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"

m (→‎Solution 1)
(→‎Solution 3)
 
(3 intermediate revisions by 2 users not shown)
Line 16: Line 16:
  
 
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>.
 
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>.
 +
 +
 +
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>
  
 
--------------
 
--------------
Line 28: Line 31:
 
Using the above method, we can derive that <math>a^{2}r^{11} = 2^{1003}</math>.
 
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>.
 
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>.
 +
 +
== 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>.
 +
 +
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>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.
 +
 +
-Phunsukh Wangdu
  
 
== See also ==
 
== See also ==

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"