View source for 2010 AMC 10B Problems/Problem 9

== Problem ==

Lucky Larry's teacher asked him to substitute numbers for <math>a</math>, <math>b</math>, <math>c</math>, <math>d</math>, and <math>e</math> in the expression <math>a-(b-(c-(d+e)))</math> and evaluate the result. Larry ignored the parentheses but added and subtracted correctly and obtained the correct result by coincidence. The number Larry substituted for <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> were <math>1</math>, <math>2</math>, <math>3</math>, and <math>4</math>, respectively. What number did Larry substitute for <math>e</math>?

<math>\textbf{(A)}\ -5 \qquad \textbf{(B)}\ -3 \qquad \textbf{(C)}\ 0 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 5</math>

==Solution 1==

Simplify the expression <math> a-(b-(c-(d+e))) </math>. 

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 \Rightarrow -2e = -6 \Rightarrow e=3</math>

So, Henry must have used the value <math>3</math> for <math>e</math>.

Our answer is <math>3 \Rightarrow \boxed{\textbf{(D)}}</math>

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

The correct way he should have done the operations is:
<cmath>1-(2-(3-(4+e))= 1-(2-(3-4-e)= 1-(2-(-1-e) = 1-(3+e) =1-3-e=-e-2</cmath>

Therefore we have the equation <math>e-8=-e-2\implies 2e=6\implies e=3 \Rightarrow \boxed{\textbf{(D)}}</math>

==See Also==
{{AMC10 box|year=2010|ab=B|num-b=8|num-a=10}}
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