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Difference between revisions of "2010 AMC 10B Problems/Problem 9"

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==Solution 1==
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== Problem ==
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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 parenthese but added and subtracted correctly and obtained the correct result by coincidence. The number Larry sustitued 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 substitude for <math>e</math>?
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<math>\textbf{(A)}\ -5 \qquad \textbf{(B)}\ -3 \qquad \textbf{(C)}\ 0 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 5</math>
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==Solution==
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===Solution 1===
  
 
Simplify the expression <math> a-(b-(c-(d+e))) </math>. I recommend to start with the innermost parenthesis and work your way out.
 
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 Henry must have used the value <math>3</math> for <math>e</math>.
 
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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Our answer is <math>3 \Rightarrow \boxed{E}</math>
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<math> \boxed{\mathrm{(D)}= 3} </math>
 
  
 
==Solution 2==
 
==Solution 2==
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<math>-e-2</math>
 
<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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Therefore we have the equation <math>e-8=-e-2\implies 2e=6\implies e=3 \Rightarrow \boxed{E}</math>
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==See Also==
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{{AMC10 box|year=2010|ab=B|num-b=8|num-a=10}}

Revision as of 14:14, 7 June 2011

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

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

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

Our answer is $3 \Rightarrow \boxed{E}$

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 $e-8=-e-2\implies 2e=6\implies e=3 \Rightarrow \boxed{E}$

See Also

2010 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions
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