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Difference between revisions of "2022 AMC 12B Problems/Problem 3"

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== Problem ==
 
== Problem ==
 
How many of the first ten numbers of the sequence <math>121</math>, <math>11211</math>, <math>1112111</math>, ... are prime numbers?
 
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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<math>\textbf{(A) } 0 \qquad \textbf{(B) }1 \qquad \textbf{(C) }2 \qquad \textbf{(D) }3 \qquad \textbf{(E) }4</math>
  
 
== Solution 1 ==
 
== Solution 1 ==
 
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>.
 
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>.
 +
 
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>
 
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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 +
Both terms are integers larger than <math>1</math> since <math>n \geq 1</math>, so <math>\textbf{(A) } 0</math> of the numbers of the sequence are prime.
  
 
~[[User:Bxiao31415|Bxiao31415]]
 
~[[User:Bxiao31415|Bxiao31415]]

Revision as of 20:47, 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

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 $\textbf{(A) } 0$ 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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