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Difference between revisions of "Mock AIME 2 2010 Problems"

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== Problem 10 ==
 
== Problem 10 ==
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How many positive integers <math>n \le 2010 </math> satisfy <math>\phi(n) | n</math>, where <math>\phi(n)</math> is the number of positive integers less than or equal to <math>n</math> relatively prime to <math>n</math>?
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How many positive integers <math>n \le 2010 </math> satisfy <math>\varphi(n) | n</math>, where <math>\varphi(n)</math> is the number of positive integers less than or equal to <math>n</math> relatively prime to <math>n</math>?
  
 
[[Mock AIME 2 2010 Problems/Problem 10|Solution]]
 
[[Mock AIME 2 2010 Problems/Problem 10|Solution]]
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== Problem 14 ==
 
== Problem 14 ==
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Alex and Mitchell decide to play a game. In this game, there are 2010 pieces of candy on a table, and starting with Alex, the two take turns eating some positive integer number of pieces of candy. Since it is bad manners to eat the last candy, whoever eats the last candy loses. The two decide that the amount of candy a person can pick will be a set equal to the positive divisors of a number less than 2010 that each person picks (individually) from the beginning. For example, if Alex picks 19 and Mitchell picks 20, then on each turn, Alex must eat either 1 or 19 pieces, and Mitchell must eat 1, 2, 4, 5, 10, or 20 pieces. Mitchell knows Alex well enough to determine with certainty that Alex will either be immature and pick 69, or be clich\'ed and pick 42. How many integers can Mitchell pick to guarantee that he will not \emph{lose the game}?
+
Alex and Mitchell decide to play a game. In this game, there are 2010 pieces of candy on a table, and starting with Alex, the two take turns eating some positive integer number of pieces of candy. Since it is bad manners to eat the last candy, whoever eats the last candy loses. The two decide that the amount of candy a person can pick will be a set equal to the positive divisors of a number less than 2010 that each person picks (individually) from the beginning. For example, if Alex picks 19 and Mitchell picks 20, then on each turn, Alex must eat either 1 or 19 pieces, and Mitchell must eat 1, 2, 4, 5, 10, or 20 pieces. Mitchell knows Alex well enough to determine with certainty that Alex will either be immature and pick 69, or be clichéd and pick 42. How many integers can Mitchell pick to guarantee that he will not ''lose the game''?
  
 
[[Mock AIME 2 2010 Problems/Problem 14|Solution]]
 
[[Mock AIME 2 2010 Problems/Problem 14|Solution]]

Latest revision as of 08:10, 8 October 2018

Mock AIME 2 2010 (Answer Key)
Printable version | AoPS Contest Collections • PDF

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

There are $2010$ lemmings. At each step, we may separate the lemmings into groups of $5$ and purge the remainder, separate them into groups of $3$ and purge the remainder, or pick one lemming and purge it. Find the smallest number of steps necessary to remove all $2010$ lemmings.

Solution

Problem 2

Let $a_1, a_2, \ldots, a_{10}$ be nonnegative integers such that $a_1 + a_2 + \ldots + a_{10} = 2010$, and define $f$ so that $f((a_1, a_2, \ldots, a_{10})) = (b_1, b_2, \ldots, b_{10})$, with $0 \le b_i \le 2, 3|a_i-b_i$ for $1 \le i \le 10$. Given that $f$ can take on $K$ distinct values, find the remainder when $K$ is divided by 1000.

Solution

Problem 3

Five gunmen are shooting each other. At the same moment, each randomly chooses one of the other four to shoot. The probability that there are some two people shooting each other can be expressed in the form $\frac{a}{b}$, where $a, b$ are relatively prime positive integers. Find $a+b$.

Solution

Problem 4

Anderson is bored in physics class. His favorite numbers are $1, 7$, and $33$. He writes $0.$, and randomly appends one of his favorite numbers to the end of the decimal he has already written. Since physics class is infinitely long, Anderson writes an infinitely long decimal number. (An example of such a number is $0.1337173377133733 \ldots$) If the expected value of the number Anderson wrote down is of the form $\frac{a}{b}$, where $a$ and $b$ are relatively prime positive integers, find $a + b$.

Solution

Problem 5

Let $P(x)=(1+x)(1+2x^2)(1+3x^4)(1+4x^8)(1+5x^{16})$. Find the three rightmost nonzero digits of the product of the coefficients of $P(x)$.

