Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2021 AMC 12A Problems/Problem 3"

m (→‎Solution 3 (Vertical Addition and Logic))
(Undo revision 232422 by Rawr3507 (talk))
(Tag: Undo)
 
(30 intermediate revisions by 11 users not shown)
Line 1: Line 1:
−
{{duplicate|[[2021 AMC 10A Problems#Problem 3|2021 AMC 10A #3]] and [[2021 AMC 12A Problems#Problem 3|2021 AMC 12A #3]]}}
+
{{duplicate|[[2021 AMC 10A Problems/Problem 3|2021 AMC 10A #3]] and [[2021 AMC 12A Problems/Problem 3|2021 AMC 12A #3]]}}
  
 
==Problem==
 
==Problem==
Line 6: Line 6:
 
<math>\textbf{(A)} ~10{,}272\qquad\textbf{(B)} ~11{,}700\qquad\textbf{(C)} ~13{,}362\qquad\textbf{(D)} ~14{,}238\qquad\textbf{(E)} ~15{,}426</math>
 
<math>\textbf{(A)} ~10{,}272\qquad\textbf{(B)} ~11{,}700\qquad\textbf{(C)} ~13{,}362\qquad\textbf{(D)} ~14{,}238\qquad\textbf{(E)} ~15{,}426</math>
  
−
==Solution 1==
+
==Solution 1 (Algebra)==
  
 
The units digit of a multiple of <math>10</math> will always be <math>0</math>. We add a <math>0</math> whenever we multiply by <math>10</math>. So, removing the units digit is equal to dividing by <math>10</math>.
 
The units digit of a multiple of <math>10</math> will always be <math>0</math>. We add a <math>0</math> whenever we multiply by <math>10</math>. So, removing the units digit is equal to dividing by <math>10</math>.
Line 12: Line 12:
 
Let the smaller number (the one we get after removing the units digit) be <math>a</math>. This means the bigger number would be <math>10a</math>.
 
Let the smaller number (the one we get after removing the units digit) be <math>a</math>. This means the bigger number would be <math>10a</math>.
  
−
We know the sum is <math>10a+a = 11a</math> so <math>11a=17402</math>. So <math>a=1582</math>. The difference is <math>10a-a = 9a</math>. So, the answer is <math>9(1582) = 14238 = \boxed{\textbf{(D)}}</math>.
+
We know the sum is <math>10a+a = 11a</math> so <math>11a=17402</math>. So <math>a=1582</math>. The difference is <math>10a-a = 9a</math>. So, the answer is <math>9(1582) = \boxed{\textbf{(D)} ~14{,}238}</math>.
  
−
--abhinavg0627
+
~abhinavg0627
  
−
==Solution 2 (Lazy Speed)==
+
==Solution 2 (Arithmetic)==
  
−
Since the ones place of a multiple of <math>10</math> is <math>0</math>, this implies the other integer has to end with a <math>2</math> since both integers sum up to a number that ends with a <math>2</math>. Thus, the ones place of the difference has to be <math>10-2=8</math>, and the only answer choice that ends with an <math>8</math> is <math>\boxed{\textbf{(D)}~14238}</math>
+
Since the unit's place of a multiple of <math>10</math> is <math>0</math>, the other integer must end with a <math>2</math>, for both integers sum up to a number ending in a <math>2</math>. Thus, the unit's place of the difference must be <math>10-2=8</math>, and the only answer choice that ends with an <math>8</math> is <math>\boxed{\textbf{(D)} ~14{,}238}</math>.
 +
 
 +
Another quick solution is to realize that the sum represents a number <math>n</math> added to <math>10n</math>. The difference is <math>9n</math>, which is <math>\frac{9}{11}</math> of the given sum.
  
 
~CoolJupiter 2021
 
~CoolJupiter 2021
  
−
Another quick solution is to realize that the sum is represents a number <math>n</math> added to <math>10n</math>. The difference is <math>9n</math>, which is <math>\frac{9}{11}</math> of the given sum.
+
~Extremelysupercooldude (Minor grammar edits)
  
 
==Solution 3 (Vertical Addition and Logic)==
 
==Solution 3 (Vertical Addition and Logic)==
−
Let the larger number be <math>\underline{AB,CD0}.</math> It follows that the smaller number is <math>\underline{A,BCD}.</math> Adding vertically, we have  
+
Let the larger number be <math>\underline{AB{,}CD0}.</math> It follows that the smaller number is <math>\underline{A{,}BCD}.</math> Adding vertically, we have  
 
<cmath>\begin{array}{cccccc}
 
<cmath>\begin{array}{cccccc}
 
   & A & B & C & D & 0 \\
 
   & A & B & C & D & 0 \\
−
+ &  & A & B & C & D \\
+
+\quad &  & A & B & C & D \\
 
\hline
 
\hline
 +
  &  &  &  &  &  \\ [-2.5ex]
 
