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Difference between revisions of "2021 Fall AMC 12A Problems/Problem 4"
(Combined solutions. Credits given to both authors.)
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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 1 ==
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== Solution ==
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First, modulo <math>2</math> or <math>5</math>, <math>\underline{20210A} \equiv A</math>.
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By divisibility rules, when <math>A=1,</math> the number <math>202101</math> is divisible by <math>3.</math> When <math>A=3,</math> the number <math>202103</math> is divisible by <math>11.</math> When <math>A=5,</math> the number <math>202105</math> is divisible by <math>5.</math> When <math>A=7,</math> the number <math>202107</math> is divisible by <math>3.</math> Thus, by the process of elimination we have that the answer is <math>\boxed{\textbf{(E)}\ 9}.</math>
 
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~NH14
 
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== Solution 2 ==
 
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First, modulo 2 or 5, <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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Second modulo 3, <math>\underline{20210A} \equiv 2 + 0 + 2 + 1 + 0 + A \equiv 5 + A</math>.
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Second modulo <math>3</math>, <math>\underline{20210A} \equiv 2 + 0 + 2 + 1 + 0 + A \equiv 5 + A</math>.
 
Hence, <math>A \neq 1, 4, 7</math>.
 
Hence, <math>A \neq 1, 4, 7</math>.
  
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Third, modulo 11, <math>\underline{20210A} \equiv A + 1 + 0 - 0 - 2 - 2 \equiv A - 3</math>.
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Third, modulo <math>11</math>, <math>\underline{20210A} \equiv A + 1 + 0 - 0 - 2 - 2 \equiv A - 3</math>.
 
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)
 
~Steven Chen (www.professorchenedu.com)

Revision as of 00:43, 26 November 2021

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

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)

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

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