Difference between revisions of "2004 AMC 12B Problems/Problem 12"

Revision as of 18:57, 3 July 2013

The following problem is from both the 2004 AMC 12B #12 and 2004 AMC 10B #19, so both problems redirect to this page.

Problem

In the sequence $2001$ , $2002$ , $2003$ , $\ldots$ , each term after the third is found by subtracting the previous term from the sum of the two terms that precede that term. For example, the fourth term is $2001 + 2002 - 2003 = 2000$ . What is the $2004^\textrm{th}$ term in this sequence?

$\mathrm{(A) \ } -2004 \qquad \mathrm{(B) \ } -2 \qquad \mathrm{(C) \ } 0 \qquad \mathrm{(D) \ } 4003 \qquad \mathrm{(E) \ } 6007$

Solution

Solution 1

We already know that $a_1=2001$ , $a_2=2002$ , $a_3=2003$ , and $a_4=2000$ . Let's compute the next few terms to get the idea how the sequence behaves. We get $a_5 = 2002+2003-2000 = 2005$ , $a_6=2003+2000-2005=1998$ , $a_7=2000+2005-1998=2007$ , and so on.

We can now discover the following pattern: $a_{2k+1} = 1999+2k$ and $a_{2k}=2004-2k$ . This is easily proved by induction. It follows that $a_{2004}=a_{2\cdot 1002} = 2004 - 2\cdot 1002 = \boxed{0}$ .

Solution 2

Note that the recurrence $a_n+a_{n+1}-a_{n+2}~=~a_{n+3}$ can be rewritten as $a_n+a_{n+1} ~=~ a_{n+2}+a_{n+3}$ .

Hence we get that $a_1+a_2 ~=~ a_3+a_4 ~=~ a_5+a_6 ~= \cdots$ and also $a_2+a_3 ~=~ a_4+a_5 ~=~ a_6+a_7 ~= \cdots$

From the values given in the problem statement we see that $a_3=a_1+2$ .

From $a_1+a_2 = a_3+a_4$ we get that $a_4=a_2-2$ .

From $a_2+a_3 = a_4+a_5$ we get that $a_5=a_3+2$ .

Following this pattern, we get $a_{2004} = a_{2002} - 2 = a_{2000} - 4 = \cdots = a_2 - 2002 = \boxed{0}$ .

2004 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

2004 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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 {{AMC12 box|year=2004|ab=B|num-b=11|num-a=13}}
 {{AMC10 box|year=2004|ab=B|num-b=18|num-a=20}}
+{{MAA Notice}}

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