Found problems: 248
2014 USAMTS Problems, 3:
Let $a_1,a_2,a_3,...$ be a sequence of positive real numbers such that:
(i) For all positive integers $m,n$, we have $a_{mn}=a_ma_n$
(ii) There exists a positive real number $B$ such that for all positive integers $m,n$ with $m<n$, we have $a_m < Ba_n$
Find all possible values of $\log_{2015}(a_{2015}) - \log_{2014}(a_{2014})$
2013 USAMTS Problems, 2
Let $a_1,a_2,a_3,\dots$ be a sequence of positive real numbers such that $a_ka_{k+2}=a_{k+1}+1$ for all positive integers $k$. If $a_1$ and $a_2$ are positive integers, find the maximum possible value of $a_{2014}$.
1999 USAMTS Problems, 2
The Fibonacci numbers are defined by $F_1=F_2=1$ and $F_n=F_{n-1}+F_{n-2}$ for $n>2$. It is well-known that the sum of any $10$ consecutive Fibonacci numbers is divisible by $11$. Determine the smallest integer $N$ so that the sum of any $N$ consecutive Fibonacci numbers is divisible by $12$.
2013 Princeton University Math Competition, 3
Consider the shape formed from taking equilateral triangle $ABC$ with side length $6$ and tracing out the arc $BC$ with center $A$. Set the shape down on line $l$ so that segment $AB$ is perpendicular to $l$, and $B$ touches $l$. Beginning from arc $BC$ touching $l$, we roll $ABC$ along $l$ until both points $A$ and $C$ are on the line. The area traced out by the roll can be written in the form $n\pi$, where $n$ is an integer. Find $n$.
2009 USAMTS Problems, 4
Let $S$ be a set of $10$ distinct positive real numbers. Show that there exist $x,y \in S$ such that
\[0 < x - y < \frac{(1 + x)(1 + y)}{9}.\]
2018 USAMTS Problems, 1:
Fill in each space of the grid with one of the numbers $1,2,\dots,30$, using each number once. For $1\le{}n\le29$, the two spaces containing $n$ and $n+1$ must be in either the same row or the same column. Some numbers have been given to you.
[asy]
unitsize(32);
int[][] a = {
{29, 000, 000, 000, 000, 000},
{000, 19, 000, 000, 17, 000},
{13, 000, 000, 21, 000, 8},
{000, 4, 000, 15, 000, 24},
{10, 000, 000, 000, 26, 000}};
for (int i = 0; i < 6; ++i) {
for (int j = 0; j < 5; ++j) {
draw((i, -j)--(i+1, -j)--(i+1, -j-1)--(i, -j-1)--cycle);
if (a[j][i] > 0 && a[j][i] < 999) label(string(a[j][i]), (i+0.5, -j-0.5), fontsize(30pt));
}
}
[/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.)
1999 USAMTS Problems, 2
Let $N=111...1222...2$, where there are $1999$ digits of $1$ followed by $1999$ digits of $2$. Express $N$ as the product of four integers, each of them greater than $1$.
2012 USAMTS Problems, 2
Find all triples $(a, b, c)$ of positive integers with $a\le b\le c$ such that\[\left(1+\dfrac1{a}\right)\left(1+\dfrac1{b}\right)\left(1+\dfrac1{c}\right)=3.\]
2002 USAMTS Problems, 2
We define the number $s$ as
\[s=\sum^{\infty}_{i=1} \dfrac{1}{10^i-1}=\dfrac{1}{9}+\dfrac{1}{99}+\dfrac{1}{999}+\dfrac{1}{9999}+\ldots=0.12232424...\]
We can determine the $n$th digit right of the decimal point of $s$ without summing the entire infinite series because after summing the first $n$ terms of the series, the rest of the series sums to less than $\dfrac{2}{10^{n+1}}$. Determine the smallest prime number $p$ for which the $p$th digit right of the decimal point of $s$ is greater than 2. Justify your answer.
2017 USAMTS Problems, 5
Does there exist a set $S$ consisting of rational numbers with the following property:
for every integer $n$ there is a unique nonempty, finite subset of $S$, whose elements sum to $n$?
2017 USAMTS Problems, 3
Let $ABC$ be an equilateral triangle with side length $1$. Let $A_1$ and $A_2$ be the trisection points of $AB$ with $A_1$ closer to $A$, $B_1$ and $B_2$ be the trisection points of $BC$ with $B_1$ closer to $B$, and $C_1$ and $C_2$ be the trisection points of $CA$ with $C_1$ closer to $C$. Grogg has an orange equilateral triangle the size of triangle $A_1B_1C_1$. He puts the orange triangle over triangle $A_1B_1C_1$ and then rotates it about its center in the shortest direction until its vertices are over $A_2B_2C_2$. Find the area of the region that the orange triangle traveled over during its rotation.
