University of South Carolina High School Math Contest/1993 Exam/Problem 2

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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University of South Carolina High School Math Contest/1993 Exam/Problem 2

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