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

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~wamofan
~wamofan
==Sidenote==
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>.
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)==
==Video Solution (Simple and Quick)==

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, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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