Difference between revisions of "1969 AHSME Problems/Problem 4"

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== Solution ==
 
== Solution ==
<math>\fbox{E}</math>
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Performing the operation based on the definition, <math>(3,3)\star(0,0) = (3,3)</math> and <math>(x,y)\star(3,2)=(x-3,y+2)</math>.  Because the outputs are identical pairs, they must equal each other, so <math>3 = x-3</math>.  Solving for x yields <math>x = 6</math>, which is answer choice <math>\boxed{\textbf{(E)}}</math>.
  
 
== See also ==
 
== See also ==
{{AHSME box|year=1969|num-b=3|num-a=5}}   
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{{AHSME 35p box|year=1969|num-b=3|num-a=5}}   
  
 
[[Category: Introductory Algebra Problems]]
 
[[Category: Introductory Algebra Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 03:13, 7 June 2018

Problem

Let a binary operation $\star$ on ordered pairs of integers be defined by $(a,b)\star (c,d)=(a-c,b+d)$. Then, if $(3,3)\star (0,0)$ and $(x,y)\star (3,2)$ represent identical pairs, $x$ equals:

$\text{(A) } -3\quad \text{(B) } 0\quad \text{(C) } 2\quad \text{(D) } 3\quad \text{(E) } 6$

Solution

Performing the operation based on the definition, $(3,3)\star(0,0) = (3,3)$ and $(x,y)\star(3,2)=(x-3,y+2)$. Because the outputs are identical pairs, they must equal each other, so $3 = x-3$. Solving for x yields $x = 6$, which is answer choice $\boxed{\textbf{(E)}}$.

See also

1969 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
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