Difference between revisions of "2006 AIME I Problems/Problem 9"
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== 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 == | == Solution == | ||
− | < | + | <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>. | |
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− | So our question is equivalent to solving <math>\log_8 (a^{12}r^{66})=2006</math> for <math>a, r</math> [[positive integer]]s. | ||
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− | <math>a^{12}r^{66}=8^{2006} = (2^3)^{2006} = (2^6)^{1003}</math> so | ||
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− | <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>: | 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> | |
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− | + | 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>. | |
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− | For <math>y</math> to be an integer, the [[numerator]] must be [[divisible]] by 11. This occurs when <math>x=1</math> because <math>1001=91*11</math>. | ||
== See also == | == See also == | ||
{{AIME box|year=2006|n=I|num-b=8|num-a=10}} | {{AIME box|year=2006|n=I|num-b=8|num-a=10}} | ||
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[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] | ||
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[[Category:Intermediate Number Theory Problems]] | [[Category:Intermediate Number Theory Problems]] |
Revision as of 17:52, 28 November 2007
Problem
The sequence is geometric with and common ratio where and are positive integers. Given that find the number of possible ordered pairs
Solution
So our question is equivalent to solving for positive integers. so .
The product of and is a power of 2. Since both numbers have to be integers, this means that and are themselves powers of 2. Now, let and :
For to be an integer, the numerator must be divisible by . This occurs when because . Because only even integers are being subtracted from , the numerator never equals an even multiple of . Therefore, the numerator takes on the value of every odd multiple of from to . Since the odd multiples are separated by a distance of , the number of ordered pairs that work is . (We must add 1 because both endpoints are being included.) So the answer is .
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 |