Difference between revisions of "2020 AMC 10B Problems/Problem 1"
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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> ~quacker88 | + | The equation becomes <math>1+2-3+4-5+6 = \boxed{\textbf{(D)}\ 5}</math>. |
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+ | ~quacker88 | ||
==Solution 2== | ==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>. | |
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− | 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 | ~BakedPotato66 |
Revision as of 16:18, 27 August 2021
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
Video Solution
Check It Out! :) Education, the Study of Everything (wow!) https://www.youtube.com/watch?v=NpDVTLSi-Ik
~IceMatrix
~savannahsolver
https://www.youtube.com/watch?v=GNPAgQ8fSP0&t=1s
~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.