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

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

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.

@@ Line 6: / Line 6: @@
 <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>.
-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>.
-==Video Solution==
+~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
+~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==

Page

Toolbox

Search