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

Revision as of 16:32, 19 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 $\overline{P(1,n) \: 2 \: P(1,n)} = \overline{P(1,n+1) \: P(0,n)} + P(1,n+1).$

Since $\overline{k \: P(0,n)} = k \cdot 10^n$ for all positive integers $k$ and $n$ , $\overline{P(1,n+1) \: P(0,n)} + P(1,n+1)$ is equal to $P(1,n+1) \cdot 10^n + P(1,n+1) = (P(1,n+1))(10^n + 1).$

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

Solution 3 (Not Rigorous)

Because $121 = 11*11$ , because $11211$ has sum of digits $6$ (and therefore is divisible by $3$ ), and because $1112111 = 1100011 + 121$ (a multiple of $11$ ), none of the first three numbers in the sequence are prime. Therefore, if the answer were anything other than $\boxed{\textbf{(A) } 0}$ , determining the answer conclusively would require proving that some positive integer $111121111$ or greater is prime, an extremely time-consuming task given the conditions of the test.

@@ Line 19: / Line 19: @@
 ~[[User:Bxiao31415|Bxiao31415]]
+== Solution 3 (Not Rigorous) ==
+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.
 == See Also ==
 {{AMC12 box|year=2022|ab=B|num-b=2|num-a=4}}
 {{MAA Notice}}

2022 AMC 12B (Problems • Answer Key • Resources)
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