Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 2)
m
Line 17: Line 17:
  
 
Substituting and simplifying we get:
 
Substituting and simplifying we get:
−
<math>-2-e = -8+e \Leftrightarrow -2e = -6 \Leftrightarrow e=3</math>
+
<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>.
 
So Henry must have used the value <math>3</math> for <math>e</math>.
  
−
Our answer is <math>3 \Rightarrow \boxed{D}</math>
+
Our answer is <math>3 \Rightarrow \boxed{\textbf{(D)}}</math>
  
−
==Solution 2==
+
===Solution 2===
 
Lucky Larry had not been aware of the parenthesis and would have done the following operations:
 
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>
 
<math>1-2-3-4+e=e-8</math>
  
 
The correct way he should have done the operations is:
 
The correct way he should have done the operations is:
−
<math> 1-(2-(3-(4+e))</math>
+
<cmath>1-(2-(3-(4+e))= 1-(2-(3-4-e)= 1-(2-(-1-e) = 1-(3+e) =1-3-e=-e-2</cmath>
  
−
<math> 1-(2-(3-4-e)</math>
+
Therefore we have the equation <math>e-8=-e-2\implies 2e=6\implies e=3 \Rightarrow \boxed{\textbf{(D)}}</math>
−
 
 
−
<math> 1-(2-(-1-e) </math>
 
−
 
 
−
<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=3 \Rightarrow \boxed{D}</math>
 
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2010|ab=B|num-b=8|num-a=10}}
 
{{AMC10 box|year=2010|ab=B|num-b=8|num-a=10}}

Revision as of 00:48, 26 November 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: $-2-e = -8+e \Rightarrow -2e = -6 \Rightarrow e=3$

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

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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10B_Problems/Problem_9&oldid=43410"