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

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<math>\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\ 5 \qquad\textbf{(D)}\ 7 \qquad\textbf{(E)}\ 9</math>
 
<math>\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\ 5 \qquad\textbf{(D)}\ 7 \qquad\textbf{(E)}\ 9</math>
  
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== Solution ==
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== Solution 1==
 
First, modulo <math>2</math> or <math>5</math>, <math>\underline{20210A} \equiv A</math>.
 
First, modulo <math>2</math> or <math>5</math>, <math>\underline{20210A} \equiv A</math>.
 
Hence, <math>A \neq 0, 2, 4, 5, 6, 8</math>.
 
Hence, <math>A \neq 0, 2, 4, 5, 6, 8</math>.
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Hence, <math>A \neq 3</math>.
 
Hence, <math>A \neq 3</math>.
  
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Therefore, the answer is <math>\boxed{\textbf{(E) }9}</math>.
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Therefore, the answer is <math>\boxed{\textbf{(E)}\ 9}</math>.
  
−
~NH14
+
~NH14 ~Steven Chen (www.professorchenedu.com)
  
−
~Steven Chen (www.professorchenedu.com)
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==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>.
  
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==Solution 2==
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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 (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==
 
==See Also==

Latest revision as of 20:13, 12 July 2023

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)

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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