Difference between revisions of "2022 AMC 10B Problems/Problem 6"

Revision as of 12:24, 30 December 2022

Problem

How many of the first ten numbers of the sequence $121, 11211, 1112111, \ldots$ are prime numbers?

$\textbf{(A) } 0 \qquad \textbf{(B) }1 \qquad \textbf{(C) }2 \qquad \textbf{(D) }3 \qquad \textbf{(E) }4$

Solution 1 (Generalization)

The $n$ th term of this sequence is $\sum_{k=n}^{2n}10^k + \sum_{k=0}^{n}10^k = 10^n\sum_{k=0}^{n}10^k + \sum_{k=0}^{n}10^k = \left(10^n+1\right)\sum_{k=0}^{n}10^k.$ It follows that the terms are $\begin{align*} 121 &= 11\cdot11, \\ 11211 &= 101\cdot111, \\ 1112111 &= 1001\cdot1111, \\ & \ \vdots \end{align*}$ Therefore, there are $\boxed{\textbf{(A) } 0}$ prime numbers in this sequence.

~MRENTHUSIASM

Solution 2 (Simple Sums)

Observe how $121 = 110+11$ and $11211 = 11100 + 111$ and $1112111 =1111000 + 1111$ all take the form of $\overbrace{111...}^{n}\overbrace{00...}^{n-1} + \overbrace{111...}^{n}$ which factors as $\overbrace{111...}^{n}(10^{n-1} + 1).$ Factoring each of the sums, we have $11(10+1), 111(100+1),$ and $1111(1000+1)$ respectively. With each number factored, there are $\boxed{\textbf{(A) } 0}$ primes in the set.

~ab2024

Solution 3 (Educated Guesses)

Note that it's obvious that $121$ is divisible by $11$ and $11211$ is divisible by $3;$ therefore, since this an AMC 10 problem 6, we may safely assume that we do not need to check two-digit prime divisibility or use obscure theorems. So, the answer is $\boxed{\textbf{(A) } 0}.$

~Dhillonr25

2022 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 18: / Line 18: @@
 ~MRENTHUSIASM
-==Solution 2 (Educated Guesses)==
+==Solution 2 (Simple Sums)==
-Note that it's obvious that <math>121</math> is divisible by <math>11</math> and <math>11211</math> is divisible by <math>3;</math> therefore, since this an AMC 10 problem 6, we may safely assume that we do not need to check two-digit prime divisibility or use obscure theorems. So, the answer is <math>\boxed{\textbf{(A) } 0}.</math>
+Observe how <math>121 = 110+11</math> and <math>11211 = 11100 + 111</math> and <math>1112111 =1111000 + 1111</math> all take the form of <math>\overbrace{111...}^{n}\overbrace{00...}^{n-1} + \overbrace{111...}^{n}</math> which factors as <math>\overbrace{111...}^{n}(10^{n-1} + 1).</math>
+Factoring each of the sums, we have <math>11(10+1), 111(100+1),</math> and <math>1111(1000+1)</math> respectively. With each number factored, there are <math>\boxed{\textbf{(A) } 0}</math> primes in the set.
-~Dhillonr25
+~ab2024
-==Solution 3 (Simple Sums)==
+==Solution 3 (Educated Guesses)==
-Observe how <math>121 = 110+11</math> and <math>11211 = 11100 + 111</math> and <math>1112111 =1111000 + 1111</math> all take the form of <math>\overbrace{111...}^{n}\overbrace{00...}^{n-1} + \overbrace{111...}^{n}</math> which factors as <math>\overbrace{111...}^{n}(10^{n-1} + 1).</math>
+Note that it's obvious that <math>121</math> is divisible by <math>11</math> and <math>11211</math> is divisible by <math>3;</math> therefore, since this an AMC 10 problem 6, we may safely assume that we do not need to check two-digit prime divisibility or use obscure theorems. So, the answer is <math>\boxed{\textbf{(A) } 0}.</math>
-Factoring each of the sums, we have <math>11(10+1), 111(100+1),</math> and <math>1111(1000+1)</math> respectively. With each number factored, there are <math>\boxed{\textbf{(A) } 0}</math> primes in the set.
-~ab2024
+~Dhillonr25
 == See Also ==
 {{AMC10 box|year=2022|ab=B|num-b=5|num-a=7}}
 {{MAA Notice}}

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