2010 AMC 10B Problems/Problem 9: Difference between revisions

Latest revision as of 19:38, 26 May 2020

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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.
+#redirect [[2010 AMC 12B Problems/Problem 5]]
-So you get:
-<math>a-(b-(c-(d+e))) = a-(b-(c-d-e)) = a-(b-c+d+e)) = a-b+c-d-e</math>
-Henry substituted <math>a, b, c, d</math> with <math>1, 2, 3, 4</math> respectively.
-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).
-Substituting and simplifying we get:
-<math>-2-e = -8+e \Leftrightarrow -2e = -6 \Leftrightarrow e=3</math>
-So Henry must have used the value <math>3</math> for <math>e</math>.
-Our answer is:
-<math> \boxed{\mathrm{(D)}= 3} </math>