2013 USAMTS Problems, 2
In the $5\times6$ grid shown, fill in all of the grid cells with the digits $0\textendash9$ so that the following conditions are satisfied:
[list=1][*] Each digit gets used exactly $3$ times.
[*] No digit is greater than the digit directly above it.
[*] In any four cells that form a $2\times2$ subgrid, the sum of the four digits must be a multiple of $3$.[/list]
You do not need to prove that your configuration is the only one possible; you merely need to find a configuration that works. (Note: In any other USAMTS problem, you need to provide a full proof. Only in this problem is an answer without justification acceptable.)
\[\begin{Large}\begin{array}{|c|c|c|c|c|c|}\hline&&&&\,7\,&\\ \hline&\,8\,&&&&\,6\,\\\hline&&\,2\,&\,4\,&&\\ \hline\,5\,&&&&1&\\ \hline&3&&&&\\ \hline\end{array}\end{Large}\]
2002 USAMTS Problems, 3
Determine, with proof, the rational number $\dfrac{m}{n}$ that equals
\[\tfrac{1}{1\sqrt2+2\sqrt1}+\tfrac{1}{2\sqrt3+3\sqrt2}+\tfrac{1}{3\sqrt4+4\sqrt3}+\ldots+\tfrac{1}{4012008\sqrt{4012009}+4012009\sqrt{4012008}}\]
2011 USAMTS Problems, 3
You have $14$ coins, dated $1901$ through $1914$. Seven of these coins are real and weigh $1.000$ ounce each. The other seven are counterfeit and weigh $0.999$ ounces each. You do not know which coins are real or counterfeit. You also cannot tell which coins are real by look or feel.
Fortunately for you, Zoltar the Fortune-Weighing Robot is capable of making very precise measurements. You may place any number of coins in each of Zoltar's two hands and Zoltar will do the following:
[list][*] If the weights in each hand are equal, Zoltar tells you so and returns all of the coins.
[*] If the weight in one hand is heavier than the weight in the other, then Zoltar takes one coin, at random, from the heavier hand as tribute. Then Zoltar tells you which hand was heavier, and returns the remaining coins to you.[/list]
Your objective is to identify a single real coin that Zoltar has not taken as tribute. Is there a strategy that guarantees this? If so, then describe the strategy and why it works. If not, then prove that no such strategy exists.
2013 USAMTS Problems, 3
An infinite sequence of positive real numbers $a_1,a_2,a_3,\dots$ is called [i]territorial[/i] if for all positive integers $i,j$ with $i<j$, we have $|a_i-a_j|\ge\tfrac1j$. Can we find a territorial sequence $a_1,a_2,a_3,\dots$ for which there exists a real number $c$ with $a_i<c$ for all $i$?
2016 USAMTS Problems, 4:
On Binary Island, residents communicate using special paper. Each piece of paper is a $1 \times n$ row of initially uncolored squares. To send a message, each square on the paper must either be colored either red or green. Unfortunately the paper on the island has become damaged, and each sheet of paper has $10$ random consecutive squares each of which is randomly colored red or green.
Malmer and Weven would like to develop a scheme that allows them to send messages of length $2016$ between one another. They would like to be able to send any message of length $2016$, and they want their scheme to work with perfect accuracy. What is the smallest value of $n$ for which they can develop such a strategy?
[i]Note that when sending a message, one can see which 10 squares are colored and what colors they are. One also knows on which square the message begins, and on which square the message ends.[/i]
2004 USAMTS Problems, 3
Find, with proof, a polynomial $f(x,y,z)$ in three variables, with integer coefficients, such that for all $a,b,c$ the sign of $f(a,b,c)$ (that is, positive, negative, or zero) is the same as the sign of $a+b\sqrt[3]{2}+c\sqrt[3]{4}$.
2013 USAMTS Problems, 5
For any positive integer $b\ge2$, we write the base-$b$ numbers as follows:
\[(d_kd_{k-1}\dots d_0)_b=d_kb^k+d_{k-1}b^{k-1}+\dots+d_1b^1+d_0b^0,\]where each digit $d_i$ is a member of the set $S=\{0,1,2,\dots,b-1\}$ and either $d_k\not=0$ or $k=0$. There is a unique way to write any nonnegative integer in the above form. If we select the digits from a di fferent set $S$ instead, we may obtain new representations of all positive integers or, in some cases, all integers. For example, if $b=3$ and the digits are selected from $S=\{-1,0,1\}$, we obtain a way to uniquely represent all integers, known as a $\emph{balanced ternary}$ representation. As further examples, the balanced ternary representation of numbers $5$, $-3$, and $25$ are:
\[5=(1\ {-1}\ {-1})_3,\qquad{-3}=({-1}\ 0)_3,\qquad25=(1\ 0\ {-1}\ 1)_3.\]However, not all digit sets can represent all integers. If $b=3$ and $S=\{-2,0,2\}$, then no odd number can be represented. Also, if $b=3$ and $S=\{0,1,2\}$ as in the usual base-$3$ representation, then no negative number can be represented.
