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 2 (Lazy Speed))
Line 22: Line 22:
  
 
~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.
  
 
==Solution 3 (Vertical Addition and Logic)==
 
==Solution 3 (Vertical Addition and Logic)==

Revision as of 13:14, 24 February 2021

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

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 $9(1582) = 14238 = \boxed{\textbf{(D)}}$.


--abhinavg0627

Solution 2 (Lazy Speed)

Since the ones place of a multiple of $10$ is $0$, this implies the other integer has to end with a $2$ since both integers sum up to a number that ends with a $2$. Thus, the ones place of the difference has to be $10-2=8$, and the only answer choice that ends with an $8$ is $\boxed{\textbf{(D)}~14238}$

~CoolJupiter 2021

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

Solution 3 (Vertical Addition and Logic)

Let the larger number be $\overline{AB,CD0}.$ It follows that the smaller number is $\overline{A,BCD}.$ Adding vertically, we have \[\begin{array}{cccccc} & A & B & C & D & 0 \\ + & & A & B & C & D \\ \hline & 1 & 7 & 4 & 0 & 2 \\ \end{array}\] Working from right to left, we have \[D=2\Rightarrow C=8 \Rightarrow B=5 \Rightarrow A=1.\] 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

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 (Using Algebra and Meta-solving)

https://youtu.be/d2musztzDjw

-pi_is_3.14

Video Solution (Simple)

https://youtu.be/SEp9flDYm2c

~ Education, the study Of Everything

Video Solution by WhyMath

https://youtu.be/VpYmQEKcBpA

~savannahsolver

Video Solution by TheBeautyofMath

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

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

~IceMatrix

Video Solution (Problems 1-3)

https://youtu.be/CupJpUzKPB0

~MathWithPi

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=147860"