Difference between revisions of "University of South Carolina High School Math Contest/1993 Exam/Problem 20"
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== Solution == | == Solution == | ||
We have <math>(1+1)(1+2)(1+3)(1+4)(1+5)(1+6)-1</math> (The <math>-1</math> since we have one less set). This is <math>7!-1=5039</math>. | We have <math>(1+1)(1+2)(1+3)(1+4)(1+5)(1+6)-1</math> (The <math>-1</math> since we have one less set). This is <math>7!-1=5039</math>. | ||
+ | If you don't understand what the above means, just think about the simplifying of the brackets. Open the brackets and you should notice why it is correct. | ||
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Revision as of 11:02, 19 October 2021
Problem
Let be the 63 nonempty subsets of
. For each of these sets
, let
denote the product of all the elements in
. Then what is the value of
?
![$\mathrm{(A) \ }5003 \qquad \mathrm{(B) \ }5012 \qquad \mathrm{(C) \ }5039 \qquad \mathrm{(D) \ }5057 \qquad \mathrm{(E) \ }5093$](http://latex.artofproblemsolving.com/4/b/b/4bb3af1414663cc5a3d37b49c5ccc4c0bb56273b.png)
Solution
We have (The
since we have one less set). This is
.
If you don't understand what the above means, just think about the simplifying of the brackets. Open the brackets and you should notice why it is correct.