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| − | == Problem ==
| + | #REDIRECT [[2022_AMC_10B_Problems/Problem_6]] |
| − | How many of the first ten numbers of the sequence <math>121</math>, <math>11211</math>, <math>1112111</math>, ... are prime numbers?
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| − | <math>\text{(A) } 0 \qquad \text{(B) }1 \qquad \text{(C) }2 \qquad \text{(D) }3 \qquad \text{(E) }4</math>
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| − | == Solution 1 ==
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| − | Let <math>P(a,b)</math> denote the digit <math>a</math> written <math>b</math> times and let <math>\overline{a_1a_2\cdots a_n}</math> denote the concatenation of <math>a_1</math>, <math>a_2</math>, ..., <math>a_n</math>.
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| − | Observe that <cmath>\overline{P(1,n) 2 P(1,n)} = \overline{P(1,n+1)P(0,n)} + P(1,n+1) = P(1,n+1) \cdot 10^n + P(1,n+1) = (P(1,n+1))(10^n + 1).</cmath>
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| − | Clearly, both terms are larger than <math>1</math> since <math>n \geq 1</math>, hence all the numbers of the sequence are <math>\fbox{0(A)}</math>, and we're done!
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| − | ~[[User:Bxiao31415|Bxiao31415]]
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| − | == See Also ==
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| − | {{AMC12 box|year=2022|ab=B|num-b=2|num-a=4}}
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| − | {{MAA Notice}}
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