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| − | The six-digit number <math>\underline{2}\,\underline{0}\,\underline{2}\,\underline{1}\,\underline{0}\,\underline{A}</math> is prime for only one digit <math>A.</math> What is <math>A?</math>
| + | #REDIRECT [[2021_Fall_AMC_12A_Problems/Problem_4]] |
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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>
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| − | == 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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| − | ~NH14
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| − | ==See Also==
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| − | {{AMC10 box|year=2021 Fall|ab=A|num-b=4|num-a=6}}
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| − | {{MAA Notice}}
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