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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).</cmath>
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| − | Since <math>\overline{k \: P(0,n)} = k \cdot 10^n</math> for all positive integers <math>k</math> and <math>n</math>, <math>\overline{P(1,n+1) \: P(0,n)} + P(1,n+1)</math> is equal to
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| − | <cmath>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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| − | == Solution 3 (Not Rigorous) ==
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| − | Because <math>121 = 11*11</math>, because <math>11211</math> has sum of digits <math>6</math> (and therefore is divisible by <math>3</math>), and because <math>1112111 = 1100011 + 121</math> (a multiple of <math>11</math>), none of the first three numbers in the sequence are prime. Therefore, if the answer were anything other than <math>\boxed{\textbf{(A) } 0}</math>, determining the answer conclusively would require proving that some positive integer <math>111121111</math> or greater is prime, an extremely time-consuming task given the conditions of the test.
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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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