2016 AIME I Problems/Problem 10
Contents
Problem
A strictly increasing sequence of positive integers , , , has the property that for every positive integer , the subsequence , , is geometric and the subsequence , , is arithmetic. Suppose that . Find .
Solution 1
We first create a similar sequence where and . Continuing the sequence,
Here we can see a pattern; every second term (starting from the first) is a square, and every second term (starting from the third) is the end of a geometric sequence. This can be proven by induction. Similarly, would also need to be the end of a geometric sequence (divisible by a square). We see that is , so the squares that would fit in are , , , , , and . By simple inspection is the only plausible square, since the other squares in the sequence don't have enough elements before them to go all the way back to while still staying as positive integers. , so .
~IYN~
Solution 2
Setting and , the sequence becomes:
and so forth, with . Then, . Keep in mind, need not be an integer, only etc. does. , so only the squares and are plausible for . But when that is anything other than , is not an integer. Therefore, .
Thanks for reading, Rowechen Zhong.
Solution 3
Instead of setting equal to something and equal to something, note that it is rather easier to set and so that and so on until you reached (Or simply notice the pattern), so . Note that since each of the terms has degree 2 so if you multiply and by you multiply each term by so each term is still a integer if the terms are already integers before you multiply and by , so let and so . Then, for the sequence to be strictly increasing positive integers we have so and ~Ddk001
Solution 4(very risky and very stupid)
The thirteenth term of the sequence is , which makes that fourteenth term of the sequence and the term . We note that is an integer so that means is an integer. Thus, we assume the smallest value of , which is . We bash all the way back to the first term and get our answer of .
-Pleaseletmewin
Solution 5
Let where . Then, . Continuing on, we get . Moreover, .
It is clear now that and . Plugging in , . The prime factorization of so we look for perfect squares.
if is an integer, and if is not an integer and is rational, so . This forces . Assuming is an integer, it can only be , .
If , . If , . Note that the latter cannot work since but in this scenario. Therefore, . Plugging back ,
Solution 6 (Headache)
Bash all the way from and to to get to the equation and then (direct consequence). Notice that from the two equations we can deduce that:
1) ,
2) , and
3) .
Since , the only possible values of that satisfy the conditions are , , and itself. Obviously cannot be a correct answer choice (from AIME rules), and if , , so is the solution.
~eevee9406
Video Solution
~MathProblemSolvingSkills.com
See also
2016 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
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