Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2006 AIME I Problems/Problem 9"

(→‎Solution)
(→‎Solution 3)
 
(19 intermediate revisions by 11 users not shown)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
−
The sequence <math> a_1, a_2, \ldots </math> is geometric with <math> a_1=a </math> and common ratio <math> r, </math> where <math> a </math> and <math> r </math> are positive integers. Given that <math> \log_8 a_1+\log_8 a_2+\cdots+\log_8 a_{12} = 2006, </math> find the number of possible ordered pairs <math> (a,r). </math>
+
The [[sequence]] <math> a_1, a_2, \ldots </math> is [[geometric sequence|geometric]] with <math> a_1=a </math> and common [[ratio]] <math> r, </math> where <math> a </math> and <math> r </math> are positive integers. Given that <math> \log_8 a_1+\log_8 a_2+\cdots+\log_8 a_{12} = 2006, </math> find the number of possible ordered pairs <math> (a,r). </math>
  
 +
== Solution 1 ==
 +
<cmath>\log_8 a_1+\log_8 a_2+\ldots+\log_8 a_{12}= \log_8 a+\log_8 (ar)+\ldots+\log_8 (ar^{11}) \\
 +
= \log_8(a\cdot ar\cdot ar^2\cdot \cdots \cdot ar^{11}) = \log_8  (a^{12}r^{66}) </cmath>
  
 +
So our question is equivalent to solving <math>\log_8 (a^{12}r^{66})=2006</math> for <math>a, r</math> [[positive integer]]s. <math>a^{12}r^{66}=8^{2006} = (2^3)^{2006} = (2^6)^{1003}</math> so <math>a^{2}r^{11}=2^{1003}</math>.
  
 +
The product of <math>a^2</math> and <math>r^{11}</math> is a power of 2.  Since both numbers have to be integers, this means that <math>a</math> and <math>r</math> are themselves powers of 2.  Now, let <math>a=2^x</math> and <math>r=2^y</math>:
  
 +
<cmath>\begin{eqnarray*}(2^x)^2\cdot(2^y)^{11}&=&2^{1003}\\
 +
2^{2x}\cdot 2^{11y}&=&2^{1003}\\
 +
2x+11y&=&1003\\
 +
y&=&\frac{1003-2x}{11} \end{eqnarray*}</cmath>
  
−
== Solution ==
+
For <math>y</math> to be an integer, the [[numerator]] must be [[divisible]] by <math>11</math>.  This occurs when <math>x=1</math> because <math>1001=91*11</math>. Because only [[even integer]]s are being subtracted from <math>1003</math>, the numerator never equals an even [[multiple]] of <math>11</math>. Therefore, the numerator takes on the value of every [[odd integer | odd]] multiple of <math>11</math> from <math>11</math> to <math>1001</math>.  Since the odd multiples are separated by a distance of <math>22</math>, the number of ordered pairs that work is <math>1 + \frac{1001-11}{22}=1 + \frac{990}{22}=46</math>.  (We must add 1 because both endpoints are being included.) So the answer is <math>\boxed{046}</math>.
−
<math>\log_8 a_1+\log_8 a_2+\ldots+\log_8 a_{12}=
 
−
    \log_8 a+log_8 (ar)+\ldots+\log_8 (ar^{11})</math>
 
  
−
<math>\log_8 a_1+\log_8 a_2+\ldots+\log_8 a_{12}=
 
−
    \log_8 (a*ar*ar^2*\cdots*ar^{11})</math>
 
  
−
<math>\log_8 a_1+\log_8 a_2+\ldots+\log_8 a_{12}= \log_8 (a^{12}r^{66})</math>
+
For the step above, you may also simply do <math>1001/11 + 1 = 91 + 1 = 92</math> to find how many multiples of <math>11</math> there are in between <math>11</math> and <math>1001</math>. Then, divide <math>92/2</math> = <math>\boxed{046}</math> to find only the odd solutions. <math>-XxHalo711</math>
  
−
<math>\log_8 (a^{12}r^{66})=2006</math>
+
--------------
 +
Another way is to write
  
