2010 AMC 10B Problems/Problem 9

Problem

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

$\textbf{(A)}\ -5 \qquad \textbf{(B)}\ -3 \qquad \textbf{(C)}\ 0 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 5$

Solution 1

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: $-2-e = -8+e \Rightarrow -2e = -6 \Rightarrow e=3$

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

Our answer is $3 \Rightarrow \boxed{\textbf{(D)}}$

Solution 2

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 $e-8=-e-2\implies 2e=6\implies e=3 \Rightarrow \boxed{\textbf{(D)}}$

2010 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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2010 AMC 10B Problems/Problem 9

Contents

Problem

Solution 1

Solution 2

See Also