# 2021 Fall AMC 12A Problems/Problem 4

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 (Elimination)

Any number ending in $5$ is divisible by $5$. So we can eliminate option $\textbf{(C)}$.

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 $\textbf{(A)}$ and $\textbf{(D)}$.

So the correct option is either $\textbf{(B)}$ or $\textbf{(E)}$. 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 $\textbf{(B)}$ 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)

~Education, the Study of Everything

~savannahsolver

~Charles3829

## Video Solution by TheBeautyofMath

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

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

~IceMatrix

~Lucas

## Video Solution

~Education, the Study of Everything