−
<math>a^{12}r^{66}=8^{2006}</math>
+
<math>x = \frac{1003-11y}2</math>
  
−
<math>a^{12}r^{66}=(2^3)^{2006}</math>
+
Since <math>1003/11 = 91 + 2/11</math>, the answer is just the number of odd integers in <math>[1,91]</math>, which is, again,  <math>\boxed{046}</math>.
 +
----------------
  
−
<math>a^{2}r^{11}=2^{1003}</math>
+
== Solution 2 ==
 +
Using the above method, we can derive that <math>a^{2}r^{11} = 2^{1003}</math>.
 +
Now, think about what happens when r is an even power of 2. Then <math>a^{2}</math> must be an odd power of 2 in order to satisfy the equation which is clearly not possible. Thus the only restriction r has is that it must be an odd power of 2, so <math>2^{1}</math>, <math>2^{3}</math>, <math>2^{5}</math> .... all work for r, until r hits <math>2^{93}</math>, when it gets greater than <math>2^{1003}</math>, so the greatest value for r is <math>2^{91}</math>. All that's left is to count the number of odd integers between 1 and 91 (inclusive), which yields <math>\boxed{046}</math>.
  
−
The product of <math>a^2</math> and <math>r^{11}</math> is a power of 2.  Since both numbers have to be integers, this means that a and r are also powers of 2.  Now, let <math>a=2^x</math> and <math>r=2^y</math>:
+
== Solution 3 ==
 +
Using the method from Solution 1, we get <math>\log_8a^{12}r^{66}=2006 \implies a^{12}r^{66}=8^{2006}=2^{6018}</math>.
  
−
<math>(2^x)^2*(2^y)^{11}=2^{1003}</math>
+
Since <math>a</math> and <math>r</math> both have to be powers of <math>2</math>, we can rewrite this as <math>12x+66y=6018</math>.
  
−
<math>2^{2x}*2^{11y}=2^{1003}</math>
+
<math>6018 \equiv 66 \equiv 6\pmod{12}</math>. So, when we subtract <math>12</math> from <math>6018</math>, the result is divisible by <math>66</math>. Evaluating that, we get <math>(1,91)</math> as a valid solution. Since <math>66 \cdot 2 = 12 \cdot 11</math>, when we add <math>11</math> to the value of <math>a</math>, we can subtract <math>2</math> from the value of <math>r</math> to keep the equation valid. Using this, we get <math>(1,91),(12,89),(23,87), \cdots (541,1)</math>. In order to count the number of ordered pairs, we can simply count the number of <math>y</math> values. Every odd number from <math>1</math> to <math>91</math> is included, so we have <math>\boxed{046}</math> solutions.
  
−
<math>2^{2x+11y}=2^{1003}</math>
+
-Phunsukh Wangdu
−
 
 
−
<math>2x+11y=1003</math>
 
−
 
 
−
<math>11y=1003-2x</math>
 
−
 
 
−
<math>y=\frac{1003-2x}{11}</math>
 
−
 
 
−
For y to be an integer, the numerator must be divisible by 2.  This occurs when <math>x=1</math> because <math>1001=91*11</math>.  Because only even numbers are being subtracted from 1003, the numerator never equals an even multiple of 11.  Therefore, the numerator takes on the value of every odd multiple of 11 from 11 to 1001.  Since the odd multiples are separated by a distance of 22, the number of ordered pairs that work is <math>\frac{1001-11}{22}=\frac{990}{22}=45</math>.  We must add 1 because both endpoints are being included, so the answer is therefore 46.
 
