Difference between revisions of "1985 AIME Problems/Problem 5"
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Finally, <math>\sum_{i = 1}^{2001}a_i = a_1 + a_2 + a_3 + \sum_{i = 1}^{1998} a_i = a + b + (b - a) = 2b</math>. | Finally, <math>\sum_{i = 1}^{2001}a_i = a_1 + a_2 + a_3 + \sum_{i = 1}^{1998} a_i = a + b + (b - a) = 2b</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 = 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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== See also == | == See also == | ||
Revision as of 19:11, 1 May 2016
Problem
A sequence of integers 
is chosen so that 
for each 
. 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 
and 
. Then 
, 
, 
, 
, 
and 
. Since the sequence is recursively defined by the first 2 terms, after this point it must continue to repeat. Thus, in particular 
for all 
, and so repeating this 
times, 
for all integers and .
Because of this, the sum of the first 1492 terms can be greatly simplified: 
is the largest multiple of 6 less than 1492, so 
, where we can make this last step because 
and so the entire second term of our expression is zero.
Similarly, since 
, 
.
Finally, 
.
Then by the givens, 
and 
so 
and so the answer is 
.
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 | ||