Given a set $S$ of four integers, one of which is $0$, call $S$ a $\emph{4-basis}$ if every integer $n$ has at least one representation in the form
\[n=(d_kd_{k-1}\dots d_0)_4=d_k4^k+d_{k-1}4^{k-1}+\dots+d_14^1+d_04^0,\]where $d_k,d_{k-1},\dots,d_0$ are all elements of $S$ and either $d_k\not=0$ or $k=0$.
[list=a]
[*]Show that there are infinitely many integers $a$ such that $\{-1,0,1,4a+2\}$ is not a $4$-basis.
[*]Show that there are infinitely many integers $a$ such that $\{-1,0,1,4a+2\}$ is a $4$-basis.[/list]
1999 USAMTS Problems, 1
The digits of the three-digit integers $a, b,$ and $c$ are the nine nonzero digits $1,2,3,\cdots 9$ each of them appearing exactly once. Given that the ratio $a:b:c$ is $1:3:5$, determine $a, b,$ and $c$.
2011 USAMTS Problems, 1
Find all the ways of placing the integers $1,2,3,\cdots,16$ in the boxes below, such that each integer appears in exactly one box, and the sum of every pair of neighboring integers is a perfect square.
[asy]
import graph;
real r=10;
size(r*cm);
picture square1;
draw(square1, (0,0)--(0,1)--(1,1)--(1,0)--cycle);
add(scale(r/31*cm)*square1,(0,0));
picture square2;
draw(square2, (-1,0.5)--(0,0.5)--(0,1)--(1,1)--(1,0)--(0,0)--(0,0.5));
for(int i=1; i<16; ++i)
{
add(scale(r/31*cm)*square2,(i,0));
}[/asy]
2004 USAMTS Problems, 5
Consider an isosceles triangle $ABC$ with side lengths $AB = AC = 10\sqrt{2}$ and $BC =10\sqrt{3}$. Construct semicircles $P$, $Q$, and $R$ with diameters $AB$, $AC$, $BC$ respectively, such that the plane of each semicircle is perpendicular to the plane of $ABC$, and all semicircles are on the same side of plane $ABC$ as shown. There exists a plane above triangle $ABC$ that is tangent to all three semicircles $P$, $Q$, $R$ at the points $D$, $E$, and $F$ respectively, as shown in the diagram. Calculate, with proof, the area of triangle $DEF$.
[asy]
size(200);
import three; defaultpen(linewidth(0.7)+fontsize(10)); currentprojection = orthographic(0,4,2.5);
// 1.15 x-scale distortion factor
triple A = (0,0,0), B = (75^.5/1.15,-125^.5,0), C = (-75^.5/1.15,-125^.5,0), D = (A+B)/2 + (0,0,abs((B-A)/2)), E = (A+C)/2 + (0,0,abs((C-A)/2)), F = (C+B)/2 + (0,0,abs((B-C)/2));
draw(D--E--F--cycle); draw(B--A--C);
// approximate guess for r
real r = 1.38; draw(B--(r*B+C)/(1+r)^^(B+r*C)/(1+r)--C,linetype("4 4")); draw((B+r*C)/(1+r)--(r*B+C)/(1+r));
// lazy so I'll draw six arcs
draw(arc((A+B)/2,A,D)); draw(arc((A+B)/2,D,B)); draw(arc((A+C)/2,E,A)); draw(arc((A+C)/2,E,C)); draw(arc((C+B)/2,F,B)); draw(arc((C+B)/2,F,C));
label("$A$",A,S); label("$B$",B,W); label("$C$",C,plain.E);
label("$D$",D,SW); label("$E$",E,SE); label("$F$",F,N);[/asy]
2013 USAMTS Problems, 1
In the $3\times5$ grid shown, fill in each empty box with a two-digit positive integer such that:
[list][*]no number appears in more than one box, and
[*] for each of the $9$ lines in the grid consisting of three boxes connected by line segments, the box in the middle of the line contains the least common multiple of the numbers in the two boxes on the line.[/list]
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]
import graph; size(7cm);
real labelscalefactor = 0.5;
pen dps = linewidth(0.8) + fontsize(14);
defaultpen(dps);
draw((0,0)--(1,0)--(1,1)--(0,1)--cycle);
draw((2,0)--(3,0)--(3,1)--(2,1)--cycle);
draw((4,0)--(5,0)--(5,1)--(4,1)--cycle);
draw((6,0)--(7,0)--(7,1)--(6,1)--cycle);
draw((8,0)--(9,0)--(9,1)--(8,1)--cycle);
draw((0,2)--(1,2)--(1,3)--(0,3)--cycle);
draw((0,4)--(1,4)--(1,5)--(0,5)--cycle);
draw((2,2)--(3,2)--(3,3)--(2,3)--cycle);
