Difference between revisions of "2006 AIME I Problems/Problem 9"
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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> | 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 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}) \ | <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> | = \log_8(a\cdot ar\cdot ar^2\cdot \cdots \cdot ar^{11}) = \log_8 (a^{12}r^{66}) </cmath> | ||
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Since <math>1003/11 = 91 + 2/11</math>, the answer is just the number of odd integers in <math>[1,91]</math>, which is <math>46</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 <math>46</math>. | ||
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== Solution 2 == | == Solution 2 == | ||
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>. |
Revision as of 18:34, 29 December 2015
Contents
[hide]Problem
The sequence is geometric with and common ratio where and are positive integers. Given that find the number of possible ordered pairs
Solution 1
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 .
Another way is to write
Since , the answer is just the number of odd integers in , which is .
Solution 2
Using the above method, we can derive that . Now, think about what happens when r is an even power of 2. Then 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 , , .... all work for r, until r hits , when it gets greater than , so the greatest value for r is . All that's left is to count the number of odd integers between 1 and 91 (inclusive), which yields .
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 |
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