Difference between revisions of "2021 Fall AMC 12A Problems/Problem 4"
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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>C</math>. | + | 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>A</math> and <math>D</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>B</math> or <math>E</math>. Let's try dividing the number with some integers. | + | 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/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>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>. | + | <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 | ~ZoBro23 | ||
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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 | ||
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~savannahsolver | ~savannahsolver | ||
| + | ==Video Solution== | ||
https://youtu.be/AgzDyKlmNAo | https://youtu.be/AgzDyKlmNAo | ||
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.
Contents
Problem
The six-digit number 
is prime for only one digit 
What is 
Solution 1
First, modulo 
or 
, 
.
Hence, 
.
Second modulo 
, 
.
Hence, 
.
Third, modulo 
, 
.
Hence, 
.
Therefore, the answer is 
.
~NH14 ~Steven Chen (www.professorchenedu.com)
Solution 2 (Elimination)
Any number ending in is divisible by . So we can eliminate option 
.
If the sum of the digits of a number is divisible by , the number is divisible by . The sum of the digits of this number is 
. If 
is divisible by , the number is divisible by . Thus we can eliminate options 
and 
.
So the correct option is either 
or 
. Let's try dividing the number with some integers.
, where 
is 
. Since 
and 
are both indivisible by 
, this does not help us narrow the choices down.
, where is 
. Since 
, option would make 
divisible by . Thus, by elimination, the correct choice must be option .
~ZoBro23
Solution 3
divisible by .
divisible by .
divisible by .
divisible by .
divisible by .
divisible by .
divisible by .
divisible by .
divisible by .
This leaves only 
.
~wamofan
Video Solution (Simple and Quick)
~Education, the Study of Everything
Video Solution
~savannahsolver
Video Solution
~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
~Lucas
Video Solution
~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. 
