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Difference between revisions of "University of South Carolina High School Math Contest/1993 Exam/Problem 2"

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<math>3 \star x = 23 \Longrightarrow 3+x+3x=23 \Longrightarrow 4x = 20 \Longrightarrow x=5</math>, so the answer is <math>\mathrm{(D) \ }</math>.
 
<math>3 \star x = 23 \Longrightarrow 3+x+3x=23 \Longrightarrow 4x = 20 \Longrightarrow x=5</math>, so the answer is <math>\mathrm{(D) \ }</math>.
  
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Alternatively, note the resemblence to [[Simon's Favorite Factoring Trick]].  <math>(a\star b) + 1 = ab + a + b + 1 = (a + 1)(b + 1)</math> so <math>24 = (x \star 3) + 1 = (x + 1)(3 + 1)</math> so <math>x + 1 = \frac{24}4 = 6</math> and <math>x = 5</math>.
 
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Latest revision as of 11:34, 31 July 2006

Problem

Suppose the operation $\star$ is defined by $a \star b = a+b+ab.$ If $3\star x = 23,$ then $x =$

$\mathrm{(A) \ } 2 \qquad \mathrm{(B) \ }3\qquad \mathrm{(C) \ }4 \qquad \mathrm{(D) \ }5 \qquad \mathrm{(E) \ }6$

Solution

$3 \star x = 23 \Longrightarrow 3+x+3x=23 \Longrightarrow 4x = 20 \Longrightarrow x=5$, so the answer is $\mathrm{(D) \ }$.


Alternatively, note the resemblence to Simon's Favorite Factoring Trick. $(a\star b) + 1 = ab + a + b + 1 = (a + 1)(b + 1)$ so $24 = (x \star 3) + 1 = (x + 1)(3 + 1)$ so $x + 1 = \frac{24}4 = 6$ and $x = 5$.

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