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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>\textbf{(A) } 0 \qquad \textbf{(B) }1 \qquad \textbf{(C) }2 \qquad \textbf{(D) }3 \qquad \textbf{(E) }4</math>
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| − | == Solution 1 ==
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| − | Write <math>121 = 110 + 11 = 11(10+1)</math>, <math>11211 = 11100 + 111 = 111(100+1)</math>, <math>1112111 = 1111000 + 1111 = 1111(1000+1)</math>. It becomes clear that <math>\boxed{\textbf{(A) } 0}</math> of these numbers are prime.
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| − | In general, <math>11...121...1</math> (where there are <math>k</math> <math>1</math>'s on either side of the <math>2</math>) can be written as <math>(11...11)10^k + 11...11 = 11...11(10^k + 1)</math>, where the first term has <math>(k + 1)</math> <math>1</math>'s.
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| − | == Solution 2 ==
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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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| − | Both terms are integers larger than <math>1</math> since <math>n \geq 1</math>, so <math>\boxed{\textbf{(A) } 0}</math> of the numbers of the sequence are prime.
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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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