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

Revision as of 20:06, 22 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 $\boxed{\textbf{(E)}.}$

~NH14

2021 Fall AMC 10A (Problems • Answer Key • Resources)
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All AMC 10 Problems and Solutions

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 ~NH14
+==See Also==
+{{AMC10 box|year=2021 Fall|ab=A|num-b=4|num-a=6}}
+{{MAA Notice}}

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

Revision as of 20:06, 22 November 2021

Solution 1

See Also