Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
m (edited the wording of the solution)
Line 7: Line 7:
 
<cmath>1, 2,4,6,9,12,16,20,25,30,36,42,49,\cdots</cmath>
 
<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 in the sequence don't have enough elements before them to go all the way back to <math>a_1</math> while staying positive. <math>a_{13}=2016=14\cdot 144</math>, so <math>a_1=14\cdot 36=\fbox{504}</math>.
  
 
~Iy31n~
 
~Iy31n~

Revision as of 17:53, 4 March 2016

A strictly increasing sequence of positive integers $a_1$, $a_2$, $a_3$, $\cdots$ has the property that for every positive integer $k$, the subsequence $a_{2k-1}$, $a_{2k}$, $a_{2k+1}$ is geometric and the subsequence $a_{2k}$, $a_{2k+1}$, $a_{2k+2}$ is arithmetic. Suppose that $a_{13} = 2016$. Find $a_1$.

Solution

We first create a similar sequence where $a_1=1$ and $a_2=2$. Continuing the sequence,

\[1, 2,4,6,9,12,16,20,25,30,36,42,49,\cdots\]

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, $a_{13}$ would also need to be the end of a geometric sequence (divisible by a square). We see that $2016$ is $2^5 \cdot 3^2 \cdot 7$, so the squares that would fit in $2016$ are $1^2=1$, $2^2=4$, $3^2=9$, $2^4=16$, $2^2 \cdot 3^2 = 36$, and $2^4 \cdot 3^2 = 144$. By simple inspection $144$ 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 $a_1$ while staying positive. $a_{13}=2016=14\cdot 144$, so $a_1=14\cdot 36=\fbox{504}$.

~Iy31n~

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2016_AIME_I_Problems/Problem_10&oldid=77325"