draw((2,4)--(3,4)--(3,5)--(2,5)--cycle);
draw((4,4)--(5,4)--(5,5)--(4,5)--cycle);
draw((4,2)--(5,2)--(5,3)--(4,3)--cycle);
draw((6,2)--(7,2)--(7,3)--(6,3)--cycle);
draw((6,4)--(7,4)--(7,5)--(6,5)--cycle);
draw((8,4)--(9,4)--(9,5)--(8,5)--cycle);
draw((8,2)--(9,2)--(9,3)--(8,3)--cycle);
draw((0.5,1)--(0.5,2));
draw((0.5,3)--(0.5,4));
draw((1,4)--(2,3));
draw((2.5,1)--(2.5,2));
draw((2.5,3)--(2.5,4));
draw((3,4)--(4,3));
draw((3,2)--(4,1));
draw((4.5,1)--(4.5,2));
draw((4.5,3)--(4.5,4));
draw((5,4.5)--(6,4.5));
draw((7,4.5)--(8,4.5));
draw((5,4)--(6,3));
draw((7,2)--(8,1));
draw((5,2)--(6,1));
draw((5,0.5)--(6,0.5));
draw((7,0.5)--(8,0.5));
draw((8.5,1)--(8.5,2));
draw((8.5,3)--(8.5,4));
label("$4$",(4.5, 0.5));
label("$9$",(8.5, 4.5));
[/asy]
2015 USAMTS Problems, 2
Fames is playing a computer game with falling two-dimensional blocks. The playing field is $7$ units wide and infinitely tall with a bottom border. Initially the entire field is empty. Each turn, the computer gives Fames a $1\times 3$ solid rectangular piece of three unit squares. Fames must decide whether to orient the piece horizontally or vertically and which column(s) the piece should occupy ($3$ consecutive columns for horizontal pieces, $1$ column for vertical pieces). Once he confirms his choice, the piece is dropped straight down into the playing field in the selected columns, stopping all three of the piece's squares as soon as the piece hits either the bottom of the playing field or any square from another piece. All of the pieces must be contained completely inside the playing field after dropping and cannot partially occupy columns.
If at any time a row of $7$ spaces is all filled with squares, Fames scores a point.
Unfortunately, Fames is playing in [i]invisible mode[/i], which prevents him from seeing the state of the playing field or how many points he has, and he has already arbitrarily dropped some number of pieces without remembering what he did with them or how many there were.
For partial credit, find a strategy that will allow Fames to eventually earn at least one more point. For full credit, find a strategy for which Fames can correctly announce "I have earned at least one more point" and know that he is correct.
2009 USAMTS Problems, 3
Prove that if $a$ and $b$ are positive integers such that $a^2 + b^2$ is a multiple of $7^{2009}$, then $ab$ is a multiple of $7^{2010}$.
2015 USAMTS Problems, 1
Fill in the spaces of the grid below with positive integers so that in each $2\times 2$ square with top left number $a$, top right number $b$, bottom left number $c$, and bottom right number $d$, either $a + d = b + c$ or $ad = bc$. 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]
size(3.85cm);
for (int i=0; i<=5; ++i)
draw((i,0)--(i,5), linewidth(.5));
for (int j=0; j<=5; ++j)
draw((0,j)--(5,j), linewidth(.5));
void draw_num(pair ll_corner, int num)
{
label(string(num), ll_corner + (0.5, 0.5), p = fontsize(19pt));
}
draw_num((0,0), 20);
draw_num((1, 0), 36);
draw_num((1,4), 9);
draw_num((4, 0), 32);
draw_num((0, 1), 15);
draw_num((0, 2), 10);
draw_num((0, 4), 3);
draw_num((1,3), 11);
draw_num((3,3), 7);
draw_num((4,3), 2);
draw_num((4,2), 16);
void foo(int x, int y, string n)
{
label(n, (x+0.5,y+0.5), p = fontsize(19pt));
}
foo(2, 4, " ");
foo(3, 4, " ");
foo(4, 4, " ");
foo(0, 3, " ");
foo(2, 3, " ");
foo(1, 2, " ");
foo(2, 2, " ");
foo(3, 2, " ");
foo(1, 1, " ");
foo(2, 1, " ");
foo(3, 1, " ");
foo(4, 1, " ");
foo(2, 0, " ");
foo(3, 0, " ");
[/asy]