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Difference between revisions of "2010 AMC 12B Problems/Problem 5"
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<cmath>2e = 6</cmath>
 
<cmath>2e = 6</cmath>
 
<cmath>e = 3 \;\;(D)</cmath>
 
<cmath>e = 3 \;\;(D)</cmath>
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==Video Solution==
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https://youtu.be/OiJ-9eGCLGM
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~Education, the Study of Everything
  
 
==Video Solution==
 
==Video Solution==
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~IceMatrix
 
~IceMatrix
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== See also ==
 
== See also ==

Latest revision as of 16:02, 1 August 2022

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 substituted for $a$, $b$, $c$, and $d$ were $1$, $2$, $3$, and $4$, respectively. What number did Larry substitute for $e$?

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

Solution

We simply plug in the numbers \[1 - 2 - 3 - 4 + e = 1 - (2 - (3 - (4 + e)))\] \[-8 + e = -2 - e\] \[2e = 6\] \[e = 3 \;\;(D)\]

Video Solution

https://youtu.be/OiJ-9eGCLGM

~Education, the Study of Everything

Video Solution

https://youtu.be/I3yihAO87CE?t=303

~IceMatrix

See also

2010 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
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 12 Problems and Solutions
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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