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Difference between revisions of "1985 AIME Problems/Problem 5"

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== Problem ==
 
== Problem ==
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A [[sequence]] of [[integer]]s <math>a_1, a_2, a_3, \ldots</math> is chosen so that <math>a_n = a_{n - 1} - a_{n - 2}</math> for each <math>n \ge 3</math>. What is the sum of the first 2001 terms of this sequence if the sum of the first 1492 terms is 1985, and the sum of the first 1985 terms is 1492?
 
A [[sequence]] of [[integer]]s <math>a_1, a_2, a_3, \ldots</math> is chosen so that <math>a_n = a_{n - 1} - a_{n - 2}</math> for each <math>n \ge 3</math>. What is the sum of the first 2001 terms of this sequence if the sum of the first 1492 terms is 1985, and the sum of the first 1985 terms is 1492?
  
 
== Solution ==
 
== Solution ==
 +
 
The problem gives us a sequence defined by a [[recursion]], so let's calculate a few values to get a feel for how it acts.  We aren't given initial values, so let <math>a_1 = a</math> and <math>a_2 = b</math>.  Then <math>a_3 = b - a</math>, <math>a_4 = (b - a) - b = -a</math>, <math>a_5 = -a - (b - a) = -b</math>, <math>a_6 = -b - (-a) = a - b</math>, <math>a_7 = (a - b) - (-b) = a</math> and <math>a_8 = a - (a - b) = b</math>.  Since the sequence is recursively defined by the first 2 terms, after this point it must continue to repeat.  Thus, in particular <math>a_{j + 6} = a_j</math> for all <math>j</math>, and so repeating this <math>n</math> times, <math>a_{j + 6n} = a_j</math> for all [[integer]]s <math>n</math> and <math>j</math>.   
 
The problem gives us a sequence defined by a [[recursion]], so let's calculate a few values to get a feel for how it acts.  We aren't given initial values, so let <math>a_1 = a</math> and <math>a_2 = b</math>.  Then <math>a_3 = b - a</math>, <math>a_4 = (b - a) - b = -a</math>, <math>a_5 = -a - (b - a) = -b</math>, <math>a_6 = -b - (-a) = a - b</math>, <math>a_7 = (a - b) - (-b) = a</math> and <math>a_8 = a - (a - b) = b</math>.  Since the sequence is recursively defined by the first 2 terms, after this point it must continue to repeat.  Thus, in particular <math>a_{j + 6} = a_j</math> for all <math>j</math>, and so repeating this <math>n</math> times, <math>a_{j + 6n} = a_j</math> for all [[integer]]s <math>n</math> and <math>j</math>.   
  
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<cmath>\sum_{i = 1}^{1492} a_i = (a_{1489} + a_{1490} + a_{1491} + a_{1492})+ \sum_{i = 1}^{1488} a_i = (a_1 + a_2 + a_3 + a_4) + \sum_{n = 0}^{247}\sum_{j = 1}^6 a_{6n + j}</cmath>
 
<cmath>\sum_{i = 1}^{1492} a_i = (a_{1489} + a_{1490} + a_{1491} + a_{1492})+ \sum_{i = 1}^{1488} a_i = (a_1 + a_2 + a_3 + a_4) + \sum_{n = 0}^{247}\sum_{j = 1}^6 a_{6n + j}</cmath>
 
<cmath>=(a + b + (b - a) + (-a)) + \sum_{n = 0}^{247}\sum_{j = 1}^6 a_j = 2b - a,</cmath>
 
<cmath>=(a + b + (b - a) + (-a)) + \sum_{n = 0}^{247}\sum_{j = 1}^6 a_j = 2b - a,</cmath>
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where we can make this last step because <math>\sum_{j = 1}^6 a_j = 0</math> and so the entire second term of our [[expression]] is [[zero (constant) | zero]].
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where we can make this last step because <math>\sum_{j = 1}^6 a_j = 0</math> and so the entire second term of our [[expression]] is [[0|zero]].
  
