Difference between revisions of "2020 AMC 10B Problems/Problem 1"

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~BakedPotato66
 
~BakedPotato66
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== Solution 3 ==
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Notice <math>1-(-2)-3-(-4)-5-(-6)</math> has three groups: <math>1-(-2)</math>, <math>-3-(-4)</math>, <math>-5-(-6)</math>
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The first group, <math>1-(-2)</math>, can be expressed as <math>a-[-(a+1)]</math>.
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Simplify,
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<math>a-[-(a+1)]</math> is <math>2a+1</math>
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The second and third groups can be expressed as <math>-c-[-(c+1)]</math>.
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Simplify,
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<math>-c-[-(c+1)]</math> is <math>1</math>
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Thus, the sum of the three groups is <math>2 \cdot 1 + 1 + 1 + 1 = 5</math> or <math>\boxed{\textbf{(D)}\ 5}</math>
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~ lovelearning999
  
 
==Video Solutions==
 
==Video Solutions==
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===Video Solution by Alex Explains===
 
===Video Solution by Alex Explains===
https://www.youtube.com/watch?v=GNPAgQ8fSP0&t=1s
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https://www.youtube.com/watch?v=GNPAgQ8fSP0&t
  
 
~AlexExplains
 
~AlexExplains

Latest revision as of 21:10, 2 October 2024

Problem

What is the value of \[1-(-2)-3-(-4)-5-(-6)?\]

$\textbf{(A)}\ -20 \qquad\textbf{(B)}\ -3 \qquad\textbf{(C)}\  3 \qquad\textbf{(D)}\ 5 \qquad\textbf{(E)}\ 21$

Solution 1

We know that when we subtract negative numbers, $a-(-b)=a+b$.

The equation becomes $1+2-3+4-5+6 = \boxed{\textbf{(D)}\ 5}$.

~quacker88

Solution 2

Like Solution 1, we know that when we subtract $a-(-b)$, that will equal $a+b$ as the opposite/negative of a negative is a positive. Thus, $1-(-2)-3-(-4)-5-(-6)=1+2-3+4-5+6$. We can group together a few terms to make our computation a bit simpler. $1+(2-3)+4+(-5+6)= 1+(-1)+4+1=\boxed{\textbf{(D)}\ 5}$.

~BakedPotato66

Solution 3

Notice $1-(-2)-3-(-4)-5-(-6)$ has three groups: $1-(-2)$, $-3-(-4)$, $-5-(-6)$

The first group, $1-(-2)$, can be expressed as $a-[-(a+1)]$.

Simplify,

$a-[-(a+1)]$ is $2a+1$

The second and third groups can be expressed as $-c-[-(c+1)]$.

Simplify,

$-c-[-(c+1)]$ is $1$

Thus, the sum of the three groups is $2 \cdot 1 + 1 + 1 + 1 = 5$ or $\boxed{\textbf{(D)}\ 5}$

~ lovelearning999

Video Solutions

Video Solution by Education, the study of everything

https://www.youtube.com/watch?v=NpDVTLSi-Ik

~Education, the Study of Everything

Video Solution by TheBeautyofMath

https://youtu.be/Gkm5rU5MlOU

~IceMatrix

Video Solution by WhyMath

https://youtu.be/-wciFhP5h3I

~savannahsolver

Video Solution by Alex Explains

https://www.youtube.com/watch?v=GNPAgQ8fSP0&t

~AlexExplains

See Also

2020 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
First Problem
Followed by
Problem 2
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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