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 "1985 AIME Problems/Problem 5"

(→‎Solution: 360 is changed to 330 since 6 * 360 = 2160, not 1980.)
(→‎Solution)
 
Line 15: Line 15:
  
 
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>.
 +
Minor edit by: PlainOldNumberTheory
  
 
== See also ==
 
== See also ==

Latest revision as of 12:37, 1 May 2022

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 $2\cdot 493 = \boxed{986}$. Minor edit by: PlainOldNumberTheory

See also

1985 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1985_AIME_Problems/Problem_5&oldid=173698"