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

Revision as of 13:03, 7 June 2011

Simplify the expression $a-(b-(c-(d+e)))$ . I recommend to start with the innermost parenthesis and work your way out.

So you get: $a-(b-(c-(d+e))) = a-(b-(c-d-e)) = a-(b-c+d+e)) = a-b+c-d-e$

Henry substituted $a, b, c, d$ with $1, 2, 3, 4$ respectively.

We have to find the value of $e$ , such that $a-b+c-d-e = a-b-c-d+e$ (the same expression without parenthesis).

Substituting and simplifying we get: $-2-e = -8+e \Leftrightarrow -2e = -6 \Leftrightarrow e=3$

So Henry must have used the value $3$ for $e$ .

Our answer is:

$\boxed{\mathrm{(D)}= 3}$

Lucky Larry had not been aware of the parenthesis and would have done the following operations: $1-2-3-4+e=e-8$

The correct way he should have done the operations is: $1-(2-(3-(4+e))$

$1-(2-(3-4-e)$

$1-(2-(-1-e)$

$1-(3+e)$

$1-3-e$

$-e-2$

Therefore we have the equation $e-8=-e-2\implies 2e=6\implies e=\boxed{\textbf{D)3}}$

@@ Line 24: / Line 24: @@
 The correct way he should have done the operations is:
-<nowiki>$ 1-(2-(3-(4+e))$
+<math> 1-(2-(3-(4+e))</math>
-$ 1-(2-(3-4-e)$
-$ 1-(2-(-1-e) $
+<math> 1-(2-(3-4-e)</math>
-$ 1-(3+e)$
-$1-3-e$
+<math> 1-(2-(-1-e) </math>
-$-e-2$</nowiki>
+<math> 1-(3+e)</math>
+<math>1-3-e</math>
+<math>-e-2</math>
 Therefore we have the equation <math>e-8=-e-2\implies 2e=6\implies e=\boxed{\textbf{D)3}}</math>