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 Fall AMC 12A Problems/Problem 4"

m
Line 60: Line 60:
  
 
~Education, the Study of Everything
 
~Education, the Study of Everything
−
 
−
==Video Solution==
 
−
https://youtu.be/jxnTkY3eb5Y
 
−
 
−
~savannahsolver
 
−
 
−
https://youtu.be/AgzDyKlmNAo
 
−
 
−
~Charles3829
 
  
 
==Video Solution by TheBeautyofMath==
 
==Video Solution by TheBeautyofMath==

Revision as of 11:59, 12 November 2022

The following problem is from both the 2021 Fall AMC 10A #5 and 2021 Fall AMC 12A #4, so both problems redirect to this page.

Problem

The six-digit number $\underline{2}\,\underline{0}\,\underline{2}\,\underline{1}\,\underline{0}\,\underline{A}$ is prime for only one digit $A.$ What is $A?$

$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\ 5 \qquad\textbf{(D)}\ 7 \qquad\textbf{(E)}\ 9$

Solution 1

First, modulo $2$ or $5$, $\underline{20210A} \equiv A$. Hence, $A \neq 0, 2, 4, 5, 6, 8$.

Second modulo $3$, $\underline{20210A} \equiv 2 + 0 + 2 + 1 + 0 + A \equiv 5 + A$. Hence, $A \neq 1, 4, 7$.

Third, modulo $11$, $\underline{20210A} \equiv A + 1 + 0 - 0 - 2 - 2 \equiv A - 3$. Hence, $A \neq 3$.

Therefore, the answer is $\boxed{\textbf{(E)}\ 9}$.

~NH14 ~Steven Chen (www.professorchenedu.com)

Solution 2

Any number ending in $5$ is divisible by $5$. So we can eliminate option $C$.

If the sum of the digits of a number is divisible by $3$, the number is divisible by $3$. The sum of the digits of this number is $2 + 0 + 2 + 1 + 0 + A = 5 + A$. If $5 + A$ is divisible by $3$, the number is divisible by $3$. Thus we can eliminate options $A$ and $D$.

So the correct option is either $B$ or $E$. Let's try dividing the number with some integers.

$20210A/7 = 2887x$, where $x$ is $1A/7$. Since $13$ and $19$ are both indivisible by $7$, this does not help us narrow the choices down.

$20210A/11 = 1837x$, where $x$ is $3A/11$. Since $33/11 = 3$, option $B$ would make $20210A$ divisible by $11$. Thus, by elimination, the correct choice must be option $\boxed{\textbf{(E)}\ 9}$.

~ZoBro23

Solution 3

$202100 \implies$ divisible by $2$.

$202101 \implies$ divisible by $3$.

$202102 \implies$ divisible by $2$.

$202103 \implies$ divisible by $11$.

$202104 \implies$ divisible by $2$.

$202105 \implies$ divisible by $5$.

$202106 \implies$ divisible by $2$.

$202107 \implies$ divisible by $3$.

$202108 \implies$ divisible by $2$.

This leaves only $A=\boxed{\textbf{(E)}\ 9}$.

~wamofan

Video Solution (Simple and Quick)

https://youtu.be/7_Dg9b2hQ5U

~Education, the Study of Everything

Video Solution by TheBeautyofMath

for AMC 10: https://youtu.be/o98vGHAUYjM?t=623

for AMC 12: https://youtu.be/jY-17W6dA3c?t=392

~IceMatrix

Video Solution

https://youtu.be/7TnFYSJ8i14

~Lucas

Video Solution

https://youtu.be/7_Dg9b2hQ5U

~Education, the Study of Everything

See Also

2021 Fall AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
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
2021 Fall AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_Fall_AMC_12A_Problems/Problem_4&oldid=181006"