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Difference between revisions of "Mock AIME 5 Pre 2005 Problems"

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== Problem 1 ==
 
== Problem 1 ==
−
1. Function <math>g(y)</math> is given such way that for all <math>y</math>,
+
Function <math>g(y)</math> is given such way that for all <math>y</math>,
  
−
<math>6g(1 + \frac {1}{y}) + 12g(y + 1) = \log_{10} y</math>
+
<cmath>6g(1 + (1/y)) + 12g(y + 1) = \log_{10} y</cmath>
  
 
If <math>g(9) + g(26) + g(126) + g(401) = \frac {m}{n}</math> where <math>m</math> and <math>n</math> are relatively prime integers, computer <math>m + n</math>.
 
If <math>g(9) + g(26) + g(126) + g(401) = \frac {m}{n}</math> where <math>m</math> and <math>n</math> are relatively prime integers, computer <math>m + n</math>.
Line 9: Line 9:
  
 
== Problem 2 ==
 
== Problem 2 ==
−
2. Two 5-digit numbers are called "responsible" if they are:
+
Two 5-digit numbers are called "responsible" if they are:
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
−
&\text {i. In form of abcde and fghij such that fghij = 2(abcde)}
+
&\text {i. In form of abcde and fghij such that fghij = 2(abcde)}\\
−
&\text {ii. all ten digits, a through j are all distinct.}
+
&\text {ii. all ten digits, a through j are all distinct.}\\
 
&\text {iii.} a + b + c + d + e + f + g + h + i + j = 45\end{align*}</cmath>
 
&\text {iii.} a + b + c + d + e + f + g + h + i + j = 45\end{align*}</cmath>
  
 
If two "responsible" numbers are small as possible, what is the sum of the three middle digits of <math>\text {abcde}</math> and last two digits on the <math>\text {fghij}</math>? That is, <math>b + c + d + i + j</math>.
 
If two "responsible" numbers are small as possible, what is the sum of the three middle digits of <math>\text {abcde}</math> and last two digits on the <math>\text {fghij}</math>? That is, <math>b + c + d + i + j</math>.
 +
 +
[[Mock AIME 5 Pre 2005 Problems/Problem 2|Solution]]
  
 
== Problem 3 ==
 
== Problem 3 ==
−
3. The triangle is called Heronian if its perimeter equals the area of the triangle. Heronian triangle COP has sides 29,6, and <math>x</math>. If the distance from its incenter to the circumcenter is exprssed as <math>\frac {\sqrt b}{a}</math> where <math>b</math> is not divisible by squares of any prime, find the remainder when <math>b</math> is divided by <math>a</math>.
+
A triangle is called ''Heronian'' if its perimeter equals the area of the triangle. ''Heronian'' triangle <math>COP</math> has sides <math>29,6,</math> and <math>x</math>. If the distance from its incenter to the circumcenter is exprssed as <math>\frac {\sqrt b}{a}</math> where <math>b</math> is not divisible by squares of any prime, find the remainder when <math>b</math> is divided by <math>a</math>.
 +
 
 +
[[Mock AIME 5 Pre 2005 Problems/Problem 3|Solution]]
  
 
== Problem 4 ==
 
== Problem 4 ==
−
4. Eight boxes are numbered A to H. The number of ways you can put 16 identical balls into the boxes such that none of them is empty is expressed as <math>\binom {a}{b}</math>. What is the remainder when <math>\binom {a}{b}</math> is divided by <math>a + b</math>? (Just use <math>\binom {a}{b}</math>, not <math>\binom {a}{a - b}</math>.
+
Eight boxes are numbered <math>A</math> to <math>H</math>. The number of ways you can put 16 identical balls into the boxes such that none of them is empty is expressed as <math>\binom {a}{b}</math>, where <math>b \le \frac{a}{2}</math>. What is the remainder when <math>\binom {a}{b}</math> is divided by <math>a + b</math>?
 +
 
 +
[[Mock AIME 5 Pre 2005 Problems/Problem 4|Solution]]
  
 
== Problem 5 ==
 
== Problem 5 ==
−
5. There are three rooms in Phil's Motel. One room for one person, one room for three people, and another one room for four people. Let's say Peter and his seven friends came to the Phil's Motel. How many ways are there to house Peter and his seven friends in these rooms?
+
There are three rooms in Phil's Motel. One room for one person, one room for three people, and another one room for four people. Let's say Peter and his seven friends came to the Phil's Motel. How many ways are there to house Peter and his seven friends in these rooms?
 +
 
 +
[[Mock AIME 5 Pre 2005 Problems/Problem 5|Solution]]
  
