Difference between revisions of "2016 AIME I Problems/Problem 10"
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− | The thirteenth term of the sequence is <math>2016</math>, which makes that fourteenth term of the sequence <math>2016+r</math> and the <math>15^{\text{th}}</math> term <math> | + | The thirteenth term of the sequence is <math>2016</math>, which makes that fourteenth term of the sequence <math>2016+r</math> and the <math>15^{\text{th}}</math> term <math>\frac{(2016+r)^2}{2016}</math>. We note that <math>r</math> is an integer so that means <math>\frac{r^2}{2016}</math> is an integer. Thus, we assume the smallest value of <math>r</math>, which is <math>168</math>. We bash all the way back to the first term and get our answer of <math>\boxed{504}</math>. |
-Pleaseletmewin | -Pleaseletmewin |
Revision as of 11:59, 15 February 2021
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(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
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 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.