Solution

Problem 6

Let $S_n$ denote the set $\{1,2,\ldots,n\}$, and define $f(S)$, where $S$ is a subset of the positive integers, to output the greatest common divisor of all elements in $S$, unless $S$ is empty, in which case it will output 0. Find the last three digits of $\sum_{S\subseteq S_{10}} f(S)$, where $S$ ranges over all subsets of $S_{10}$.

Solution

Problem 7

Find the number of functions $f$ from $\{1,2,3,4,5\}$ to itself such that $f(f(f(x))) = f(f(x))$ for all $x \in \{1,2,3,4,5\}$.

Solution

Problem 8

In triangle $ABC$, $BA = 15$, $AC = 20$, and $BC = 25$. In addition, there is a point $D$ lying on segment $BC$ such that $BD = 16$. Given that the length of the radius of the circle through $B$ and $D$ that is tangent to side $AC$ can be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are relatively prime integers, find $p + q$.

Solution

Problem 9

Given that $x,y,z$ are reals such that $16 \sin^4(x+y) + 49 \cos^4(x+y) = \sin^2(2x + 2y)(8\sin^2(x+z) + 6 \cos^2(y+z))$, the largest possible value of $3\cos^2(x+y+z)$ can be expressed in the form $\frac{a+b\sqrt{c}}{d}$, where $a$ and $b$ are integers, $c$ is a positive integer not divisible by the square of any prime, and $d$ is a positive integer such that $\gcd(a,b,d)=1$. Find $a+b+c+d$.

Solution

Problem 10

How many positive integers $n \le 2010$ satisfy $\varphi(n) | n$, where $\varphi(n)$ is the number of positive integers less than or equal to $n$ relatively prime to $n$?

Solution

Problem 11

Let $f:\mathbb{N}\to\mathbb{N}$ be a function such that \[f(n) =\sum_{abc = n | a,b,c \in \mathbb{N}} ab + bc + ca.\] For example, $f(5) = (1 \cdot 1 + 1 \cdot 5 + 5 \cdot 1) + (1 \cdot 5 + 5 \cdot 1 + 1 \cdot 1) + (5 \cdot 1 + 1 \cdot 1 + 1 \cdot 5) = 33$, where we are summing over the triples $(a,b,c) = (1,1,5), (1,5,1)$, and $(5,1,1)$. Find the last three digits of $f(30^{3})$.

Solution

Problem 12

Let \[K = \sum_{a_1=0}^8\sum_{a_2=0}^{a_1}...\sum_{a_{100}=0}^{a_{99}}\binom{8}{a_1}\binom{a_1}{a_2}...\binom{a_{99}}{a_{100}}.\] Find the sum of digits of $K$ in base-100.

Solution

Problem 13

$\triangle ABC$ is inscribed in circle $\mathcal{C}$. The radius of $\mathcal{C}$ is $285$, and $BC = 567$. When the incircle of $\triangle ABC$ is reflected across segment $BC$, it is tangent to $\mathcal{C}$. Given that the inradius of $\triangle ABC$ can be expressed in the form $a\sqrt{b}$, where $a$ and $b$ are positive integers and $b$ is not divisible by the square of any prime, find $a+b$.

Solution

Problem 14

Alex and Mitchell decide to play a game. In this game, there are 2010 pieces of candy on a table, and starting with Alex, the two take turns eating some positive integer number of pieces of candy. Since it is bad manners to eat the last candy, whoever eats the last candy loses. The two decide that the amount of candy a person can pick will be a set equal to the positive divisors of a number less than 2010 that each person picks (individually) from the beginning. For example, if Alex picks 19 and Mitchell picks 20, then on each turn, Alex must eat either 1 or 19 pieces, and Mitchell must eat 1, 2, 4, 5, 10, or 20 pieces. Mitchell knows Alex well enough to determine with certainty that Alex will either be immature and pick 69, or be clichéd and pick 42. How many integers can Mitchell pick to guarantee that he will not lose the game?

Solution

Problem 15

Given that $\sum_{a = 1}^{491}\sum_{b = 1}^{491} \frac{1}{(a+bi)^4 - 1}$ can be expressed in the form $\frac{p + qi}{r}$, where $\gcd(p, q, r) = 1$ and $r > 0$, find the unique integer $k$ between 0 and 982 inclusive such that $983$ divides $(p + q) - kr$. [Note: $983$ is a prime.]

Solution

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