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

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<math> \textbf{(A)}\ -3 \qquad\textbf{(B)}\ -1 \qquad\textbf{(C)}\ 0 \qquad\textbf{(D)}\ 1 \qquad\textbf{(E)}\ 3 </math>
 
<math> \textbf{(A)}\ -3 \qquad\textbf{(B)}\ -1 \qquad\textbf{(C)}\ 0 \qquad\textbf{(D)}\ 1 \qquad\textbf{(E)}\ 3 </math>
 
==Solution==
 
==Solution==
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We work backwards to find <math>a</math>.
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<cmath> d+0=1\implies d=1</cmath>
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<cmath>c+1=0\implies c=-1</cmath>
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<cmath> b+(-1)=1\implies b=2</cmath>
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<cmath>a+2=-1\implies a=\boxed{\textbf{(A)}-3.}</cmath>
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--Solution by [http://www.artofproblemsolving.com/Forum/memberlist.php?mode=viewprofile&u=200685 TheMaskedMagician]
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==See Also==
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{{AHSME box|year=1994|num-b=5|num-a=7}}
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{{MAA Notice}}

Latest revision as of 17:27, 9 January 2021

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 $a$.

\[d+0=1\implies d=1\] \[c+1=0\implies c=-1\] \[b+(-1)=1\implies b=2\] \[a+2=-1\implies a=\boxed{\textbf{(A)}-3.}\]

--Solution by TheMaskedMagician

See Also

1994 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
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 26 27 28 29 30
All AHSME Problems and Solutions

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