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Difference between revisions of "2022 AMC 12B Problems/Problem 3"
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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.  
 
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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Note: In general, <math>11..121...1</math> (where there are <math>k</math> "ones" 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> ones.  
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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.
  
 
== Solution 2 ==
 
== Solution 2 ==

Revision as of 22:16, 17 November 2022

Problem

How many of the first ten numbers of the sequence $121$, $11211$, $1112111$, ... are prime numbers? $\textbf{(A) } 0 \qquad \textbf{(B) }1 \qquad \textbf{(C) }2 \qquad \textbf{(D) }3 \qquad \textbf{(E) }4$

Solution 1

Write $121 = 110 + 11 = 11(10+1)$, $11211 = 11100 + 111 = 111(100+1)$, $1112111 = 1111000 + 1111 = 1111(1000+1)$. It becomes clear that $\boxed{\textbf{(A) } 0}$ of these numbers are prime.

In general, $11...121...1$ (where there are $k$ $1$'s on either side of the $2$) can be written as $(11...11)10^k + 11...11 = 11...11(10^k + 1)$, where the first term has $(k + 1)$ $1$'s.

Solution 2

Let $P(a,b)$ denote the digit $a$ written $b$ times and let $\overline{a_1a_2\cdots a_n}$ denote the concatenation of $a_1$, $a_2$, ..., $a_n$.

Observe that \[\overline{P(1,n) 2 P(1,n)} = \overline{P(1,n+1)P(0,n)} + P(1,n+1) = P(1,n+1) \cdot 10^n + P(1,n+1) = (P(1,n+1))(10^n + 1).\]

Both terms are integers larger than $1$ since $n \geq 1$, so $\boxed{\textbf{(A) } 0}$ of the numbers of the sequence are prime.

~Bxiao31415

See Also

2022 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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