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Difference between revisions of "2008 Mock ARML 2 Problems"

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= Set 3 =
 
= Set 3 =
 
== Problem 5 ==
 
== Problem 5 ==
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Al is thinking of a function, <math>f(x)</math>. He reveals to Bob that the function is a polynomial of the form <math>f(x) = ax^8 + bx^5 + cx^2 + dx + e</math>, where <math>a</math>, <math>b</math>, <math>c</math>, <math>d</math>, and <math>e</math> are [b]complex[/b] number coefficients. Bob wishes to determine the value of <math>d</math>. For any [b]complex[/b] number <math>x</math> that Bob asks about, Al will tell him the value of <math>f(x)</math>. At least how many values of <math>x</math> must Bob ask about in order to definitively determine the value of <math>d</math>.
+
Al is thinking of a function, <math>f(x)</math>. He reveals to Bob that the function is a polynomial of the form <math>f(x) = ax^8 + bx^5 + cx^2 + dx + e</math>, where <math>a</math>, <math>b</math>, <math>c</math>, <math>d</math>, and <math>e</math> are '''complex''' number coefficients. Bob wishes to determine the value of <math>d</math>. For any [b]complex[/b] number <math>x</math> that Bob asks about, Al will tell him the value of <math>f(x)</math>. At least how many values of <math>x</math> must Bob ask about in order to definitively determine the value of <math>d</math>.
  
 
[[2008 Mock ARML 2 Problems/Problem 5 | Solution]]
 
[[2008 Mock ARML 2 Problems/Problem 5 | Solution]]

Revision as of 08:57, 30 May 2008

Set 1

Problem 1

$ABCD$ is a convex quadrilateral such that $|AB| = 5$, $|BC| = 17$, $|CD| = 7$, and $|DA| = 25$. Given that $m\angle{ABC} + m\angle{BCD} = 270^{\circ}$, find the area of $ABCD$.

Solution

Problem 2

Given that the sum of all positive integers with exactly two proper divisors, each of which is less than $30$, is $2397$, find the sum of all positive integers with exactly three proper divisors, each of which is less than $30$ (a proper divisor of $n$ is a positive integer that divides but is not equal to $n$).

Solution

Set 2

Problem 3

A variation of Pascal's triangle is constructed by writing the numbers $2$ and $3$ in the top row and writing each subsequent term as the sum of the two terms above it. Find the fifth term from the left in the thirteenth row. \[\begin{tabular}{ccccccccccc} & & & & & & & & & & \\ & & & & 2 & & 3 & & & & \\ & & & 2 & & 5 & & 3 & & & \\ & & 2 & & 7 & & 8 & & 3 & & \\ & 2 & & 9 & & 15 & & 11 & & 3 & \\ 2 & & 11 & & 24 & & 26 & & 14 & & 3 \\ \end{tabular}\]

Solution

Problem 4

Equilateral triangle $ABC$ has a side length of $7$. A ball begins at vertex $A$, rolls through the interior of the triangle, bounces off side $BC$, and settles at point P. Given that $|BP| = 3$ and $|CP| = 5$, find the total distance that the ball travels.

Solution

Set 3

Problem 5

Al is thinking of a function, $f(x)$. He reveals to Bob that the function is a polynomial of the form $f(x) = ax^8 + bx^5 + cx^2 + dx + e$, where $a$, $b$, $c$, $d$, and $e$ are complex number coefficients. Bob wishes to determine the value of $d$. For any [b]complex[/b] number $x$ that Bob asks about, Al will tell him the value of $f(x)$. At least how many values of $x$ must Bob ask about in order to definitively determine the value of $d$.

Solution

Problem 6

John has a pile of $63$ blocks. On top of the pile is one block. Below this block are two smaller blocks. Below each of these two blocks are two even smaller blocks. Below each of these blocks are two still smaller blocks, and so on until the last row, which contains $32$ blocks. John removes blocks one at a time, removing only blocks that currently have no blocks on top of them. Find the number of ways in which John can remove exactly seven blocks.

Solution

Set 4

Problem 7

Let $f(x)$ equal the number of zeroes to the right of the rightmost non-zero digit in the decimal form of $x!$, and let $n = \frac {5^{2008} + 2}{3}$. Given that $f(n)$ can be written as $\frac {5^{a} - b}{c}$, where $b$ and $c$ are relatively prime positive integers, $b$ is less than $10^5$, and $c$ is less than $10^2$, find $a + b + c$.

Solution

Problem 8

Given that $\sum_{i = 0}^{n}a_ia_{n - i} = 1$ and $a_n > 0$ for all non-negative integers $n$, evaluate $\sum_{j = 0}^{\infty}\frac {a_j}{2^j}$.

Solution

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