Difference between revisions of "University of South Carolina High School Math Contest/1993 Exam/Problem 2"
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− | Alternatively, note the resemblence to [[Simon's Favorite Factoring Trick]]. <math>a\star 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>. | + | 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
.