 
== Problem 6 ==
 
== Problem 6 ==
−
6. Larry and his two friends toss a die four times. From all possible outcomes, the probability that outcome has at least one occurrence of 2 is <math>\frac {p}{q}</math> where <math>p</math> and <math>q</math> are relatively prime integers. Find <math>|p - q|</math>.
+
Larry and his two friends toss a die four times. From all possible outcomes, the probability that outcome has at least one occurrence of 2 is <math>\frac {p}{q}</math> where <math>p</math> and <math>q</math> are relatively prime integers. Find <math>|p - q|</math>.
 +
 
 +
[[Mock AIME 5 Pre 2005 Problems/Problem 6|Solution]]
  
 
== Problem 7 ==
 
== Problem 7 ==
−
7. In <math>\triangle RST</math>, X is on <math>\overline {RT}</math>, dividing <math>RX:XT = 1:2</math>. Y is on <math>\overline {ST}</math>, dividing <math>SY:YT = 2:1</math>. V is on <math>\overline {XY}</math>, dividing <math>XV:VY = 1:2</math>. It is found that ray VT intersects <math>\overline {RS}</math> at Z. Find:
+
In <math>\triangle RST</math>, X is on <math>\overline {RT}</math>, dividing <math>RX:XT = 1:2</math>. Y is on <math>\overline {ST}</math>, dividing <math>SY:YT = 2:1</math>. V is on <math>\overline {XY}</math>, dividing <math>XV:VY = 1:2</math>. It is found that ray VT intersects <math>\overline {RS}</math> at Z. Find
  
−
<math>128 ({\frac {TV}{VZ} + \frac {RZ}{ZS})</math>
+
<cmath>128 (\frac {TV}{VZ} + \frac {RZ}{ZS})</cmath>
 +
 
 +
[[Mock AIME 5 Pre 2005 Problems/Problem 7|Solution]]
  
 
== Problem 8 ==
 
== Problem 8 ==
−
8. Let <math>x,y, \in \mathbb {Z}</math> and that:
+
Let <math>x,y, \in \mathbb {Z}</math> and that:
  
−
<math>x + \sqrt y = \sqrt {22 + \sqrt {384}}</math>.
+
<cmath>x + \sqrt y = \sqrt {22 + \sqrt {384}}.</cmath>
  
 
If <math>\frac {x}{y} = \frac {p}{q}</math> with <math>(p,q) = 1</math>, then what is <math>p + q</math>?
 
If <math>\frac {x}{y} = \frac {p}{q}</math> with <math>(p,q) = 1</math>, then what is <math>p + q</math>?
 +
 +
[[Mock AIME 5 Pre 2005 Problems/Problem 8|Solution]]
  
 
== Problem 9 ==
 
== Problem 9 ==
−
9. On a circle with center <math>\zeta</math>, two points, <math>X</math> and <math>Y</math> are on a circle and <math>Z</math> is outside the circle but in ray XY. <math>\overline {XY} = 33.6, \overline {YZ} = 39.2,</math> and <math>\overline {\zeta X} = 21</math>. If <math>\overline {\zeta Z} = \frac {j}{k}</math> where <math>j</math> and <math>k</math> are relatively prime integers, and such that <math>\frac {j + k}{j - k} = \frac {o}{p}</math>, find <math>o + p.</math>
+
On a circle with center <math>\zeta</math>, two points, <math>X</math> and <math>Y</math> are on a circle and <math>Z</math> is outside the circle but in ray XY. <math>\overline {XY} = 33.6, \overline {YZ} = 39.2,</math> and <math>\overline {\zeta X} = 21</math>. If <math>\overline {\zeta Z} = \frac {j}{k}</math> where <math>j</math> and <math>k</math> are relatively prime integers, and such that <math>\frac {j + k}{j - k} = \frac {o}{p}</math>, find <math>o + p.</math>
 +
 
 +
[[Mock AIME 5 Pre 2005 Problems/Problem 9|Solution]]
  
 
== Problem 10 ==
 
== Problem 10 ==
−
10. Given that:
+
Given that:
  
−
<math>(\frac {1}{r})(\frac {1}{s})(\frac {1}{t}) = \frac {3}{391} \\
+
<cmath>\begin{align*}(\frac {1}{r})(\frac {1}{s})(\frac {1}{t}) &= \frac {3}{391} \\
−
r + \frac {1}{s} = \frac {35}{46} \\
+
r + \frac {1}{s} &= \frac {35}{46} \\
−
s + \frac {1}{t} = \frac {1064}{23} \\
+
s + \frac {1}{t} &= \frac {1064}{23} \\
−
t + \frac {1}{r} = \frac {529}{102}</math>.
+
t + \frac {1}{r} &= \frac {529}{102}.\end{align*}</cmath>
  
 
Then what is the smallest integer that is divisible <math>rs</math> and <math>12t</math>?  
 
