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

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==Problem==
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A strictly increasing sequence of positive integers <math>a_1</math>, <math>a_2</math>, <math>a_3</math>, <math>\cdots</math> has the property that for every positive integer <math>k</math>, the subsequence <math>a_{2k-1}</math>, <math>a_{2k}</math>, <math>a_{2k+1}</math> is geometric and the subsequence <math>a_{2k}</math>, <math>a_{2k+1}</math>, <math>a_{2k+2}</math> is arithmetic. Suppose that <math>a_{13} = 2016</math>. Find <math>a_1</math>.
 
A strictly increasing sequence of positive integers <math>a_1</math>, <math>a_2</math>, <math>a_3</math>, <math>\cdots</math> has the property that for every positive integer <math>k</math>, the subsequence <math>a_{2k-1}</math>, <math>a_{2k}</math>, <math>a_{2k+1}</math> is geometric and the subsequence <math>a_{2k}</math>, <math>a_{2k+1}</math>, <math>a_{2k+2}</math> is arithmetic. Suppose that <math>a_{13} = 2016</math>. Find <math>a_1</math>.
  
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==Solution==
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==Solution 1==
  
 
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,
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<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>
  
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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>.
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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, <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 still staying as positive integers. <math>a_{13}=2016=14\cdot 144</math>, so <math>a_1=14\cdot 36=\fbox{504}</math>.
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~IYN~
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==Solution 2==
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Setting <math>a_1 = a</math> and <math>a_2 = ka</math>, the sequence becomes:
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<cmath>a, ka, k^2a, k(2k-1)a, (2k-1)^2a, (2k-1)(3k-2)a, (3k-2)^2a, \cdots</cmath>
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and so forth, with <math>a_{2n+1} = (nk-(n-1))^2a</math>. Then, <math>a_{13} = (6k-5)^2a = 2016</math>. Keep in mind, <math>k</math> need not be an integer, only <math>k^2a, (k+1)^2a,</math> etc. does. <math>2016 = 2^5*3^2*7</math>, so only the squares <math>1, 4, 9, 16, 36,</math> and <math>144</math> are plausible for <math>(6k-5)^2</math>. But when that is anything other than <math>2</math>, <math>k^2a</math> is not an integer. Therefore, <math>a = 2016/2^2 = 504</math>.
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Thanks for reading, Rowechen Zhong.
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==Solution 3(Very Risky and Very Stupid)==
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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>\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>.
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-Pleaseletmewin
  
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~Iy31n~
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== See also ==
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{{AIME box|year=2016|n=I|num-b=9|num-a=11}}
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{{MAA Notice}}

Revision as of 12:59, 15 February 2021

Problem

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 1

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. This can be proven by induction. 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 still staying as positive integers. $a_{13}=2016=14\cdot 144$, so $a_1=14\cdot 36=\fbox{504}$.

~IYN~

Solution 2

Setting $a_1 = a$ and $a_2 = ka$, the sequence becomes:

\[a, ka, k^2a, k(2k-1)a, (2k-1)^2a, (2k-1)(3k-2)a, (3k-2)^2a, \cdots\] and so forth, with $a_{2n+1} = (nk-(n-1))^2a$. Then, $a_{13} = (6k-5)^2a = 2016$. Keep in mind, $k$ need not be an integer, only $k^2a, (k+1)^2a,$ etc. does. $2016 = 2^5*3^2*7$, so only the squares $1, 4, 9, 16, 36,$ and $144$ are plausible for $(6k-5)^2$. But when that is anything other than $2$, $k^2a$ is not an integer. Therefore, $a = 2016/2^2 = 504$.

Thanks for reading, Rowechen Zhong.


Solution 3(Very Risky and Very Stupid)

The thirteenth term of the sequence is $2016$, which makes that fourteenth term of the sequence $2016+r$ and the $15^{\text{th}}$ term $\frac{(2016+r)^2}{2016}$. We note that $r$ is an integer so that means $\frac{r^2}{2016}$ is an integer. Thus, we assume the smallest value of $r$, which is $168$. We bash all the way back to the first term and get our answer of $\boxed{504}$.

-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

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