Difference between revisions of "2006 AIME I Problems/Problem 9"
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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>. | ||
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+ | == 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>. | ||
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+ | 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>. | ||
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+ | <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. | ||
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+ | -Phunsukh Wangdu | ||
== See also == | == See also == |
Latest revision as of 17:27, 22 May 2021
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 .
For the step above, you may also simply do to find how many multiples of there are in between and . Then, divide = to find only the odd solutions.
Another way is to write
Since , the answer is just the number of odd integers in , which is, again, .
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 .
Solution 3
Using the method from Solution 1, we get .
Since and both have to be powers of , we can rewrite this as .
. So, when we subtract from , the result is divisible by . Evaluating that, we get as a valid solution. Since , when we add to the value of , we can subtract from the value of to keep the equation valid. Using this, we get . In order to count the number of ordered pairs, we can simply count the number of values. Every odd number from to is included, so we have 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 |
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