Difference between revisions of "2020 AMC 10B Problems/Problem 1"
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<math>\textbf{(A)}\ -20 \qquad\textbf{(B)}\ -3 \qquad\textbf{(C)}\ 3 \qquad\textbf{(D)}\ 5 \qquad\textbf{(E)}\ 21</math> | <math>\textbf{(A)}\ -20 \qquad\textbf{(B)}\ -3 \qquad\textbf{(C)}\ 3 \qquad\textbf{(D)}\ 5 \qquad\textbf{(E)}\ 21</math> | ||
| − | ==Solution== | + | ==Solution 1== |
We know that when we subtract negative numbers, <math>a-(-b)=a+b</math>. | We know that when we subtract negative numbers, <math>a-(-b)=a+b</math>. | ||
| − | The equation becomes <math>1+2-3+4-5+6 = \boxed{\textbf{(D)}\ 5}</math> | + | The equation becomes <math>1+2-3+4-5+6 = \boxed{\textbf{(D)}\ 5}</math>. |
| − | + | ~quacker88 | |
| − | |||
| − | == | + | ==Solution 2== |
| − | + | Like Solution 1, we know that when we subtract <math>a-(-b)</math>, that will equal <math>a+b</math> as the opposite/negative of a negative is a positive. Thus, <math>1-(-2)-3-(-4)-5-(-6)=1+2-3+4-5+6</math>. We can group together a few terms to make our computation a bit simpler. <math>1+(2-3)+4+(-5+6)= 1+(-1)+4+1=\boxed{\textbf{(D)}\ 5}</math>. | |
| − | + | ||
| + | ~BakedPotato66 | ||
| + | |||
| + | == Solution 3 == | ||
| + | Notice <math>1-(-2)-3-(-4)-5-(-6)</math> has three groups: <math>1-(-2)</math>, <math>-3-(-4)</math>, <math>-5-(-6)</math> | ||
| + | |||
| + | The first group, <math>1-(-2)</math>, can be expressed as <math>a-[-(a+1)]</math>. | ||
| + | |||
| + | Simplify, | ||
| + | |||
| + | <math>a-[-(a+1)]</math> is <math>2a+1</math> | ||
| + | |||
| + | The second and third groups can be expressed as <math>-c-[-(c+1)]</math>. | ||
| + | |||
| + | Simplify, | ||
| + | |||
| + | <math>-c-[-(c+1)]</math> is <math>1</math> | ||
| + | |||
| + | Thus, the sum of the three groups is <math>2 \cdot 1 + 1 + 1 + 1 = 5</math> or <math>\boxed{\textbf{(D)}\ 5}</math> | ||
| + | |||
| + | ~ lovelearning999 | ||
| + | |||
| + | ==Video Solutions== | ||
| + | |||
| + | ===Video Solution by Education, the study of everything=== | ||
https://www.youtube.com/watch?v=NpDVTLSi-Ik | https://www.youtube.com/watch?v=NpDVTLSi-Ik | ||
| + | ~Education, the Study of Everything | ||
| + | ===Video Solution by TheBeautyofMath=== | ||
https://youtu.be/Gkm5rU5MlOU | https://youtu.be/Gkm5rU5MlOU | ||
~IceMatrix | ~IceMatrix | ||
| − | + | ===Video Solution by WhyMath=== | |
https://youtu.be/-wciFhP5h3I | https://youtu.be/-wciFhP5h3I | ||
~savannahsolver | ~savannahsolver | ||
| + | ===Video Solution by Alex Explains=== | ||
| + | https://www.youtube.com/watch?v=GNPAgQ8fSP0&t | ||
| − | + | ~AlexExplains | |
| − | |||
| − | ~ | ||
==See Also== | ==See Also== | ||
Latest revision as of 22:10, 2 October 2024
Contents
Problem
What is the value of
![\[1-(-2)-3-(-4)-5-(-6)?\]](http://latex.artofproblemsolving.com/9/4/c/94cdfebdc087637cc5450da458e0db822c81da88.png)
Solution 1
We know that when we subtract negative numbers, 
.
The equation becomes 
.
~quacker88
Solution 2
Like Solution 1, we know that when we subtract 
, that will equal 
as the opposite/negative of a negative is a positive. Thus, 
. We can group together a few terms to make our computation a bit simpler. 
.
~BakedPotato66
Solution 3
Notice 
has three groups: 
, 
, 
The first group, , can be expressed as ![$a-[-(a+1)]$](http://latex.artofproblemsolving.com/0/0/5/00500b0987c4218e2cb1c9f2c18ea6fba730cb6e.png)
.
Simplify,
is 
The second and third groups can be expressed as ![$-c-[-(c+1)]$](http://latex.artofproblemsolving.com/7/7/a/77a1ce88b47f9f7ec6bc604b0950cf00cdc53d53.png)
.
Simplify,
is 
Thus, the sum of the three groups is 
or 
~ 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
~IceMatrix
Video Solution by WhyMath
~savannahsolver
Video Solution by Alex Explains
https://www.youtube.com/watch?v=GNPAgQ8fSP0&t
~AlexExplains
See Also
| 2020 AMC 10B (Problems • Answer Key • Resources) | ||
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
