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| − | ==Solution 1==
| + | #redirect [[2010 AMC 12B Problems/Problem 5]] |
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| − | Simplify the expression <math> a-(b-(c-(d+e))) </math>. I recommend to start with the innermost parenthesis and work your way out.
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| − | So you get:
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| − | <math>a-(b-(c-(d+e))) = a-(b-(c-d-e)) = a-(b-c+d+e)) = a-b+c-d-e</math>
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| − | Henry substituted <math>a, b, c, d</math> with <math>1, 2, 3, 4</math> respectively.
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| − | We have to find the value of <math>e</math>, such that <math> a-b+c-d-e = a-b-c-d+e</math> (the same expression without parenthesis).
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| − | Substituting and simplifying we get:
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| − | <math>-2-e = -8+e \Leftrightarrow -2e = -6 \Leftrightarrow e=3</math>
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| − | So Henry must have used the value <math>3</math> for <math>e</math>.
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| − | Our answer is:
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| − | <math> \boxed{\mathrm{(D)}= 3} </math>
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| − | ==Solution 2==
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| − | Lucky Larry had not been aware of the parenthesis and would have done the following operations:
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| − | <math>1-2-3-4+e=e-8</math>
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| − | The correct way he should have done the operations is:
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| − | <math> 1-(2-(3-(4+e))</math>
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| − | <math> 1-(2-(3-4-e)</math>
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| − | <math> 1-(2-(-1-e) </math>
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| − | <math> 1-(3+e)</math>
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| − | <math>1-3-e</math>
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| − | <math>-e-2</math>
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| − | Therefore we have the equation <math>e-8=-e-2\implies 2e=6\implies e=\boxed{\textbf{D)3}}</math>
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