine(1,2);
drawHLine(3,2);
drawHLine(4,3);
drawHLine(2,4);
drawHLine(3,5);
drawVLine(0,4);
drawVLine(1,3);
drawVLine(2,1);
drawVLine(2,3);
drawVLine(3,4);
drawVLine(4,1);
drawVLine(5,2);
drawNum(0, 0, 5);
drawNum(4, 0, 3);
drawNum(1, 2, 2);
drawNum(3, 3, 4);
for(int i = 0; i <= 6; i += 1) {
draw((i,0)--(i,6));
draw((0,i)--(6,i));
}
[/asy]
2005 USAMTS Problems, 2
[i]Centered hexagonal numbers[/i] are the numbers of dots used to create hexagonal arrays of dots. The first four centered hexagonal numbers are 1, 7, 19, and 37 as shown below:
[asy]
size(250);defaultpen(linewidth(0.4));
dot(origin^^shift(-12,0)*origin^^shift(-24,0)*origin^^shift(-36,0)*origin);
int i;
for(i=0; i<360; i=i+60) {
dot(1*dir(i)^^2*dir(i)^^3*dir(i));
dot(shift(1/2, sqrt(3)/2)*1*dir(i)^^shift(1/2, sqrt(3)/2)*2*dir(i));
dot(shift(1, sqrt(3))*1*dir(i));
dot(shift(-12,0)*origin+1*dir(i)^^shift(-12,0)*origin+2*dir(i));
dot(shift(-12,0)*origin+sqrt(3)*dir(i+30));
dot(shift(-24,0)*origin+1*dir(i));
}
label("$1$", (-36, -5), S);
label("$7$", (-24, -5), S);
label("$19$", (-12, -5), S);
label("$37$", (0, -5), S);
label("Centered Hexagonal Numbers", (-18,-10), S);[/asy]
Consider an arithmetic sequence 1, $a$, $b$ and a geometric sequence 1,$c$,$d$, where $a$,$b$,$c$, and $d$ are all positive integers and $a+b=c+d$. Prove that each centered hexagonal number is a possible value of $a$, and prove that each possible value of $a$ is a centered hexagonal number.
2002 USAMTS Problems, 3
Find the real-numbered solution to the equation below and demonstrate that it is unique.
\[\dfrac{36}{\sqrt{x}}+\dfrac{9}{\sqrt{y}}=42-9\sqrt{x}-\sqrt{y}\]
2016 USAMTS Problems, 1:
Another round, another diagram...
[asy]
unitsize(1cm);
defaultpen(linewidth(0.45));
real[][] arr = {
{0,0,0,0},
{0,0,0,0},
{0,0,0,0},
{0,0,0,0}};
for (int i=0; i<4; ++i){
for (int j=0; j<4; ++j){
if(arr[3-j][i] != 0){
label((string) arr[3-j][i], (i+0.5, j+0.5));
}
}
}
label("$+$", (-0.5, 4.5), dir(-45));
label("$-$", (4.5, -0.5), dir(135));
label("\Large 13", (-0.5, 3.5));
label("\Large 28", (-0.5, 2.5));
label("\Large 23", (-0.5, 0.5));
label("\Large 7", (4.5, 3.5));
label("\Large 8", (4.5, 1.5));
label("\Large 8", (4.5, 0.5));
label("\Large 12", (3.5, -0.5));
label("\Large 12", (2.5,-0.5));
label("\Large 7", (0.5, -0.5));
label("\Large 23", (3.5, 4.5 ));
label("\Large 25", (2.5,4.5));
label("\Large 28", (1.5,4.5));
label("\Large 13", (0.5,4.5));
for(int i = 1; i <= 3; ++i){
draw((i, 0)--(i, 4));
draw((0, i)--(4, i));
}
draw((0,0)--(0,4)--(4,4)--(4,0)--cycle, linewidth(1.5));
draw((-0.8,-0.8)--(0,0), linewidth(1.5));
draw((4,4 )--(4.8,4.8), linewidth(1.5));
[/asy]
Use [code]\begin{asy}
\end{asy}[/code]environment to render the diagram correctly in a latex document. Remember to write [code]\usepackage{asymptote}[/code] in the preamble.
And of course, replace the 0's in the array at the beginning of the code with the numbers you wish to fill it in with.
2011 USAMTS Problems, 4
Renata the robot packs boxes in a warehouse. Each box is a cube of side length $1$ foot. The warehouse floor is a square, $12$ feet on each side, and is divided into a $12$-by-$12$ grid of square tiles $1$ foot on a side. Each tile can either support one box or be empty. The warehouse has exactly one door, which opens onto one of the corner tiles.
