2016 AIME II Problems/Problem 9
Contents
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 (No trial and error)
We have and . First, implies , so .
It follows that , i.e., Moreover, since is atleast we get , i.e. . For every value of in this range, define , and define . We are looking for values of such that . Let's make a table: The only admissible values for are . However, since , we must have . This does not hold for because does not divide . This leaves as the only option.
For and , we check: implies , i.e. . Then and and ; so it works! Then .
Solution 3
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 4 (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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