Then what is the smallest integer that is divisible <math>rs</math> and <math>12t</math>?  
 +
 +
[[Mock AIME 5 Pre 2005 Problems/Problem 10|Solution]]
  
 
== Problem 11 ==
 
== Problem 11 ==
−
11.  There are <math>z</math> number of ways to represent 10000000 as product of three factors and <math>Z</math> number of ways to represent 11390625 as product of three factors.  If <math>|z - Z| = p^q</math>, find <math>p + q</math>.
+
There are <math>z</math> number of ways to represent 10000000 as product of three factors and <math>Z</math> number of ways to represent 11390625 as product of three factors.  If <math>|z - Z| = p^q</math>, find <math>p + q</math>.
 +
 
 +
[[Mock AIME 5 Pre 2005 Problems/Problem 11|Solution]]
  
 
== Problem 12 ==
 
== Problem 12 ==
−
12. Let <math>m = 101^4 + 256</math>. Find the sum of digits of <math>m</math>.  
+
Let <math>m = 101^4 + 256</math>. Find the sum of digits of <math>m</math>.  
  
 
[[Mock AIME 5 Pre 2005 Problems/Problem 12|Solution]]
 
[[Mock AIME 5 Pre 2005 Problems/Problem 12|Solution]]
  
 
== Problem 13 ==
 
== Problem 13 ==
−
13. If the reciprocal of sum of real roots of the following equation can be written in form of <math>\frac {r}{s}</math> where <math>(r,s) = 1</math>, find <math>r + s</math>.
+
If the reciprocal of sum of real roots of the following equation can be written in form of <math>\frac {r}{s}</math> where <math>(r,s) = 1</math>, find <math>r + s</math>.
  
 
<math>1000x^6 - 1900x^5 - 1400x^4 - 190x^3 - 130x^2 - 38x - 30 = 0</math>
 
<math>1000x^6 - 1900x^5 - 1400x^4 - 190x^3 - 130x^2 - 38x - 30 = 0</math>
 +
 +
[[Mock AIME 5 Pre 2005 Problems/Problem 13|Solution]]
  
 
== Problem 14 ==
 
== Problem 14 ==
−
14. The set <math>Z</math> contains complex numbers <math>\zeta_0,\zeta_1,\zeta_2...}</math> such that <math>n = 0,1,2,3.....</math>.
+
The set <math>Z</math> contains complex numbers <math>\zeta_0,\zeta_1,\zeta_2...</math> such that <math>n = 0,1,2,3.....</math>.
  
 
<math>\zeta_n</math> is defined this way:
 
<math>\zeta_n</math> is defined this way:
  
−
<math>\zeta_{n + 1} = (\frac {\zeta_n - i}{\zeta_n + i})^{ - 1}</math>  
+
<math>\zeta_{n + 1} = (\frac {\zeta_n - i}{\zeta_n + i})^{- 1}</math>
  
−
If <math>\zeta_0 = i + \frac {1}{121}</math> and 2008th number of the set is in form of <math>a + bi</math>, find <math>ab</math>.
+
If <math>\zeta_0 = i + \frac{1}{121}</math> and 2008th number of the set is in form of (<math>a + bi</math>), find <math>ab</math>.
 +
 
 +
[[Mock AIME 5 Pre 2005 Problems/Problem 14|Solution]]
  
 
== Problem 15 ==
 
== Problem 15 ==
−
15. Call the value of <math>[\log_2 1] + [\log_2 2] + [\log_2 3] + ....[\log_2 64]</math> as <math>\alpha</math>. Let the value of <math>[\log_2 1] + [\log_2 2] + [\log_2 3] + ....[\log_2 52]</math> as <math>\delta</math>. Find the absolute value of <math>\delta - \alpha</math>.
+
Call the value of <math>[\log_2 1] + [\log_2 2] + [\log_2 3] + ....[\log_2 64]</math> as <math>\alpha</math>. Let the value of <math>[\log_2 1] + [\log_2 2] + [\log_2 3] + ....[\log_2 52]</math> as <math>\delta</math>. Find the absolute value of <math>\delta - \alpha</math>.
 +
 
 +
[[Mock AIME 5 Pre 2005 Problems/Problem 15|Solution]]

Latest revision as of 06:27, 12 October 2022

Problem 1

Function $g(y)$ is given such way that for all $y$,

\[6g(1 + (1/y)) + 12g(y + 1) = \log_{10} y\]

If $g(9) + g(26) + g(126) + g(401) = \frac {m}{n}$ where $m$ and $n$ are relatively prime integers, computer $m + n$.