   & 1 & 7 & 4 & 0 & 2 \\
 
   & 1 & 7 & 4 & 0 & 2 \\
 
\end{array}</cmath>
 
\end{array}</cmath>
 
Working from right to left, we get <cmath>D=2\implies C=8 \implies B=5 \implies A=1.</cmath>
 
Working from right to left, we get <cmath>D=2\implies C=8 \implies B=5 \implies A=1.</cmath>
−
The larger number is <math>15,820</math> and the smaller number is <math>1,582.</math> Their difference is <math>15,820-1,582=\boxed{\textbf{(D)} ~14{,}238}.</math>
+
The larger number is <math>15{,}820</math> and the smaller number is <math>1{,}582.</math> Their difference is <math>15{,}820-1{,}582=\boxed{\textbf{(D)} ~14{,}238}.</math>
  
 
~MRENTHUSIASM
 
~MRENTHUSIASM
  
−
==Video Solution (Simple)==
+
==Solution 4 (Logic)==
 +
We know that the larger number has a units digit of <math>0</math> since it is divisible by 10. If <math>D</math> is the ten's digit of the larger number, then <math>D</math> is the units digit of the smaller number. Since the sum of the natural numbers has a unit's digit of <math>2</math>, <math>D=2</math>.
 +
 
 +
The units digit of the larger number is <math>0</math> and the units digit of the smaller number is <math>2</math>, so the positive difference between the numbers is 8. There is only one answer choice that has this units digit, and that is <math>\boxed{\textbf{(D)} ~14{,}238}.</math>
 +
 
 +
(Similar to MRENTHUSIASM's solution)
 +
 
 +
~AARUSHGORADIA18
 +
 
 +
==Video Solutions==
 +
 
 +
==Video Solution 1 (easily understandable)==
 +
https://www.youtube.com/watch?v=jPzcN0KeJ88&t=132s
 +
 
 +
===Video Solution (Simple)===
 
https://youtu.be/SEp9flDYm2c
 
https://youtu.be/SEp9flDYm2c
  
 
~ Education, the study Of Everything
 
~ Education, the study Of Everything
  
−
==Video Solution by North America Math Contest Go Go Go==
+
===Video Solution by North America Math Contest Go Go Go===
 
https://www.youtube.com/watch?v=hMqA8i8a2SQ&list=PLexHyfQ8DMuKqltG3cHT7Di4jhVl6L4YJ&index=3
 
https://www.youtube.com/watch?v=hMqA8i8a2SQ&list=PLexHyfQ8DMuKqltG3cHT7Di4jhVl6L4YJ&index=3
  
−
==Video Solution by Aaron He==
+
===Video Solution by Aaron He===
 
https://www.youtube.com/watch?v=xTGDKBthWsw&t=1m28s
 
https://www.youtube.com/watch?v=xTGDKBthWsw&t=1m28s
  
−
==Video Solution by Punxsutawney Phil==
+
===Video Solution by Punxsutawney Phil===
 
https://youtube.com/watch?v=MUHja8TpKGw&t=143s
 
https://youtube.com/watch?v=MUHja8TpKGw&t=143s
  
−
==Video Solution by Hawk Math==
+
===Video Solution by Hawk Math===
 
https://www.youtube.com/watch?v=P5al76DxyHY
 
https://www.youtube.com/watch?v=P5al76DxyHY
  
−
==Video Solution (Using Algebra and Meta-solving)==
+
===Video Solution by OmegaLearn (Using Algebra and Meta-solving)===
 
https://youtu.be/d2musztzDjw
 
https://youtu.be/d2musztzDjw
  
 
-pi_is_3.14
 
-pi_is_3.14
  
−
==Video Solution by WhyMath==
+
===Video Solution by WhyMath===
 
https://youtu.be/VpYmQEKcBpA
 
https://youtu.be/VpYmQEKcBpA
  
 
~savannahsolver
 
~savannahsolver
  
−
==Video Solution by TheBeautyofMath==
+
===Video Solution by TheBeautyofMath===
 
https://youtu.be/50CThrk3RcM?t=107 (for AMC 10A)
 
https://youtu.be/50CThrk3RcM?t=107 (for AMC 10A)
  
 +
===Video Solution by IceMatrix===
 
https://youtu.be/rEWS75W0Q54?t=198 (for AMC 12A)
 
https://youtu.be/rEWS75W0Q54?t=198 (for AMC 12A)
  
 
~IceMatrix
 
~IceMatrix
  
−
==Video Solution (Problems 1-3)==
+
===Video Solution (Problems 1-3)===
 
https://youtu.be/CupJpUzKPB0
 
https://youtu.be/CupJpUzKPB0
  
 
~MathWithPi
 
~MathWithPi
 +
 +
===Video Solution by The Learning Royal===
 +
https://youtu.be/slVBYmcDMOI
  
 
==See also==
 
==See also==

Latest revision as of 23:22, 8 November 2024

The following problem is from both the 2021 AMC 10A #3 and 2021 AMC 12A #3, so both problems redirect to this page.

