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 | ||
− | https://www.youtube.com/watch?v=GNPAgQ8fSP0&t | + | ===Video Solution by Alex Explains=== |
+ | https://www.youtube.com/watch?v=GNPAgQ8fSP0&t | ||
+ | |||
+ | ~AlexExplains | ||
==See Also== | ==See Also== |
Latest revision as of 21:10, 2 October 2024
Contents
Problem
What is the value of
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 .
Simplify,
is
The second and third groups can be expressed as .
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.