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"

(Created page with "== Problem == The six-digit number <math>\underline{2}\,\underline{0}\,\underline{2}\,\underline{1}\,\underline{0}\,\underline{A}</math> is prime for only one digit <math>A.</...")
 
Line 1: Line 1:
 +
{{duplicate|[[2021 Fall AMC 10A Problems#Problem 5|2021 Fall AMC 10A #5]] and [[2021 Fall AMC 10A Problems#Problem 4|2021 Fall AMC 12A #4]]}}
 +
 
== Problem ==
 
== Problem ==
 
The six-digit number <math>\underline{2}\,\underline{0}\,\underline{2}\,\underline{1}\,\underline{0}\,\underline{A}</math> is prime for only one digit <math>A.</math> What is <math>A?</math>
 
The six-digit number <math>\underline{2}\,\underline{0}\,\underline{2}\,\underline{1}\,\underline{0}\,\underline{A}</math> is prime for only one digit <math>A.</math> What is <math>A?</math>

Revision as of 18:14, 23 November 2021

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

By divisibility rules, when $A=1,$ the number $202101$ is divisible by $3.$ When $A=3,$ the number $202103$ is divisible by $11.$ When $A=5,$ the number $202105$ is divisible by $5.$ When $A=7,$ the number $202107$ is divisible by $3.$ Thus, by the process of elimination we have that the answer is $\boxed{\textbf{(E)}\ 9}.$

~NH14

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=165588"