Difference between revisions of "University of South Carolina High School Math Contest/1993 Exam/Problem 2"
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== Problem == | == Problem == | ||
− | Suppose the operation <math>\star</math> is defined by <math>a \star b = a+b+ab.</math> If <math>3\star x = 23,</math> then <math>x =</math> | + | Suppose the [[operation]] <math>\star</math> is defined by <math>a \star b = a+b+ab.</math> If <math>3\star x = 23,</math> then <math>x =</math> |
<center><math> \mathrm{(A) \ } 2 \qquad \mathrm{(B) \ }3\qquad \mathrm{(C) \ }4 \qquad \mathrm{(D) \ }5 \qquad \mathrm{(E) \ }6 </math></center> | <center><math> \mathrm{(A) \ } 2 \qquad \mathrm{(B) \ }3\qquad \mathrm{(C) \ }4 \qquad \mathrm{(D) \ }5 \qquad \mathrm{(E) \ }6 </math></center> | ||
== Solution == | == Solution == | ||
− | <math>3+x+3x=23 \Longrightarrow x=5</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 is defined by
If
then
![$\mathrm{(A) \ } 2 \qquad \mathrm{(B) \ }3\qquad \mathrm{(C) \ }4 \qquad \mathrm{(D) \ }5 \qquad \mathrm{(E) \ }6$](http://latex.artofproblemsolving.com/7/f/c/7fca7225123463e8169320ea40b76b9cb84e5c09.png)
Solution
, so the answer is
.
Alternatively, note the resemblence to Simon's Favorite Factoring Trick. so
so
and
.