Problem

The sum of two natural numbers is $17{,}402$. One of the two numbers is divisible by $10$. If the units digit of that number is erased, the other number is obtained. What is the difference of these two numbers?

$\textbf{(A)} ~10{,}272\qquad\textbf{(B)} ~11{,}700\qquad\textbf{(C)} ~13{,}362\qquad\textbf{(D)} ~14{,}238\qquad\textbf{(E)} ~15{,}426$

Solution 1 (Algebra)

The units digit of a multiple of $10$ will always be $0$. We add a $0$ whenever we multiply by $10$. So, removing the units digit is equal to dividing by $10$.

Let the smaller number (the one we get after removing the units digit) be $a$. This means the bigger number would be $10a$.

We know the sum is $10a+a = 11a$ so $11a=17402$. So $a=1582$. The difference is $10a-a = 9a$. So, the answer is $9(1582) = \boxed{\textbf{(D)} ~14{,}238}$.

~abhinavg0627

Solution 2 (Arithmetic)

Since the unit's place of a multiple of $10$ is $0$, the other integer must end with a $2$, for both integers sum up to a number ending in a $2$. Thus, the unit's place of the difference must be $10-2=8$, and the only answer choice that ends with an $8$ is $\boxed{\textbf{(D)} ~14{,}238}$.

Another quick solution is to realize that the sum represents a number $n$ added to $10n$. The difference is $9n$, which is $\frac{9}{11}$ of the given sum.

~CoolJupiter 2021

~Extremelysupercooldude (Minor grammar edits)

Solution 3 (Vertical Addition and Logic)

Let the larger number be $\underline{AB{,}CD0}.$ It follows that the smaller number is $\underline{A{,}BCD}.$ Adding vertically, we have \[\begin{array}{cccccc} & A & B & C & D & 0 \\ +\quad & & A & B & C & D \\ \hline & & & & & \\ [-2.5ex] & 1 & 7 & 4 & 0 & 2 \\ \end{array}\] Working from right to left, we get \[D=2\implies C=8 \implies B=5 \implies A=1.\] The larger number is $15{,}820$ and the smaller number is $1{,}582.$ Their difference is $15{,}820-1{,}582=\boxed{\textbf{(D)} ~14{,}238}.$

~MRENTHUSIASM

Solution 4 (Logic)

We know that the larger number has a units digit of $0$ since it is divisible by 10. If $D$ is the ten's digit of the larger number, then $D$ is the units digit of the smaller number. Since the sum of the natural numbers has a unit's digit of $2$, $D=2$.

The units digit of the larger number is $0$ and the units digit of the smaller number is $2$, so the positive difference between the numbers is 8. There is only one answer choice that has this units digit, and that is $\boxed{\textbf{(D)} ~14{,}238}.$

(Similar to MRENTHUSIASM's solution)

~AARUSHGORADIA18

Video Solutions

Video Solution 1 (easily understandable)

https://www.youtube.com/watch?v=jPzcN0KeJ88&t=132s

Video Solution (Simple)

https://youtu.be/SEp9flDYm2c

~ Education, the study Of Everything

Video Solution by North America Math Contest Go Go Go

https://www.youtube.com/watch?v=hMqA8i8a2SQ&list=PLexHyfQ8DMuKqltG3cHT7Di4jhVl6L4YJ&index=3

Video Solution by Aaron He

https://www.youtube.com/watch?v=xTGDKBthWsw&t=1m28s

Video Solution by Punxsutawney Phil

https://youtube.com/watch?v=MUHja8TpKGw&t=143s

Video Solution by Hawk Math

https://www.youtube.com/watch?v=P5al76DxyHY

Video Solution by OmegaLearn (Using Algebra and Meta-solving)

https://youtu.be/d2musztzDjw

-pi_is_3.14

Video Solution by WhyMath

https://youtu.be/VpYmQEKcBpA

~savannahsolver

Video Solution by TheBeautyofMath

https://youtu.be/50CThrk3RcM?t=107 (for AMC 10A)

Video Solution by IceMatrix

https://youtu.be/rEWS75W0Q54?t=198 (for AMC 12A)

~IceMatrix

Video Solution (Problems 1-3)

https://youtu.be/CupJpUzKPB0

~MathWithPi

Video Solution by The Learning Royal

https://youtu.be/slVBYmcDMOI

See also

2021 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
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
2021 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
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 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12A_Problems/Problem_3&oldid=232426"