Difference between revisions of "1994 AHSME Problems/Problem 6"

(Solution)
(Solution)
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We work backwards to find <math>a</math>.  
 
We work backwards to find <math>a</math>.  
  
<cmath>\begin{align*}d+0=1&\implies d=1\\ c+1=0&\implies c=-1\\ b+(-1)=1&\implies b=2\\ a+2=-1&\implies a=\boxed{\textbf{(A) }-3.}</cmath>
+
<cmath>d+0=1&\implies d=1\\ c+1=0&\implies c=-1\\ b+(-1)=1&\implies b=2\\ a+2=-1&\implies a=\boxed{\textbf{(A) }-3.}</cmath>
  
 
--Solution by [http://www.artofproblemsolving.com/Forum/memberlist.php?mode=viewprofile&u=200685 TheMaskedMagician]
 
--Solution by [http://www.artofproblemsolving.com/Forum/memberlist.php?mode=viewprofile&u=200685 TheMaskedMagician]

Revision as of 21:19, 13 February 2017

Problem

In the sequence \[..., a, b, c, d, 0, 1, 1, 2, 3, 5, 8,...\] each term is the sum of the two terms to its left. Find $a$.

$\textbf{(A)}\ -3 \qquad\textbf{(B)}\ -1 \qquad\textbf{(C)}\ 0 \qquad\textbf{(D)}\ 1 \qquad\textbf{(E)}\ 3$

Solution

We work backwards to find $a$.

\[d+0=1&\implies d=1\\ c+1=0&\implies c=-1\\ b+(-1)=1&\implies b=2\\ a+2=-1&\implies a=\boxed{\textbf{(A) }-3.}\] (Error compiling LaTeX. Unknown error_msg)

--Solution by TheMaskedMagician