Renata fits on a tile and can roll between tiles that share a side. To access a box, Renata must be able to roll along a path of empty tiles starting at the door and ending at a tile sharing a side with that box.
[list=a]
[*]Show how Renata can pack $91$ boxes into the warehouse and still be able to access any box.
[*]Show that Renata [b]cannot[/b] pack $95$ boxes into the warehouse and still be able to access any box.[/list]
2011 USAJMO, 6
Consider the assertion that for each positive integer $n\geq2$, the remainder upon dividing $2^{2^n}$ by $2^n-1$ is a power of $4$. Either prove the assertion or find (with proof) a counterexample.
2011 USAMTS Problems, 1
The grid on the right has $12$ boxes and $15$ edges connecting boxes. In each box, place one of the six integers from $1$ to $6$ such that the following conditions hold:
[list]
[*]For each possible pair of distinct numbers from $1$ to $6$, there is exactly one edge connecting two boxes with that pair of numbers.
[*]If an edge has an arrow, then it points from a box with a smaller number to a box with a larger number.[/list]
You do not need to prove that your con guration is the only one possible; you merely need to fi nd a con guration that satis es the constraints above. (Note: In any other USAMTS problem, you need to provide a full proof. Only in this problem is an answer without justi cation acceptable.)
[asy]
size(200);
defaultpen(linewidth(0.8));
int i,j;
for(i=0;i<4;i=i+1) {
for(j=0;j<3;j=j+1) {
draw((i,j)--(i,j+1/2)--(i+1/2,j+1/2)--(i+1/2,j)--cycle);
}
}
draw((1/2,1/4)--(1,1/4)^^(1/2,5/4)--(1,5/4)^^(3/2,5/4)--(2,5/4)^^(5/2,5/4)--(3,5/4)^^(5/2,9/4)--(3,9/4));
draw((1/4,1)--(1/4,1/2),Arrow(5));
draw((5/4,1)--(5/4,1/2),Arrow(5));
draw((1/4,2)--(1/4,3/2),Arrow(5));
draw((9/4,2)--(9/4,3/2),Arrow(5));
draw((13/4,2)--(13/4,3/2),Arrow(5));
draw((13/4,1)--(13/4,1/2),Arrow(5));
draw((2,1/4)--(3/2,1/4),Arrow(5));
draw((1,9/4)--(1/2,9/4),Arrow(5));
draw((5/2,1/4)--(3,1/4),Arrow(5));
draw((3/2,9/4)--(2,9/4),Arrow(5));
[/asy]
2011 USAMTS Problems, 1
Fill in the circles to the right with the numbers 1 through 16 so that each number is used once (the number 1 has been fi lled in already). The number in any non-circular region is equal to the greatest di fference between any two numbers in the circles on that region's vertices.
You do not need to prove that your con figuration is the only one possible; you merely need to find a valid con guration. (Note: In any other USAMTS problem, you need to provide a full proof. Only in this problem is an answer without justi cation acceptable.)
[asy]
size(190);
defaultpen(linewidth(0.8));
int i,j;
path p;
for(i=0;i<=3;++i){
draw((i,0)--(i,3));
draw((0,i)--(3,i));
}
draw((0,3)--(1,2)^^(0,1)--(2,3)^^(1,0)--(3,2)^^(3,0)--(2,1));
for(i=0;i<=3;++i){
for(j=0;j<=3;++j){
p=circle((i,j),1/4);
unfill(p);
draw(p);
}
}
label("$1$",(0,3));
label("$7$",(1/3,2+1/3));
label("$8$",(2/3,2+2/3));
label("$2$",(1+1/3,2+2/3));
label("$2$",(1/3,1+2/3));
label("$2$",(2+2/3,1+1/3));
label("$8$",(1+2/3,1/3));
label("$5$",(2+1/3,1/3));
label("$4$",(2+2/3,2/3));
label("$4$",(1/2,1/2));
label("$10$",(3/2,3/2));
label("$11$",(5/2,5/2));
[/asy]
2021 USAMTS Problems, 2
Let $n$ be a fixed positive integer. Which is greater?[list=1]
[*]The number of $n$-tuples of integers whose largest value is $7$ and whose smallest value
is $0$; or
[*]The number of ordered triples $(A, B, C)$ that satisfy the following property: $A$, $B$, $C$
are subsets of $\{1, 2, 3, \dots , n\}$, and neither $C\subseteq A\cup B$, nor $B\subseteq A\cup C$.
[/list]
Your answer can be: $(1)$, $(2)$, the two counts are equal, or it depends on $n$.