  
 
== See also ==
 
== See also ==
−
* [[2006 AIME I Problems]]
+
{{AIME box|year=2006|n=I|num-b=8|num-a=10}}
  
−
[[Category:Intermediate Geometry Problems]]
+
[[Category:Intermediate Algebra Problems]]
 +
[[Category:Intermediate Number Theory Problems]]
 +
{{MAA Notice}}

Latest revision as of 18:27, 22 May 2021

Problem

The sequence $a_1, a_2, \ldots$ is geometric with $a_1=a$ and common ratio $r,$ where $a$ and $r$ are positive integers. Given that $\log_8 a_1+\log_8 a_2+\cdots+\log_8 a_{12} = 2006,$ find the number of possible ordered pairs $(a,r).$

Solution 1

\[\log_8 a_1+\log_8 a_2+\ldots+\log_8 a_{12}= \log_8 a+\log_8 (ar)+\ldots+\log_8 (ar^{11}) \\ = \log_8(a\cdot ar\cdot ar^2\cdot \cdots \cdot ar^{11}) = \log_8 (a^{12}r^{66})\]

So our question is equivalent to solving $\log_8 (a^{12}r^{66})=2006$ for $a, r$ positive integers. $a^{12}r^{66}=8^{2006} = (2^3)^{2006} = (2^6)^{1003}$ so $a^{2}r^{11}=2^{1003}$.

The product of $a^2$ and $r^{11}$ is a power of 2. Since both numbers have to be integers, this means that $a$ and $r$ are themselves powers of 2. Now, let $a=2^x$ and $r=2^y$:

\begin{eqnarray*}(2^x)^2\cdot(2^y)^{11}&=&2^{1003}\\ 2^{2x}\cdot 2^{11y}&=&2^{1003}\\ 2x+11y&=&1003\\ y&=&\frac{1003-2x}{11} \end{eqnarray*}

For $y$ to be an integer, the numerator must be divisible by $11$. This occurs when $x=1$ because $1001=91*11$. Because only even integers are being subtracted from $1003$, the numerator never equals an even multiple of $11$. Therefore, the numerator takes on the value of every odd multiple of $11$ from $11$ to $1001$. Since the odd multiples are separated by a distance of $22$, the number of ordered pairs that work is $1 + \frac{1001-11}{22}=1 + \frac{990}{22}=46$. (We must add 1 because both endpoints are being included.) So the answer is $\boxed{046}$.


For the step above, you may also simply do $1001/11 + 1 = 91 + 1 = 92$ to find how many multiples of $11$ there are in between $11$ and $1001$. Then, divide $92/2$ = $\boxed{046}$ to find only the odd solutions. $-XxHalo711$

Another way is to write

$x = \frac{1003-11y}2$

Since $1003/11 = 91 + 2/11$, the answer is just the number of odd integers in $[1,91]$, which is, again, $\boxed{046}$.

Solution 2

Using the above method, we can derive that $a^{2}r^{11} = 2^{1003}$. Now, think about what happens when r is an even power of 2. Then $a^{2}$ must be an odd power of 2 in order to satisfy the equation which is clearly not possible. Thus the only restriction r has is that it must be an odd power of 2, so $2^{1}$, $2^{3}$, $2^{5}$ .... all work for r, until r hits $2^{93}$, when it gets greater than $2^{1003}$, so the greatest value for r is $2^{91}$. All that's left is to count the number of odd integers between 1 and 91 (inclusive), which yields $\boxed{046}$.

Solution 3

Using the method from Solution 1, we get $\log_8a^{12}r^{66}=2006 \implies a^{12}r^{66}=8^{2006}=2^{6018}$.

Since $a$ and $r$ both have to be powers of $2$, we can rewrite this as $12x+66y=6018$.

$6018 \equiv 66 \equiv 6\pmod{12}$. So, when we subtract $12$ from $6018$, the result is divisible by $66$. Evaluating that, we get $(1,91)$ as a valid solution. Since $66 \cdot 2 = 12 \cdot 11$, when we add $11$ to the value of $a$, we can subtract $2$ from the value of $r$ to keep the equation valid. Using this, we get $(1,91),(12,89),(23,87), \cdots (541,1)$. In order to count the number of ordered pairs, we can simply count the number of $y$ values. Every odd number from $1$ to $91$ is included, so we have $\boxed{046}$ solutions.

-Phunsukh Wangdu

See also

2006 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2006_AIME_I_Problems/Problem_9&oldid=154112"