 
Similarly, since <math>1980 = 6 \cdot 330</math>, <math>\sum_{i = 1}^{1985} a_i = (a_1 + a_2 + a_3 + a_4 + a_5) + \sum_{i = 1}^{1980}a_i = a + b + (b - a) + (-a) + (-b) = b - a</math>.
 
Similarly, since <math>1980 = 6 \cdot 330</math>, <math>\sum_{i = 1}^{1985} a_i = (a_1 + a_2 + a_3 + a_4 + a_5) + \sum_{i = 1}^{1980}a_i = a + b + (b - a) + (-a) + (-b) = b - a</math>.
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Then by the givens, <math>2b - a = 1985</math> and <math>b - a = 1492</math> so <math>b = 1985 - 1492 = 493</math> and so the answer is <math>2\cdot 493 = \boxed{986}</math>.
 
Then by the givens, <math>2b - a = 1985</math> and <math>b - a = 1492</math> so <math>b = 1985 - 1492 = 493</math> and so the answer is <math>2\cdot 493 = \boxed{986}</math>.
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Minor edit by: PlainOldNumberTheory
 
  
 
== See also ==
 
== See also ==
 +
 
{{AIME box|year=1985|num-b=4|num-a=6}}
 
{{AIME box|year=1985|num-b=4|num-a=6}}
 +
{{MAA Notice}}
 
* [[AIME Problems and Solutions]]
 
* [[AIME Problems and Solutions]]
 
* [[American Invitational Mathematics Examination]]
 
* [[American Invitational Mathematics Examination]]

Latest revision as of 10:20, 15 February 2025

Problem

A sequence of integers $a_1, a_2, a_3, \ldots$ is chosen so that $a_n = a_{n - 1} - a_{n - 2}$ for each $n \ge 3$. What is the sum of the first 2001 terms of this sequence if the sum of the first 1492 terms is 1985, and the sum of the first 1985 terms is 1492?

Solution

The problem gives us a sequence defined by a recursion, so let's calculate a few values to get a feel for how it acts. We aren't given initial values, so let $a_1 = a$ and $a_2 = b$. Then $a_3 = b - a$, $a_4 = (b - a) - b = -a$, $a_5 = -a - (b - a) = -b$, $a_6 = -b - (-a) = a - b$, $a_7 = (a - b) - (-b) = a$ and $a_8 = a - (a - b) = b$. Since the sequence is recursively defined by the first 2 terms, after this point it must continue to repeat. Thus, in particular $a_{j + 6} = a_j$ for all $j$, and so repeating this $n$ times, $a_{j + 6n} = a_j$ for all integers $n$ and $j$.

Because of this, the sum of the first 1492 terms can be greatly simplified: $1488 = 6 \cdot 248$ is the largest multiple of 6 less than 1492, so \[\sum_{i = 1}^{1492} a_i = (a_{1489} + a_{1490} + a_{1491} + a_{1492})+ \sum_{i = 1}^{1488} a_i = (a_1 + a_2 + a_3 + a_4) + \sum_{n = 0}^{247}\sum_{j = 1}^6 a_{6n + j}\] \[=(a + b + (b - a) + (-a)) + \sum_{n = 0}^{247}\sum_{j = 1}^6 a_j = 2b - a,\] where we can make this last step because $\sum_{j = 1}^6 a_j = 0$ and so the entire second term of our expression is zero.

Similarly, since $1980 = 6 \cdot 330$, $\sum_{i = 1}^{1985} a_i = (a_1 + a_2 + a_3 + a_4 + a_5) + \sum_{i = 1}^{1980}a_i = a + b + (b - a) + (-a) + (-b) = b - a$.

Finally, $\sum_{i = 1}^{2001}a_i = a_1 + a_2 + a_3 + \sum_{i = 1}^{1998} a_i = a + b + (b - a) = 2b$.

Then by the givens, $2b - a = 1985$ and $b - a = 1492$ so $b = 1985 - 1492 = 493$ and so the answer is $2\cdot 493 = \boxed{986}$.

See also

1985 AIME (Problems • Answer Key • Resources)
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Problem 6
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