Difference between revisions of "2016 AIME II Problems/Problem 9"
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==Solution 1== | ==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== |
Revision as of 13:23, 20 June 2019
Contents
[hide]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
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 |