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Difference between revisions of "2020 CIME I Problems"
Line 41: Line 41:
 
==Problem 8==
 
==Problem 8==
 
A person has been declared the first to inhabit a certain planet on day </math>N=0<math>. For each positive integer </math>N>0<math>, if there is a positive number of people on the planet, then either one of the following three occurs, each with probability </math>\frac{1}{3}<math>:
 
A person has been declared the first to inhabit a certain planet on day </math>N=0<math>. For each positive integer </math>N>0<math>, if there is a positive number of people on the planet, then either one of the following three occurs, each with probability </math>\frac{1}{3}<math>:
 +
 
:(i) the population stays the same;
 
:(i) the population stays the same;
 
:(ii) the population increases by </math>2^N<math>; or
 
:(ii) the population increases by </math>2^N<math>; or
 
:(iii) the population decreases by </math>2^{N-1}<math>. (If there are no greater than </math>2^{N-1}<math> people on the planet, the population drops to zero, and the process terminates).  
 
:(iii) the population decreases by </math>2^{N-1}<math>. (If there are no greater than </math>2^{N-1}<math> people on the planet, the population drops to zero, and the process terminates).  
 +
 
The probability that at some point there are exactly </math>2^{20}+2^{19}+2^{10}+2^9+1<math> people on the planet can be written as </math>\frac{p}{3^q}<math>, where </math>p<math> and </math>q<math> are positive integers such that </math>p<math> is not divisible by </math>3<math>. Find the remainder when </math>p+q<math> is divided by </math>1000$.
 
The probability that at some point there are exactly </math>2^{20}+2^{19}+2^{10}+2^9+1<math> people on the planet can be written as </math>\frac{p}{3^q}<math>, where </math>p<math> and </math>q<math> are positive integers such that </math>p<math> is not divisible by </math>3<math>. Find the remainder when </math>p+q<math> is divided by </math>1000$.

Revision as of 14:05, 30 August 2020

2020 CIME I (Answer Key)
Printable version | AoPS Contest Collections

Instructions

  1. This is a 15-question, 3-hour examination. All answers are integers ranging from $000$ to $999$, inclusive. Your score will be the number of correct answers; i.e., there is neither partial credit nor a penalty for wrong answers.
  2. No aids other than scratch paper, graph paper, ruler, compass, and protractor are permitted. In particular, calculators and computers are not permitted.
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

Problem 1

A knight begins on the point $(0,0)$ in the coordinate plane. From any point $(x,y)$ the knight moves to either $(x+2,y+1)$ or $(x+1,y+2)$. Find the number of ways the knight can reach $(15,15)$.

Solution

Problem 2

At the local Blast Store, there are sufficiently many items with a price of $$n.99$ for each nonnegative integer $n$. A sales tax of $7.5\%$ is applied on all items. If the total cost of a purchase, after tax, is an integer number of cents, find the minimum possible number of items in the purchase.

Solution

Problem 3

In a math competition, all teams must consist of between $12$ and $15$ members, inclusive. Mr. Beluhov has $n > 0$ students and he realizes that he cannot form teams so that each of his students is on exactly one team. Find the sum of all possible values of $n$.

Solution

Problem 4

There exists a unique positive real number $x$ satisfying \[x=\sqrt{x^2+\frac{1}{x^2}} - \sqrt{x^2-\frac{1}{x^2}}.\] Given that $x$ can be written in the form $x=2^\frac{m}{n} \cdot 3^\frac{-p}{q}$ for integers $m, n, p, q$ with $\gcd(m, n) = \gcd(p, q) = 1$, find $m+n+p+q$.

Solution

Problem 5

Let $ABCD$ be a rectangle with sides $AB>BC$ and let $E$ be the reflection of $A$ over $\overline{BD}$. If $EC=AD$ and the area of $ECBD$ is $144$, find the area of $ABCD$.

Solution

Problem 6

Find the number of complex numbers $z$ satisfying $|z|=1$ and $z^{850}$+z^{350}+1=0$.

[[2020 CIME I Problems/Problem 6 | Solution]]

==Problem 7== For every positive integer$ (Error compiling LaTeX. Unknown error_msg)n$define <cmath>f(n)=\frac{n}{1 \cdot 3 \cdot 5 \cdots (2n+1)}.</cmath> Suppose that the sum$f(1)+f(2)+\cdots+f(2020)$can be expressed as$\frac{p}{q}$for relatively prime integers$p$and$q$. Find the remainder when$p$is divided by$1000$.

[[2020 CIME I Problems/Problem 7 | Solution]]

==Problem 8== A person has been declared the first to inhabit a certain planet on day$ (Error compiling LaTeX. Unknown error_msg)N=0$. For each positive integer$N>0$, if there is a positive number of people on the planet, then either one of the following three occurs, each with probability$\frac{1}{3}$:

(i) the population stays the same;
(ii) the population increases by$ (Error compiling LaTeX. Unknown error_msg)2^N$; or :(iii) the population decreases by$2^{N-1}$. (If there are no greater than$2^{N-1}$people on the planet, the population drops to zero, and the process terminates).

The probability that at some point there are exactly$ (Error compiling LaTeX. Unknown error_msg)2^{20}+2^{19}+2^{10}+2^9+1$people on the planet can be written as$\frac{p}{3^q}$, where$p$and$q$are positive integers such that$p$is not divisible by$3$. Find the remainder when$p+q$is divided by$1000$.

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