Solution

Problem 2

Two 5-digit numbers are called "responsible" if they are: \begin{align*} &\text {i. In form of abcde and fghij such that fghij = 2(abcde)}\\ &\text {ii. all ten digits, a through j are all distinct.}\\ &\text {iii.} a + b + c + d + e + f + g + h + i + j = 45\end{align*}

If two "responsible" numbers are small as possible, what is the sum of the three middle digits of $\text {abcde}$ and last two digits on the $\text {fghij}$? That is, $b + c + d + i + j$.

Solution

Problem 3

A triangle is called Heronian if its perimeter equals the area of the triangle. Heronian triangle $COP$ has sides $29,6,$ and $x$. If the distance from its incenter to the circumcenter is exprssed as $\frac {\sqrt b}{a}$ where $b$ is not divisible by squares of any prime, find the remainder when $b$ is divided by $a$.

Solution

Problem 4

Eight boxes are numbered $A$ to $H$. The number of ways you can put 16 identical balls into the boxes such that none of them is empty is expressed as $\binom {a}{b}$, where $b \le \frac{a}{2}$. What is the remainder when $\binom {a}{b}$ is divided by $a + b$?

Solution

Problem 5

There are three rooms in Phil's Motel. One room for one person, one room for three people, and another one room for four people. Let's say Peter and his seven friends came to the Phil's Motel. How many ways are there to house Peter and his seven friends in these rooms?

Solution

Problem 6

Larry and his two friends toss a die four times. From all possible outcomes, the probability that outcome has at least one occurrence of 2 is $\frac {p}{q}$ where $p$ and $q$ are relatively prime integers. Find $|p - q|$.

Solution

Problem 7

In $\triangle RST$, X is on $\overline {RT}$, dividing $RX:XT = 1:2$. Y is on $\overline {ST}$, dividing $SY:YT = 2:1$. V is on $\overline {XY}$, dividing $XV:VY = 1:2$. It is found that ray VT intersects $\overline {RS}$ at Z. Find

\[128 (\frac {TV}{VZ} + \frac {RZ}{ZS})\]

Solution

Problem 8

Let $x,y, \in \mathbb {Z}$ and that:

\[x + \sqrt y = \sqrt {22 + \sqrt {384}}.\]

If $\frac {x}{y} = \frac {p}{q}$ with $(p,q) = 1$, then what is $p + q$?

Solution

Problem 9

On a circle with center $\zeta$, two points, $X$ and $Y$ are on a circle and $Z$ is outside the circle but in ray XY. $\overline {XY} = 33.6, \overline {YZ} = 39.2,$ and $\overline {\zeta X} = 21$. If $\overline {\zeta Z} = \frac {j}{k}$ where $j$ and $k$ are relatively prime integers, and such that $\frac {j + k}{j - k} = \frac {o}{p}$, find $o + p.$

Solution

Problem 10

Given that:

\begin{align*}(\frac {1}{r})(\frac {1}{s})(\frac {1}{t}) &= \frac {3}{391} \\ r + \frac {1}{s} &= \frac {35}{46} \\ s + \frac {1}{t} &= \frac {1064}{23} \\ t + \frac {1}{r} &= \frac {529}{102}.\end{align*}

Then what is the smallest integer that is divisible $rs$ and $12t$?

Solution

Problem 11

There are $z$ number of ways to represent 10000000 as product of three factors and $Z$ number of ways to represent 11390625 as product of three factors. If $|z - Z| = p^q$, find $p + q$.

Solution

Problem 12

Let $m = 101^4 + 256$. Find the sum of digits of $m$.

Solution

Problem 13

If the reciprocal of sum of real roots of the following equation can be written in form of $\frac {r}{s}$ where $(r,s) = 1$, find $r + s$.

$1000x^6 - 1900x^5 - 1400x^4 - 190x^3 - 130x^2 - 38x - 30 = 0$

Solution

Problem 14

The set $Z$ contains complex numbers $\zeta_0,\zeta_1,\zeta_2...$ such that $n = 0,1,2,3.....$.

$\zeta_n$ is defined this way:

$\zeta_{n + 1} = (\frac {\zeta_n - i}{\zeta_n + i})^{- 1}$

If $\zeta_0 = i + \frac{1}{121}$ and 2008th number of the set is in form of ($a + bi$), find $ab$.

Solution

Problem 15

Call the value of $[\log_2 1] + [\log_2 2] + [\log_2 3] + ....[\log_2 64]$ as $\alpha$. Let the value of $[\log_2 1] + [\log_2 2] + [\log_2 3] + ....[\log_2 52]$ as $\delta$. Find the absolute value of $\delta - \alpha$.

Solution

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