# Difference between revisions of "2016 AIME I Problems/Problem 10"

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We first create a similar sequence where <math>a_1=1</math> and <math>a_2=2</math>. Continuing the sequence, | We first create a similar sequence where <math>a_1=1</math> and <math>a_2=2</math>. Continuing the sequence, | ||

− | < | + | <cmath>1, 2,4,6,9,12,16,20,25,30,36,42,49,\cdots</cmath> |

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. Similarly, <math>a_{13}</math> would also need to be the end of a geometric sequence (divisible by a square). We see that <math>2016</math> is <math>2^5 \cdot 3^2 \cdot 7</math>, so the squares that would fit in <math>2016</math> are <math>1^2=1</math>, <math>2^2=4</math>, <math>3^2=9</math>, <math>2^4=16</math>, <math>2^2 \cdot 3^2 = 36</math>, and <math>2^4 \cdot 3^2 = 144</math>. By simple inspection <math>144</math> is the only plausible square, since the other squares don't have enough numbers before them to go all the way back to <math>a_1</math>. <math>a_{13}=2016=14\cdot 144</math>, so <math>a_1=14\cdot 36=\fbox{504}</math>. | 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. Similarly, <math>a_{13}</math> would also need to be the end of a geometric sequence (divisible by a square). We see that <math>2016</math> is <math>2^5 \cdot 3^2 \cdot 7</math>, so the squares that would fit in <math>2016</math> are <math>1^2=1</math>, <math>2^2=4</math>, <math>3^2=9</math>, <math>2^4=16</math>, <math>2^2 \cdot 3^2 = 36</math>, and <math>2^4 \cdot 3^2 = 144</math>. By simple inspection <math>144</math> is the only plausible square, since the other squares don't have enough numbers before them to go all the way back to <math>a_1</math>. <math>a_{13}=2016=14\cdot 144</math>, so <math>a_1=14\cdot 36=\fbox{504}</math>. | ||

~Iy31n~ | ~Iy31n~ |

## Revision as of 15:51, 4 March 2016

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

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. 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 don't have enough numbers before them to go all the way back to . , so .

~Iy31n~