Difference between revisions of "2021 Fall AMC 12A Problems/Problem 4"

Revision as of 18:43, 7 April 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

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

$202100 \implies$ 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

https://youtu.be/jxnTkY3eb5Y

~savannahsolver

Video Solution by TheBeautyofMath

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

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

~IceMatrix

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.

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