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

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Hence, <math>A \neq 3</math>.
Hence, <math>A \neq 3</math>.


Therefore, the answer is <math>\boxed{\textbf{(E) }9}</math>.
Therefore, the answer is <math>\boxed{\textbf{(E)}\ 9}</math>.


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


==Solution 2==
==Solution 2 (Elimination)==
Any number ending in <math>5</math> is divisible by <math>5</math>. So we can eliminate option <math>\textbf{(C)}</math>.
 
If the sum of the digits of a number is divisible by <math>3</math>, the number is divisible by <math>3</math>. The sum of the digits of this number is <math>2 + 0 + 2 + 1 + 0 + A = 5 + A</math>. If <math>5 + A</math> is divisible by <math>3</math>, the number is divisible by <math>3</math>. Thus we can eliminate options <math>\textbf{(A)}</math> and <math>\textbf{(D)}</math>.
 
So the correct option is either <math>\textbf{(B)}</math> or <math>\textbf{(E)}</math>. Let's try dividing the number with some integers.
 
<math>20210A/7 = 2887x</math>, where <math>x</math> is <math>1A/7</math>. Since <math>13</math> and <math>19</math> are both indivisible by <math>7</math>, this does not help us narrow the choices down.
 
<math>20210A/11 = 1837x</math>, where <math>x</math> is <math>3A/11</math>. Since <math>33/11 = 3</math>, option <math>\textbf{(B)}</math> would make <math>20210A</math> divisible by <math>11</math>. Thus, by elimination, the correct choice must be option <math>\boxed{\textbf{(E)}\ 9}</math>.
 
~ZoBro23
 
==Solution 3==
<math>202100 \implies</math> divisible by <math>2</math>.
<math>202100 \implies</math> divisible by <math>2</math>.


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<math>202108 \implies</math> divisible by <math>2</math>.
<math>202108 \implies</math> divisible by <math>2</math>.


This leaves only <math>A=\boxed{\textbf{(E) }9}</math>.
This leaves only <math>A=\boxed{\textbf{(E)}\ 9}</math>.


~wamofan
~wamofan


==Video Solution 1==
==Sidenote==
https://youtu.be/7_Dg9b2hQ5U
The divisibility test for <math>11</math> is if the difference between the sum of the alternating digits is <math>0</math> or <math>11</math>, then that number is divisible by <math>11</math>. For <math>202103</math>, we have <math>2</math>+<math>2</math>-<math>1</math>-<math>3</math>=<math>0</math>.
 
~Education, the Study of Everything


For <math>7</math>, the divisibility test can be slow. Take the last digit, double it, and subtract it from the rest of the number. If it is divisible by <math>7</math>, then the original number is divisible by <math>7</math>. If the number is too large to determine whether or not it is a multiple of <math>7</math>, then we repeat the process until we can determine whether or not the number is divisible by <math>7</math>.


~Yvz2900


==Video Solution (Simple and Quick)==
https://youtu.be/7_Dg9b2hQ5U


~Education, the Study of Everything


==Video Solution==
==Video Solution==
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~savannahsolver
~savannahsolver


==Video Solution==
https://youtu.be/AgzDyKlmNAo
https://youtu.be/AgzDyKlmNAo


Line 72: Line 89:


~Lucas
~Lucas
==Video Solution==
https://youtu.be/7_Dg9b2hQ5U
~Education, the Study of Everything


==See Also==
==See Also==

Latest revision as of 17:37, 1 November 2025

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

Sidenote

The divisibility test for $11$ is if the difference between the sum of the alternating digits is $0$ or $11$, then that number is divisible by $11$. For $202103$, we have $2$+$2$-$1$-$3$=$0$.

For $7$, the divisibility test can be slow. Take the last digit, double it, and subtract it from the rest of the number. If it is divisible by $7$, then the original number is divisible by $7$. If the number is too large to determine whether or not it is a multiple of $7$, then we repeat the process until we can determine whether or not the number is divisible by $7$.

~Yvz2900

Video Solution (Simple and Quick)

https://youtu.be/7_Dg9b2hQ5U

~Education, the Study of Everything

Video Solution

https://youtu.be/jxnTkY3eb5Y

~savannahsolver

Video Solution

https://youtu.be/AgzDyKlmNAo

~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

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

These problems are copyrighted © by the Mathematical Association of America.

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