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Difference between revisions of "2021 Fall AMC 10A Problems/Problem 5"

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== Solution 1==
 
== Solution 1==
  
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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)}.}</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>
  
 
~NH14
 
~NH14

Revision as of 16:51, 23 November 2021

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

By divisibility rules, when $A=1,$ the number $202101$ is divisible by $3.$ When $A=3,$ the number $202103$ is divisible by $11.$ When $A=5,$ the number $202105$ is divisible by $5.$ When $A=7,$ the number $202107$ is divisible by $3.$ Thus, by the process of elimination we have that the answer is $\boxed{\textbf{(E)} 9.}$

~NH14

See Also

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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