Difference between revisions of "2016 AIME II Problems/Problem 9"
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+ | ==Problem== | ||
The sequences of positive integers <math>1,a_2, a_3,...</math> and <math>1,b_2, b_3,...</math> are an increasing arithmetic sequence and an increasing geometric sequence, respectively. Let <math>c_n=a_n+b_n</math>. There is an integer <math>k</math> such that <math>c_{k-1}=100</math> and <math>c_{k+1}=1000</math>. Find <math>c_k</math>. | The sequences of positive integers <math>1,a_2, a_3,...</math> and <math>1,b_2, b_3,...</math> are an increasing arithmetic sequence and an increasing geometric sequence, respectively. Let <math>c_n=a_n+b_n</math>. There is an integer <math>k</math> such that <math>c_{k-1}=100</math> and <math>c_{k+1}=1000</math>. Find <math>c_k</math>. | ||
− | ==Solution== | + | ==Solution 1== |
Since all the terms of the sequences are integers, and 100 isn't very big, we should just try out the possibilities for <math>b_2</math>. When we get to <math>b_2=9</math> and <math>a_2=91</math>, we have <math>a_4=271</math> and <math>b_4=729</math>, which works, therefore, the answer is <math>b_3+a_3=81+181=\boxed{262}</math>. | Since all the terms of the sequences are integers, and 100 isn't very big, we should just try out the possibilities for <math>b_2</math>. When we get to <math>b_2=9</math> and <math>a_2=91</math>, we have <math>a_4=271</math> and <math>b_4=729</math>, which works, therefore, the answer is <math>b_3+a_3=81+181=\boxed{262}</math>. | ||
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== Solution 2== | == Solution 2== | ||
− | Using the same reasoning (<math>100</math> isn't very big), we can guess which terms will work. The first case is <math>k=3</math>, so the second and fourth terms of <math>c</math> are <math>100</math> and <math>1000</math>. We let <math>r</math> be the common ratio of the geometric sequence and write the arithmetic relationships in terms of <math>r</math>. | + | Using the same reasoning (<math>100</math> isn't very big), we can guess which terms will work. The first case is <math>k=3</math>, so we assume the second and fourth terms of <math>c</math> are <math>100</math> and <math>1000</math>. We let <math>r</math> be the common ratio of the geometric sequence and write the arithmetic relationships in terms of <math>r</math>. |
The common difference is <math>100-r - 1</math>, and so we can equate: <math>2(99-r)+100-r=1000-r^3</math>. Moving all the terms to one side and the constants to the other yields | The common difference is <math>100-r - 1</math>, and so we can equate: <math>2(99-r)+100-r=1000-r^3</math>. Moving all the terms to one side and the constants to the other yields | ||
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Solution by rocketscience | Solution by rocketscience | ||
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+ | == Solution 3 (More Robust Bash) == | ||
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+ | The reason for bashing in this context can also be justified by the fact 100 isn't very big. | ||
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+ | Let the common difference for the arithmetic sequence be <math>a</math>, and the common ratio for the geometric sequence be <math>b</math>. The sequences are now <math>1, a+1, 2a+1, \ldots</math>, and <math>1, b, b^2, \ldots</math>. We can now write the given two equations as the following: | ||
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+ | <math>1+(k-2)a+b^{k-2} = 100</math> | ||
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+ | <math>1+ka+b^k = 1000</math> | ||
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+ | Take the difference between the two equations to get <math>2a+(b^2-1)b^{k-2} = 900</math>. Since 900 is divisible by 4, we can tell <math>a</math> is even and <math>b</math> is odd. Let <math>a=2m</math>, <math>b=2n+1</math>, where <math>m</math> and <math>n</math> are positive integers. Substitute variables and divide by 4 to get: | ||
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+ | <math>m+(n+1)(n)(2n+1)^{k-2} = 225</math> | ||
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+ | Because very small integers for <math>n</math> yield very big results, we can bash through all cases of <math>n</math>. Here, we set an upper bound for <math>n</math> by setting <math>k</math> as 3. After trying values, we find that <math>n\leq 4</math>, so <math>b=9, 7, 5, 3</math>. Testing out <math>b=9</math> yields the correct answer of <math>\boxed{262}</math>. Note that even if this answer were associated with another b value like <math>b=3</math>, the value of <math>k</math> can still only be 3 for all of the cases. | ||
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+ | -Dankster42 | ||
+ | |||
+ | == See also == | ||
+ | {{AIME box|year=2016|n=II|num-b=8|num-a=10}} | ||
+ | {{MAA Notice}} |
Revision as of 20:18, 18 August 2020
Problem
The sequences of positive integers and are an increasing arithmetic sequence and an increasing geometric sequence, respectively. Let . There is an integer such that and . Find .
Solution 1
Since all the terms of the sequences are integers, and 100 isn't very big, we should just try out the possibilities for . When we get to and , we have and , which works, therefore, the answer is .
Solution 2
Using the same reasoning ( isn't very big), we can guess which terms will work. The first case is , so we assume the second and fourth terms of are and . We let be the common ratio of the geometric sequence and write the arithmetic relationships in terms of .
The common difference is , and so we can equate: . Moving all the terms to one side and the constants to the other yields
, or . Simply listing out the factors of shows that the only factor less than a square that works is . Thus and we solve from there to get .
Solution by rocketscience
Solution 3 (More Robust Bash)
The reason for bashing in this context can also be justified by the fact 100 isn't very big.
Let the common difference for the arithmetic sequence be , and the common ratio for the geometric sequence be . The sequences are now , and . We can now write the given two equations as the following:
Take the difference between the two equations to get . Since 900 is divisible by 4, we can tell is even and is odd. Let , , where and are positive integers. Substitute variables and divide by 4 to get:
Because very small integers for yield very big results, we can bash through all cases of . Here, we set an upper bound for by setting as 3. After trying values, we find that , so . Testing out yields the correct answer of . Note that even if this answer were associated with another b value like , the value of can still only be 3 for all of the cases.
-Dankster42
See also
2016 AIME II (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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